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""Definition"" -> 9|>","<|""Book"" -> 11, ""Definition"" -> 10|>","<|""Book"" -> 11, ""Definition"" -> 11|>","<|""Book"" -> 11, ""Definition"" -> 12|>","<|""Book"" -> 11, ""Definition"" -> 13|>","<|""Book"" -> 11, ""Definition"" -> 14|>","<|""Book"" -> 11, ""Definition"" -> 15|>","<|""Book"" -> 11, ""Definition"" -> 16|>","<|""Book"" -> 11, ""Definition"" -> 17|>","<|""Book"" -> 11, ""Definition"" -> 18|>","<|""Book"" -> 11, ""Definition"" -> 19|>","<|""Book"" -> 11, ""Definition"" -> 20|>","<|""Book"" -> 11, ""Definition"" -> 21|>","<|""Book"" -> 11, ""Definition"" -> 22|>","<|""Book"" -> 11, ""Definition"" -> 23|>","<|""Book"" -> 11, ""Definition"" -> 24|>","<|""Book"" -> 11, ""Definition"" -> 25|>","<|""Book"" -> 11, ""Definition"" -> 26|>","<|""Book"" -> 11, ""Definition"" -> 27|>","<|""Book"" -> 11, ""Definition"" -> 28|>","<|""Book"" -> 1, ""Theorem"" -> 1|>","<|""Book"" -> 1, ""Theorem"" -> 2|>","<|""Book"" -> 1, ""Theorem"" -> 3|>","<|""Book"" -> 1, ""Theorem"" -> 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""Theorem"" -> 30|>","<|""Book"" -> 1, ""Theorem"" -> 31|>","<|""Book"" -> 1, ""Theorem"" -> 32|>","<|""Book"" -> 1, ""Theorem"" -> 33|>","<|""Book"" -> 1, ""Theorem"" -> 34|>","<|""Book"" -> 1, ""Theorem"" -> 35|>","<|""Book"" -> 1, ""Theorem"" -> 36|>","<|""Book"" -> 1, ""Theorem"" -> 37|>","<|""Book"" -> 1, ""Theorem"" -> 38|>","<|""Book"" -> 1, ""Theorem"" -> 39|>","<|""Book"" -> 1, ""Theorem"" -> 40|>","<|""Book"" -> 1, ""Theorem"" -> 41|>","<|""Book"" -> 1, ""Theorem"" -> 42|>","<|""Book"" -> 1, ""Theorem"" -> 43|>","<|""Book"" -> 1, ""Theorem"" -> 44|>","<|""Book"" -> 1, ""Theorem"" -> 45|>","<|""Book"" -> 1, ""Theorem"" -> 46|>","<|""Book"" -> 1, ""Theorem"" -> 47|>","<|""Book"" -> 1, ""Theorem"" -> 48|>","<|""Book"" -> 2, ""Theorem"" -> 1|>","<|""Book"" -> 2, ""Theorem"" -> 2|>","<|""Book"" -> 2, ""Theorem"" -> 3|>","<|""Book"" -> 2, ""Theorem"" -> 4|>","<|""Book"" -> 2, ""Theorem"" -> 5|>","<|""Book"" -> 2, ""Theorem"" -> 6|>","<|""Book"" -> 2, ""Theorem"" -> 7|>","<|""Book"" -> 2, ""Theorem"" -> 8|>","<|""Book"" -> 2, ""Theorem"" -> 9|>","<|""Book"" -> 2, ""Theorem"" -> 10|>","<|""Book"" -> 2, ""Theorem"" -> 11|>","<|""Book"" -> 2, ""Theorem"" -> 12|>","<|""Book"" -> 2, ""Theorem"" -> 13|>","<|""Book"" -> 2, ""Theorem"" -> 14|>","<|""Book"" -> 3, ""Theorem"" -> 1|>","<|""Book"" -> 3, ""Theorem"" -> 2|>","<|""Book"" -> 3, ""Theorem"" -> 3|>","<|""Book"" -> 3, ""Theorem"" -> 4|>","<|""Book"" -> 3, ""Theorem"" -> 5|>","<|""Book"" -> 3, ""Theorem"" -> 6|>","<|""Book"" -> 3, ""Theorem"" -> 7|>","<|""Book"" -> 3, ""Theorem"" -> 8|>","<|""Book"" -> 3, ""Theorem"" -> 9|>","<|""Book"" -> 3, ""Theorem"" -> 10|>","<|""Book"" -> 3, ""Theorem"" -> 11|>","<|""Book"" -> 3, ""Theorem"" -> 12|>","<|""Book"" -> 3, ""Theorem"" -> 13|>","<|""Book"" -> 3, ""Theorem"" -> 14|>","<|""Book"" -> 3, ""Theorem"" -> 15|>","<|""Book"" -> 3, ""Theorem"" -> 16|>","<|""Book"" -> 3, ""Theorem"" -> 17|>","<|""Book"" -> 3, ""Theorem"" -> 18|>","<|""Book"" -> 3, ""Theorem"" -> 19|>","<|""Book"" -> 3, ""Theorem"" -> 20|>","<|""Book"" -> 3, ""Theorem"" -> 21|>","<|""Book"" -> 3, ""Theorem"" -> 22|>","<|""Book"" -> 3, ""Theorem"" -> 23|>","<|""Book"" -> 3, ""Theorem"" -> 24|>","<|""Book"" -> 3, ""Theorem"" -> 25|>","<|""Book"" -> 3, ""Theorem"" -> 26|>","<|""Book"" -> 3, ""Theorem"" -> 27|>","<|""Book"" -> 3, ""Theorem"" -> 28|>","<|""Book"" -> 3, ""Theorem"" -> 29|>","<|""Book"" -> 3, ""Theorem"" -> 30|>","<|""Book"" -> 3, ""Theorem"" -> 31|>","<|""Book"" -> 3, ""Theorem"" -> 32|>","<|""Book"" -> 3, ""Theorem"" -> 33|>","<|""Book"" -> 3, ""Theorem"" -> 34|>","<|""Book"" -> 3, ""Theorem"" -> 35|>","<|""Book"" -> 3, ""Theorem"" -> 36|>","<|""Book"" -> 3, ""Theorem"" -> 37|>","<|""Book"" -> 4, ""Theorem"" -> 1|>","<|""Book"" -> 4, ""Theorem"" -> 2|>","<|""Book"" -> 4, ""Theorem"" -> 3|>","<|""Book"" -> 4, ""Theorem"" -> 4|>","<|""Book"" -> 4, ""Theorem"" -> 5|>","<|""Book"" -> 4, ""Theorem"" -> 6|>","<|""Book"" -> 4, ""Theorem"" -> 7|>","<|""Book"" -> 4, ""Theorem"" -> 8|>","<|""Book"" -> 4, ""Theorem"" -> 9|>","<|""Book"" -> 4, ""Theorem"" -> 10|>","<|""Book"" -> 4, ""Theorem"" -> 11|>","<|""Book"" -> 4, ""Theorem"" -> 12|>","<|""Book"" -> 4, ""Theorem"" -> 13|>","<|""Book"" -> 4, ""Theorem"" -> 14|>","<|""Book"" -> 4, ""Theorem"" -> 15|>","<|""Book"" -> 4, ""Theorem"" -> 16|>","<|""Book"" -> 5, ""Theorem"" -> 1|>","<|""Book"" -> 5, ""Theorem"" -> 2|>","<|""Book"" -> 5, ""Theorem"" -> 3|>","<|""Book"" -> 5, ""Theorem"" -> 4|>","<|""Book"" -> 5, ""Theorem"" -> 5|>","<|""Book"" -> 5, ""Theorem"" -> 6|>","<|""Book"" -> 5, ""Theorem"" -> 7|>","<|""Book"" -> 5, ""Theorem"" -> 8|>","<|""Book"" -> 5, ""Theorem"" -> 9|>","<|""Book"" -> 5, ""Theorem"" -> 10|>","<|""Book"" -> 5, ""Theorem"" -> 11|>","<|""Book"" -> 5, ""Theorem"" -> 12|>","<|""Book"" -> 5, ""Theorem"" -> 13|>","<|""Book"" -> 5, ""Theorem"" -> 14|>","<|""Book"" -> 5, ""Theorem"" -> 15|>","<|""Book"" -> 5, ""Theorem"" -> 16|>","<|""Book"" -> 5, ""Theorem"" -> 17|>","<|""Book"" -> 5, ""Theorem"" -> 18|>","<|""Book"" -> 5, ""Theorem"" -> 19|>","<|""Book"" -> 5, ""Theorem"" -> 20|>","<|""Book"" -> 5, ""Theorem"" -> 21|>","<|""Book"" -> 5, ""Theorem"" -> 22|>","<|""Book"" -> 5, ""Theorem"" -> 23|>","<|""Book"" -> 5, ""Theorem"" -> 24|>","<|""Book"" -> 5, ""Theorem"" -> 25|>","<|""Book"" -> 6, ""Theorem"" -> 1|>","<|""Book"" -> 6, ""Theorem"" -> 2|>","<|""Book"" -> 6, ""Theorem"" -> 3|>","<|""Book"" -> 6, ""Theorem"" -> 4|>","<|""Book"" -> 6, ""Theorem"" -> 5|>","<|""Book"" -> 6, ""Theorem"" -> 6|>","<|""Book"" -> 6, ""Theorem"" -> 7|>","<|""Book"" -> 6, ""Theorem"" -> 8|>","<|""Book"" -> 6, ""Theorem"" -> 9|>","<|""Book"" -> 6, ""Theorem"" -> 10|>","<|""Book"" -> 6, ""Theorem"" -> 11|>","<|""Book"" -> 6, ""Theorem"" -> 12|>","<|""Book"" -> 6, ""Theorem"" -> 13|>","<|""Book"" -> 6, ""Theorem"" -> 14|>","<|""Book"" -> 6, ""Theorem"" -> 15|>","<|""Book"" -> 6, ""Theorem"" -> 16|>","<|""Book"" -> 6, ""Theorem"" -> 17|>","<|""Book"" -> 6, ""Theorem"" -> 18|>","<|""Book"" -> 6, ""Theorem"" -> 19|>","<|""Book"" -> 6, ""Theorem"" -> 20|>","<|""Book"" -> 6, ""Theorem"" -> 21|>","<|""Book"" -> 6, ""Theorem"" -> 22|>","<|""Book"" -> 6, ""Theorem"" -> 23|>","<|""Book"" -> 6, ""Theorem"" -> 24|>","<|""Book"" -> 6, ""Theorem"" -> 25|>","<|""Book"" -> 6, ""Theorem"" -> 26|>","<|""Book"" -> 6, ""Theorem"" -> 27|>","<|""Book"" -> 6, ""Theorem"" -> 28|>","<|""Book"" -> 6, ""Theorem"" -> 29|>","<|""Book"" -> 6, ""Theorem"" -> 30|>","<|""Book"" -> 6, ""Theorem"" -> 31|>","<|""Book"" -> 6, ""Theorem"" -> 32|>","<|""Book"" -> 6, ""Theorem"" -> 33|>","<|""Book"" -> 7, ""Theorem"" -> 1|>","<|""Book"" -> 7, ""Theorem"" -> 2|>","<|""Book"" -> 7, ""Theorem"" -> 3|>","<|""Book"" -> 7, ""Theorem"" -> 4|>","<|""Book"" -> 7, ""Theorem"" -> 5|>","<|""Book"" -> 7, ""Theorem"" -> 6|>","<|""Book"" -> 7, ""Theorem"" -> 7|>","<|""Book"" -> 7, ""Theorem"" -> 8|>","<|""Book"" -> 7, ""Theorem"" -> 9|>","<|""Book"" -> 7, ""Theorem"" -> 10|>","<|""Book"" -> 7, ""Theorem"" -> 11|>","<|""Book"" -> 7, ""Theorem"" -> 12|>","<|""Book"" -> 7, ""Theorem"" -> 13|>","<|""Book"" -> 7, ""Theorem"" -> 14|>","<|""Book"" -> 7, ""Theorem"" -> 15|>","<|""Book"" -> 7, ""Theorem"" -> 16|>","<|""Book"" -> 7, ""Theorem"" -> 17|>","<|""Book"" -> 7, ""Theorem"" -> 18|>","<|""Book"" -> 7, ""Theorem"" -> 19|>","<|""Book"" -> 7, ""Theorem"" -> 20|>","<|""Book"" -> 7, ""Theorem"" -> 21|>","<|""Book"" -> 7, ""Theorem"" -> 22|>","<|""Book"" -> 7, ""Theorem"" -> 23|>","<|""Book"" -> 7, ""Theorem"" -> 24|>","<|""Book"" -> 7, ""Theorem"" -> 25|>","<|""Book"" -> 7, ""Theorem"" -> 26|>","<|""Book"" -> 7, ""Theorem"" -> 27|>","<|""Book"" -> 7, ""Theorem"" -> 28|>","<|""Book"" -> 7, ""Theorem"" -> 29|>","<|""Book"" -> 7, ""Theorem"" -> 30|>","<|""Book"" -> 7, ""Theorem"" -> 31|>","<|""Book"" -> 7, ""Theorem"" -> 32|>","<|""Book"" -> 7, ""Theorem"" -> 33|>","<|""Book"" -> 7, ""Theorem"" -> 34|>","<|""Book"" -> 7, ""Theorem"" -> 35|>","<|""Book"" -> 7, ""Theorem"" -> 36|>","<|""Book"" -> 7, ""Theorem"" -> 37|>","<|""Book"" -> 7, ""Theorem"" -> 38|>","<|""Book"" -> 7, ""Theorem"" -> 39|>","<|""Book"" -> 8, ""Theorem"" -> 1|>","<|""Book"" -> 8, ""Theorem"" -> 2|>","<|""Book"" -> 8, ""Theorem"" -> 3|>","<|""Book"" -> 8, ""Theorem"" -> 4|>","<|""Book"" -> 8, ""Theorem"" -> 5|>","<|""Book"" -> 8, ""Theorem"" -> 6|>","<|""Book"" -> 8, ""Theorem"" -> 7|>","<|""Book"" -> 8, ""Theorem"" -> 8|>","<|""Book"" -> 8, ""Theorem"" -> 9|>","<|""Book"" -> 8, ""Theorem"" -> 10|>","<|""Book"" -> 8, ""Theorem"" -> 11|>","<|""Book"" -> 8, ""Theorem"" -> 12|>","<|""Book"" -> 8, ""Theorem"" -> 13|>","<|""Book"" -> 8, ""Theorem"" -> 14|>","<|""Book"" -> 8, ""Theorem"" -> 15|>","<|""Book"" -> 8, ""Theorem"" -> 16|>","<|""Book"" -> 8, ""Theorem"" -> 17|>","<|""Book"" -> 8, ""Theorem"" -> 18|>","<|""Book"" -> 8, ""Theorem"" -> 19|>","<|""Book"" -> 8, ""Theorem"" -> 20|>","<|""Book"" -> 8, ""Theorem"" -> 21|>","<|""Book"" -> 8, ""Theorem"" -> 22|>","<|""Book"" -> 8, ""Theorem"" -> 23|>","<|""Book"" -> 8, ""Theorem"" -> 24|>","<|""Book"" -> 8, ""Theorem"" -> 25|>","<|""Book"" -> 8, ""Theorem"" -> 26|>","<|""Book"" -> 8, ""Theorem"" -> 27|>","<|""Book"" -> 9, ""Theorem"" -> 1|>","<|""Book"" -> 9, ""Theorem"" -> 2|>","<|""Book"" -> 9, ""Theorem"" -> 3|>","<|""Book"" -> 9, ""Theorem"" -> 4|>","<|""Book"" -> 9, ""Theorem"" -> 5|>","<|""Book"" -> 9, ""Theorem"" -> 6|>","<|""Book"" -> 9, ""Theorem"" -> 7|>","<|""Book"" -> 9, ""Theorem"" -> 8|>","<|""Book"" -> 9, ""Theorem"" -> 9|>","<|""Book"" -> 9, ""Theorem"" -> 10|>","<|""Book"" -> 9, ""Theorem"" -> 11|>","<|""Book"" -> 9, ""Theorem"" -> 12|>","<|""Book"" -> 9, ""Theorem"" -> 13|>","<|""Book"" -> 9, ""Theorem"" -> 14|>","<|""Book"" -> 9, ""Theorem"" -> 15|>","<|""Book"" -> 9, ""Theorem"" -> 16|>","<|""Book"" -> 9, ""Theorem"" -> 17|>","<|""Book"" -> 9, ""Theorem"" -> 18|>","<|""Book"" -> 9, ""Theorem"" -> 19|>","<|""Book"" -> 9, ""Theorem"" -> 20|>","<|""Book"" -> 9, ""Theorem"" -> 21|>","<|""Book"" -> 9, ""Theorem"" -> 22|>","<|""Book"" -> 9, ""Theorem"" -> 23|>","<|""Book"" -> 9, ""Theorem"" -> 24|>","<|""Book"" -> 9, ""Theorem"" -> 25|>","<|""Book"" -> 9, ""Theorem"" -> 26|>","<|""Book"" -> 9, ""Theorem"" -> 27|>","<|""Book"" -> 9, ""Theorem"" -> 28|>","<|""Book"" -> 9, ""Theorem"" -> 29|>","<|""Book"" -> 9, ""Theorem"" -> 30|>","<|""Book"" -> 9, ""Theorem"" -> 31|>","<|""Book"" -> 9, ""Theorem"" -> 32|>","<|""Book"" -> 9, ""Theorem"" -> 33|>","<|""Book"" -> 9, ""Theorem"" -> 34|>","<|""Book"" -> 9, ""Theorem"" -> 35|>","<|""Book"" -> 9, ""Theorem"" -> 36|>","<|""Book"" -> 10, ""Theorem"" -> 1|>","<|""Book"" -> 10, ""Theorem"" -> 2|>","<|""Book"" -> 10, ""Theorem"" -> 3|>","<|""Book"" -> 10, ""Theorem"" -> 4|>","<|""Book"" -> 10, ""Theorem"" -> 5|>","<|""Book"" -> 10, ""Theorem"" -> 6|>","<|""Book"" -> 10, ""Theorem"" -> 7|>","<|""Book"" -> 10, ""Theorem"" -> 8|>","<|""Book"" -> 10, ""Theorem"" -> 9|>","<|""Book"" -> 10, ""Theorem"" -> 10|>","<|""Book"" -> 10, ""Theorem"" -> 11|>","<|""Book"" -> 10, ""Theorem"" -> 12|>","<|""Book"" -> 10, ""Theorem"" -> 13|>","<|""Book"" -> 10, ""Theorem"" -> 14|>","<|""Book"" -> 10, ""Theorem"" -> 15|>","<|""Book"" -> 10, ""Theorem"" -> 16|>","<|""Book"" -> 10, ""Theorem"" -> 17|>","<|""Book"" -> 10, ""Theorem"" -> 18|>","<|""Book"" -> 10, ""Theorem"" -> 19|>","<|""Book"" -> 10, ""Theorem"" -> 20|>","<|""Book"" -> 10, ""Theorem"" -> 21|>","<|""Book"" -> 10, ""Theorem"" -> 22|>","<|""Book"" -> 10, ""Theorem"" -> 23|>","<|""Book"" -> 10, ""Theorem"" -> 24|>","<|""Book"" -> 10, ""Theorem"" -> 25|>","<|""Book"" -> 10, ""Theorem"" -> 26|>","<|""Book"" -> 10, ""Theorem"" -> 27|>","<|""Book"" -> 10, ""Theorem"" -> 28|>","<|""Book"" -> 10, ""Theorem"" -> 29|>","<|""Book"" -> 10, ""Theorem"" -> 30|>","<|""Book"" -> 10, ""Theorem"" -> 31|>","<|""Book"" -> 10, ""Theorem"" -> 32|>","<|""Book"" -> 10, ""Theorem"" -> 33|>","<|""Book"" -> 10, ""Theorem"" -> 34|>","<|""Book"" -> 10, ""Theorem"" -> 35|>","<|""Book"" -> 10, ""Theorem"" -> 36|>","<|""Book"" -> 10, ""Theorem"" -> 37|>","<|""Book"" -> 10, ""Theorem"" -> 38|>","<|""Book"" -> 10, ""Theorem"" -> 39|>","<|""Book"" -> 10, ""Theorem"" -> 40|>","<|""Book"" -> 10, ""Theorem"" -> 41|>","<|""Book"" -> 10, ""Theorem"" -> 42|>","<|""Book"" -> 10, ""Theorem"" -> 43|>","<|""Book"" -> 10, ""Theorem"" -> 44|>","<|""Book"" -> 10, ""Theorem"" -> 45|>","<|""Book"" -> 10, ""Theorem"" -> 46|>","<|""Book"" -> 10, ""Theorem"" -> 47|>","<|""Book"" -> 10, ""Theorem"" -> 48|>","<|""Book"" -> 10, ""Theorem"" -> 49|>","<|""Book"" -> 10, ""Theorem"" -> 50|>","<|""Book"" -> 10, ""Theorem"" -> 51|>","<|""Book"" -> 10, ""Theorem"" -> 52|>","<|""Book"" -> 10, ""Theorem"" -> 53|>","<|""Book"" -> 10, ""Theorem"" -> 54|>","<|""Book"" -> 10, ""Theorem"" -> 55|>","<|""Book"" -> 10, ""Theorem"" -> 56|>","<|""Book"" -> 10, ""Theorem"" -> 57|>","<|""Book"" -> 10, ""Theorem"" -> 58|>","<|""Book"" -> 10, ""Theorem"" -> 59|>","<|""Book"" -> 10, ""Theorem"" -> 60|>","<|""Book"" -> 10, ""Theorem"" -> 61|>","<|""Book"" -> 10, ""Theorem"" -> 62|>","<|""Book"" -> 10, ""Theorem"" -> 63|>","<|""Book"" -> 10, ""Theorem"" -> 64|>","<|""Book"" -> 10, ""Theorem"" -> 65|>","<|""Book"" -> 10, ""Theorem"" -> 66|>","<|""Book"" -> 10, ""Theorem"" -> 67|>","<|""Book"" -> 10, ""Theorem"" -> 68|>","<|""Book"" -> 10, ""Theorem"" -> 69|>","<|""Book"" -> 10, ""Theorem"" -> 70|>","<|""Book"" -> 10, ""Theorem"" -> 71|>","<|""Book"" -> 10, ""Theorem"" -> 72|>","<|""Book"" -> 10, ""Theorem"" -> 73|>","<|""Book"" -> 10, ""Theorem"" -> 74|>","<|""Book"" -> 10, ""Theorem"" -> 75|>","<|""Book"" -> 10, ""Theorem"" -> 76|>","<|""Book"" -> 10, ""Theorem"" -> 77|>","<|""Book"" -> 10, ""Theorem"" -> 78|>","<|""Book"" -> 10, ""Theorem"" -> 79|>","<|""Book"" -> 10, ""Theorem"" -> 80|>","<|""Book"" -> 10, ""Theorem"" -> 81|>","<|""Book"" -> 10, ""Theorem"" -> 82|>","<|""Book"" -> 10, ""Theorem"" -> 83|>","<|""Book"" -> 10, ""Theorem"" -> 84|>","<|""Book"" -> 10, ""Theorem"" -> 85|>","<|""Book"" -> 10, ""Theorem"" -> 86|>","<|""Book"" -> 10, ""Theorem"" -> 87|>","<|""Book"" -> 10, ""Theorem"" -> 88|>","<|""Book"" -> 10, ""Theorem"" -> 89|>","<|""Book"" -> 10, ""Theorem"" -> 90|>","<|""Book"" -> 10, ""Theorem"" -> 91|>","<|""Book"" -> 10, ""Theorem"" -> 92|>","<|""Book"" -> 10, ""Theorem"" -> 93|>","<|""Book"" -> 10, ""Theorem"" -> 94|>","<|""Book"" -> 10, ""Theorem"" -> 95|>","<|""Book"" -> 10, ""Theorem"" -> 96|>","<|""Book"" -> 10, ""Theorem"" -> 97|>","<|""Book"" -> 10, ""Theorem"" -> 98|>","<|""Book"" -> 10, ""Theorem"" -> 99|>","<|""Book"" -> 10, ""Theorem"" -> 100|>","<|""Book"" -> 10, ""Theorem"" -> 101|>","<|""Book"" -> 10, ""Theorem"" -> 102|>","<|""Book"" -> 10, ""Theorem"" -> 103|>","<|""Book"" -> 10, ""Theorem"" -> 104|>","<|""Book"" -> 10, ""Theorem"" -> 105|>","<|""Book"" -> 10, ""Theorem"" -> 106|>","<|""Book"" -> 10, ""Theorem"" -> 107|>","<|""Book"" -> 10, ""Theorem"" -> 108|>","<|""Book"" -> 10, ""Theorem"" -> 109|>","<|""Book"" -> 10, ""Theorem"" -> 110|>","<|""Book"" -> 10, ""Theorem"" -> 111|>","<|""Book"" -> 10, ""Theorem"" -> 112|>","<|""Book"" -> 10, ""Theorem"" -> 113|>","<|""Book"" -> 10, ""Theorem"" -> 114|>","<|""Book"" -> 10, ""Theorem"" -> 115|>","<|""Book"" -> 11, ""Theorem"" -> 1|>","<|""Book"" -> 11, ""Theorem"" -> 2|>","<|""Book"" -> 11, ""Theorem"" -> 3|>","<|""Book"" -> 11, ""Theorem"" -> 4|>","<|""Book"" -> 11, ""Theorem"" -> 5|>","<|""Book"" -> 11, ""Theorem"" -> 6|>","<|""Book"" -> 11, ""Theorem"" -> 7|>","<|""Book"" -> 11, ""Theorem"" -> 8|>","<|""Book"" -> 11, ""Theorem"" -> 9|>","<|""Book"" -> 11, ""Theorem"" -> 10|>","<|""Book"" -> 11, ""Theorem"" -> 11|>","<|""Book"" -> 11, ""Theorem"" -> 12|>","<|""Book"" -> 11, ""Theorem"" -> 13|>","<|""Book"" -> 11, ""Theorem"" -> 14|>","<|""Book"" -> 11, ""Theorem"" -> 15|>","<|""Book"" -> 11, ""Theorem"" -> 16|>","<|""Book"" -> 11, ""Theorem"" -> 17|>","<|""Book"" -> 11, ""Theorem"" -> 18|>","<|""Book"" -> 11, ""Theorem"" -> 19|>","<|""Book"" -> 11, ""Theorem"" -> 20|>","<|""Book"" -> 11, ""Theorem"" -> 21|>","<|""Book"" -> 11, ""Theorem"" -> 22|>","<|""Book"" -> 11, ""Theorem"" -> 23|>","<|""Book"" -> 11, ""Theorem"" -> 24|>","<|""Book"" -> 11, ""Theorem"" -> 25|>","<|""Book"" -> 11, ""Theorem"" -> 26|>","<|""Book"" -> 11, ""Theorem"" -> 27|>","<|""Book"" -> 11, ""Theorem"" -> 28|>","<|""Book"" -> 11, ""Theorem"" -> 29|>","<|""Book"" -> 11, ""Theorem"" -> 30|>","<|""Book"" -> 11, ""Theorem"" -> 31|>","<|""Book"" -> 11, ""Theorem"" -> 32|>","<|""Book"" -> 11, ""Theorem"" -> 33|>","<|""Book"" -> 11, ""Theorem"" -> 34|>","<|""Book"" -> 11, ""Theorem"" -> 35|>","<|""Book"" -> 11, ""Theorem"" -> 36|>","<|""Book"" -> 11, ""Theorem"" -> 37|>","<|""Book"" -> 11, ""Theorem"" -> 38|>","<|""Book"" -> 11, ""Theorem"" -> 39|>","<|""Book"" -> 12, ""Theorem"" -> 1|>","<|""Book"" -> 12, ""Theorem"" -> 2|>","<|""Book"" -> 12, ""Theorem"" -> 3|>","<|""Book"" -> 12, ""Theorem"" -> 4|>","<|""Book"" -> 12, ""Theorem"" -> 5|>","<|""Book"" -> 12, ""Theorem"" -> 6|>","<|""Book"" -> 12, ""Theorem"" -> 7|>","<|""Book"" -> 12, ""Theorem"" -> 8|>","<|""Book"" -> 12, ""Theorem"" -> 9|>","<|""Book"" -> 12, ""Theorem"" -> 10|>","<|""Book"" -> 12, ""Theorem"" -> 11|>","<|""Book"" -> 12, ""Theorem"" -> 12|>","<|""Book"" -> 12, ""Theorem"" -> 13|>","<|""Book"" -> 12, ""Theorem"" -> 14|>","<|""Book"" -> 12, ""Theorem"" -> 15|>","<|""Book"" -> 12, ""Theorem"" -> 16|>","<|""Book"" -> 12, ""Theorem"" -> 17|>","<|""Book"" -> 12, ""Theorem"" -> 18|>","<|""Book"" -> 13, ""Theorem"" -> 1|>","<|""Book"" -> 13, ""Theorem"" -> 2|>","<|""Book"" -> 13, ""Theorem"" -> 3|>","<|""Book"" -> 13, ""Theorem"" -> 4|>","<|""Book"" -> 13, ""Theorem"" -> 5|>","<|""Book"" -> 13, ""Theorem"" -> 6|>","<|""Book"" -> 13, ""Theorem"" -> 7|>","<|""Book"" -> 13, ""Theorem"" -> 8|>","<|""Book"" -> 13, ""Theorem"" -> 9|>","<|""Book"" -> 13, ""Theorem"" -> 10|>","<|""Book"" -> 13, ""Theorem"" -> 11|>","<|""Book"" -> 13, ""Theorem"" -> 12|>","<|""Book"" -> 13, ""Theorem"" -> 13|>","<|""Book"" -> 13, ""Theorem"" -> 14|>","<|""Book"" -> 13, ""Theorem"" -> 15|>","<|""Book"" -> 13, ""Theorem"" -> 16|>","<|""Book"" -> 13, ""Theorem"" -> 17|>","<|""Book"" -> 13, ""Theorem"" -> 18|>"
"<|""VertexLabel"" -> ""P1"", ""Text"" -> ""To draw a straight line from any point to any point."", ""TextWordCount"" -> 11, ""GreekText"" -> ""Ἠιτήσθω ἀπὸ παντὸς σημείου ἐπὶ πᾶν σημεῖον εὐθεῖαν γραμμὴν ἀγαγεῖν."", ""GreekTextWordCount"" -> 10, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""P2"", ""Text"" -> ""To produce a finite straight line continuously in a straight line."", ""TextWordCount"" -> 11, ""GreekText"" -> ""καὶ πεπερασμένην εὐθεῖαν κατὰ τὸ συνεχὲς ἐπ᾽ εὐθείας ἐκβαλεῖν."", ""GreekTextWordCount"" -> 10, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""P3"", ""Text"" -> ""To describe a circle with any centre and distance."", ""TextWordCount"" -> 9, ""GreekText"" -> ""καὶ παντὶ κέντρῳ καὶ διαστήματι κύκλον γράφεσθαι."", ""GreekTextWordCount"" -> 7, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""P4"", ""Text"" -> ""That all right angles are equal to one another."", ""TextWordCount"" -> 9, ""GreekText"" -> ""καὶ πάσας τὰς ὀρθὰς γωνίας ἴσας ἀλλήλαις εἶναι."", ""GreekTextWordCount"" -> 8, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""P5"", ""Text"" -> ""That, if a straight line falling on two straight lines make the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles."", ""TextWordCount"" -> 45, ""GreekText"" -> ""καὶ ἐὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη γωνίας δύο ὀρθῶν ἐλάσσονας ποιῇ, ἐκβαλλομένας τὰς δύο εὐθείας ἐπ᾽ ἄπειρον συμπίπτειν, ἐφ᾽ ἃ μέρη εἰσὶν αἱ τῶν δύο ὀρθῶν ἐλάσσονες."", ""GreekTextWordCount"" -> 37, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""CN1"", ""Text"" -> ""Things which are equal to the same thing are also equal to one another."", ""TextWordCount"" -> 14, ""GreekText"" -> ""Τὰ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα."", ""GreekTextWordCount"" -> 8, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""CN2"", ""Text"" -> ""If equals be added to equals, the wholes are equal."", ""TextWordCount"" -> 10, ""GreekText"" -> ""Καὶ ἐὰν ἴσοις ἴσα προστεθῇ, τὰ ὅλα ἐστὶν ἴσα."", ""GreekTextWordCount"" -> 9, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""CN3"", ""Text"" -> ""If equals be subtracted from equals, the remainders are equal."", ""TextWordCount"" -> 10, ""GreekText"" -> ""Καὶ ἐὰν ἀπὸ ἴσων ἴσα ἀφαιρεθῇ, τὰ καταλειπόμενά ἐστιν ἴσα."", ""GreekTextWordCount"" -> 10, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""CN4"", ""Text"" -> ""Things which coincide with one another are equal to one another."", ""TextWordCount"" -> 11, ""GreekText"" -> ""Καὶ τὰ ἐφαρμόζοντα ἐπ ̓ ἀλλήλα ἴσα ἀλλήλοις ἐστίν."", ""GreekTextWordCount"" -> 9, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""CN5"", ""Text"" -> ""The whole is greater than the part."", ""TextWordCount"" -> 7, ""GreekText"" -> ""Καὶ τὸ ὅλον τοῦ μέρους μεῖζόν [ἐστιν]."", ""GreekTextWordCount"" -> 7, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.1"", ""Text"" -> ""A point is that which has no part."", ""TextWordCount"" -> 8, ""GreekText"" -> ""σημεῖόν ἐστιν, οὗ μέρος οὐθέν."", ""GreekTextWordCount"" -> 5, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.2"", ""Text"" -> ""A line is breadthless length."", ""TextWordCount"" -> 5, ""GreekText"" -> ""γραμμὴ δὲ μῆκος ἀπλατές."", ""GreekTextWordCount"" -> 4, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.3"", ""Text"" -> ""The extremities of a line are points."", ""TextWordCount"" -> 7, ""GreekText"" -> ""γραμμῆς δὲ πέρατα σημεῖα."", ""GreekTextWordCount"" -> 4, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.4"", ""Text"" -> ""A straight line is a line which lies evenly with the points on itself."", ""TextWordCount"" -> 14, ""GreekText"" -> ""εὐθεῖα γραμμή ἐστιν, ἥτις ἐξ ἴσου τοῖς ἐφ᾽ ἑαυτῆς σημείοις κεῖται."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.5"", ""Text"" -> ""A surface is that which has length and breadth only."", ""TextWordCount"" -> 10, ""GreekText"" -> ""ἐπιφάνεια δέ ἐστιν, ὃ μῆκος καὶ πλάτος μόνον ἔχει."", ""GreekTextWordCount"" -> 9, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.6"", ""Text"" -> ""The extremities of a surface are lines."", ""TextWordCount"" -> 7, ""GreekText"" -> ""ἐπιφανείας δὲ πέρατα γραμμαί."", ""GreekTextWordCount"" -> 4, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.7"", ""Text"" -> ""A plane surface is a surface which lies evenly with the straight lines on itself."", ""TextWordCount"" -> 15, ""GreekText"" -> ""ἐπίπεδος ἐπιφάνειά ἐστιν, ἥτις ἐξ ἴσου ταῖς ἐφ᾽ ἑαυτῆς εὐθείαις κεῖται."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.8"", ""Text"" -> ""A plane angle is the inclination to one another of two lines in a plane which meet one another and do not lie in a straight line."", ""TextWordCount"" -> 27, ""GreekText"" -> ""ἐπίπεδος δὲ γωνία ἐστὶν ἡ ἐν ἐπιπέδῳ δύο γραμμῶν ἁπτομένων ἀλλήλων καὶ μὴ ἐπ᾽ εὐθείας κειμένων πρὸς ἀλλήλας τῶν γραμμῶν κλίσις."", ""GreekTextWordCount"" -> 22, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.9"", ""Text"" -> ""And when the lines containing the angle are straight, the angle is called rectilineal."", ""TextWordCount"" -> 14, ""GreekText"" -> ""ὅταν δὲ αἱ περιέχουσαι τὴν γωνίαν γραμμαὶ εὐθεῖαι ὦσιν, εὐθύγραμμος καλεῖται ἡ γωνία."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.10"", ""Text"" -> ""When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands."", ""TextWordCount"" -> 43, ""GreekText"" -> ""ὅταν δὲ εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστι, καὶ ἡ ἐφεστηκυῖα εὐθεῖα κάθετος καλεῖται, ἐφ᾽ ἣν ἐφέστηκεν."", ""GreekTextWordCount"" -> 29, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.11"", ""Text"" -> ""An obtuse angle is an angle greater than a right angle."", ""TextWordCount"" -> 11, ""GreekText"" -> ""ἀμβλεῖα γωνία ἐστὶν ἡ μείζων ὀρθῆς."", ""GreekTextWordCount"" -> 6, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.12"", ""Text"" -> ""An acute angle is an angle less than a right angle."", ""TextWordCount"" -> 11, ""GreekText"" -> ""ὀξεῖα δὲ ἡ ἐλάσσων ὀρθῆς."", ""GreekTextWordCount"" -> 5, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.13"", ""Text"" -> ""A boundary is that which is an extremity of anything."", ""TextWordCount"" -> 10, ""GreekText"" -> ""ὅρος ἐστίν, ὅ τινός ἐστι πέρας."", ""GreekTextWordCount"" -> 6, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.14"", ""Text"" -> ""A figure is that which is contained by any boundary or boundaries."", ""TextWordCount"" -> 12, ""GreekText"" -> ""σχῆμά ἐστι τὸ ὑπό τινος ἤ τινων ὅρων περιεχόμενον."", ""GreekTextWordCount"" -> 9, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.15"", ""Text"" -> ""A circle is a plane figure contained by one line such that all the straight lines falling upon it from one point among those lying within the figure are equal to one another;"", ""TextWordCount"" -> 33, ""GreekText"" -> ""κύκλος ἐστὶ σχῆμα ἐπίπεδον ὑπὸ μιᾶς γραμμῆς περιεχόμενον ἣ καλεῖται περιφέρεια, πρὸς ἣν ἀφ᾽ ἑνὸς σημείου τῶν ἐντὸς τοῦ σχήματος κειμένων πᾶσαι αἱ προσπίπτουσαι εὐθεῖαι πρὸς τὴν τοῦ κύκλου περιφέρειαν ἴσαι ἀλλήλαις εἰσίν."", ""GreekTextWordCount"" -> 34, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.16"", ""Text"" -> ""And the point is called the centre of the circle."", ""TextWordCount"" -> 10, ""GreekText"" -> ""κέντρον δὲ τοῦ κύκλου τὸ σημεῖον καλεῖται."", ""GreekTextWordCount"" -> 7, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.17"", ""Text"" -> ""A diameter of the circle is any straight line drawn through the centre and terminated in both directions by the circumference of the circle, and such a straight line also bisects the circle."", ""TextWordCount"" -> 33, ""GreekText"" -> ""διάμετρος δὲ τοῦ κύκλου ἐστὶν εὐθεῖά τις διὰ τοῦ κέντρου ἠγμένη καὶ περατουμένη ἐφ᾽ ἑκάτερα τὰ μέρη ὑπὸ τῆς τοῦ κύκλου περιφερείας, ἥτις καὶ δίχα τέμνει τὸν κύκλον."", ""GreekTextWordCount"" -> 29, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.18"", ""Text"" -> ""A semicircle is the figure contained by the diameter and the circumference cut off by it. And the centre of the semicircle is the same as that of the circle."", ""TextWordCount"" -> 30, ""GreekText"" -> ""ἡμικύκλιον δέ ἐστι τὸ περιεχόμενον σχῆμα ὑπό τε τῆς διαμέτρου καὶ τῆς ἀπολαμβανομένης ὑπ᾽ αὐτῆς περιφερείας. κέντρον δὲ τοῦ ἡμικυκλίου τὸ αὐτό, ὃ καὶ τοῦ κύκλου ἐστίν."", ""GreekTextWordCount"" -> 28, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.19"", ""Text"" -> ""Rectilineal figures are those which are contained by straight lines, trilateral figures being those contained by three, quadrilateral those contained by four, and multilateral those contained by more than four straight lines."", ""TextWordCount"" -> 32, ""GreekText"" -> ""σχήματα εὐθύγραμμά ἐστι τὰ ὑπὸ εὐθειῶν περιεχόμενα, τρίπλευρα μὲν τὰ ὑπὸ τριῶν, τετράπλευρα δὲ τὰ ὑπὸ τεσσάρων, πολύπλευρα δὲ τὰ ὑπὸ πλειόνων ἢ τεσσάρων εὐθειῶν περιεχόμενα."", ""GreekTextWordCount"" -> 26, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.20"", ""Text"" -> ""Of trilateral figures, an equilateral triangle is that which has its three sides equal, an isosceles triangle that which has two of its sides alone equal, and a scalene triangle that which has its three sides unequal."", ""TextWordCount"" -> 37, ""GreekText"" -> ""τῶν δὲ τριπλεύρων σχημάτων ἰσόπλευρον μὲν τρίγωνόν ἐστι τὸ τὰς τρεῖς ἴσας ἔχον πλευράς, ἰσοσκελὲς δὲ τὸ τὰς δύο μόνας ἴσας ἔχον πλευράς, σκαληνὸν δὲ τὸ τὰς τρεῖς ἀνίσους ἔχον πλευράς."", ""GreekTextWordCount"" -> 31, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.21"", ""Text"" -> ""Further, of trilateral figures, a right-angled triangle is that which has a right angle, an obtuse-angled triangle that which has an obtuse angle, and an acuteangled triangle that which has its three angles acute."", ""TextWordCount"" -> 34, ""GreekText"" -> ""ἔτι δὲ τῶν τριπλεύρων σχημάτων ὀρθογώνιον μὲν τρίγωνόν ἐστι τὸ ἔχον ὀρθὴν γωνίαν, ἀμβλυγώνιον δὲ τὸ ἔχον ἀμβλεῖαν γωνίαν, ὀξυγώνιον δὲ τὸ τὰς τρεῖς ὀξείας ἔχον γωνίας."", ""GreekTextWordCount"" -> 27, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.22"", ""Text"" -> ""Of quadrilateral figures, a square is that which is both equilateral and right-angled; an oblong that which is right-angled but not equilateral; a rhombus that which is equilateral but not right-angled; and a rhomboid that which has its opposite sides and angles equal to one another but is neither equilateral nor right-angled. And let quadrilaterals other than these be called trapezia."", ""TextWordCount"" -> 61, ""GreekText"" -> ""τῶν δὲ τετραπλεύρων σχημάτων τετράγωνον μέν ἐστιν, ὃ ἰσόπλευρόν τέ ἐστι καὶ ὀρθογώνιον, ἑτερόμηκες δέ, ὃ ὀρθογώνιον μέν, οὐκ ἰσόπλευρον δέ, ῥόμβος δέ, ὃ ἰσόπλευρον μέν, οὐκ ὀρθογώνιον δέ, ῥομβοειδὲς δὲ τὸ τὰς ἀπεναντίον πλευράς τε καὶ γωνίας ἴσας ἀλλήλαις ἔχον, ὃ οὔτε ἰσόπλευρόν ἐστιν οὔτε ὀρθογώνιον: τὰ δὲ παρὰ ταῦτα τετράπλευρα τραπέζια καλείσθω."", ""GreekTextWordCount"" -> 54, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D1.23"", ""Text"" -> ""Parallel straight lines are straight lines which, being in the same plane and being produced indefinitely in both directions, do not meet one another in either direction."", ""TextWordCount"" -> 27, ""GreekText"" -> ""παράλληλοί εἰσιν εὐθεῖαι, αἵτινες ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι καὶ ἐκβαλλόμεναι εἰς ἄπειρον ἐφ᾽ ἑκάτερα τὰ μέρη ἐπὶ μηδέτερα συμπίπτουσιν ἀλλήλαις."", ""GreekTextWordCount"" -> 22, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D2.1"", ""Text"" -> ""Any rectangular parallelogram is said to be contained by the two straight lines containing the right angle."", ""TextWordCount"" -> 17, ""GreekText"" -> ""πᾶν παραλληλόγραμμον ὀρθογώνιον περιέχεσθαι λέγεται ὑπὸ δύο τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν εὐθειῶν."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D2.2"", ""Text"" -> ""And in any parallelogrammic area let any one whatever of the parallelograms about its diameter with the two complements be called a gnomon."", ""TextWordCount"" -> 23, ""GreekText"" -> ""παντὸς δὲ παραλληλογράμμου χωρίου τῶν περὶ τὴν διάμετρον αὐτοῦ παραλληλογράμμων ἓν ὁποιονοῦν σὺν τοῖς δυσὶ παραπληρώμασι γνώμων καλείσθω."", ""GreekTextWordCount"" -> 18, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.1"", ""Text"" -> ""Equal circles are those the diameters of which are equal, or the radii of which are equal."", ""TextWordCount"" -> 17, ""GreekText"" -> ""ἴσοι κύκλοι εἰσίν, ὧν αἱ διάμετροι ἴσαι εἰσίν, ἢ ὧν αἱ ἐκ τῶν κέντρων ἴσαι εἰσίν."", ""GreekTextWordCount"" -> 16, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.2"", ""Text"" -> ""A straight line is said to touch a circle which, meeting the circle and being produced, does not cut the circle."", ""TextWordCount"" -> 21, ""GreekText"" -> ""εὐθεῖα κύκλου ἐφάπτεσθαι λέγεται, ἥτις ἁπτομένη τοῦ κύκλου καὶ ἐκβαλλομένη οὐ τέμνει τὸν κύκλον."", ""GreekTextWordCount"" -> 14, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.3"", ""Text"" -> ""Circles are said to touch one another which, meeting one another, do not cut one another."", ""TextWordCount"" -> 16, ""GreekText"" -> ""κύκλοι ἐφάπτεσθαι ἀλλήλων λέγονται οἵτινες ἁπτόμενοι ἀλλήλων οὐ τέμνουσιν ἀλλήλους."", ""GreekTextWordCount"" -> 10, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.4"", ""Text"" -> ""In a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal."", ""TextWordCount"" -> 25, ""GreekText"" -> ""ἐν κύκλῳ ἴσον ἀπέχειν ἀπὸ τοῦ κέντρου εὐθεῖαι λέγονται, ὅταν αἱ ἀπὸ τοῦ κέντρου ἐπ᾽ αὐτὰς κάθετοι ἀγόμεναι ἴσαι ὦσιν."", ""GreekTextWordCount"" -> 21, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.5"", ""Text"" -> ""And that straight line is said to be at a greater distance on which the greater perpendicular falls."", ""TextWordCount"" -> 18, ""GreekText"" -> ""μεῖζον δὲ ἀπέχειν λέγεται, ἐφ᾽ ἣν ἡ μείζων κάθετος πίπτει."", ""GreekTextWordCount"" -> 11, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.6"", ""Text"" -> ""A segment of a circle is the figure contained by a straight line and a circumference of a circle."", ""TextWordCount"" -> 19, ""GreekText"" -> ""τμῆμα κύκλου ἐστὶ τὸ περιεχόμενον σχῆμα ὑπό τε εὐθείας καὶ κύκλου περιφερείας."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.7"", ""Text"" -> ""An angle of a segment is that contained by a straight line and a circumference of a circle."", ""TextWordCount"" -> 18, ""GreekText"" -> ""τμήματος δὲ γωνία ἐστὶν ἡ περιεχομένη ὑπό τε εὐθείας καὶ κύκλου περιφερείας."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.8"", ""Text"" -> ""An angle in a segment is the angle which, when a point is taken on the circumference of the segment and straight lines are joined from it to the extremities of the straight line which is the base of the segment, is contained by the straight lines so joined."", ""TextWordCount"" -> 49, ""GreekText"" -> ""ἐν τμήματι δὲ γωνία ἐστίν, ὅταν ἐπὶ τῆς περιφερείας τοῦ τμήματος ληφθῇ τι σημεῖον καὶ ἀπ᾽ αὐτοῦ ἐπὶ τὰ πέρατα τῆς εὐθείας, ἥ ἐστι βάσις τοῦ τμήματος, ἐπιζευχθῶσιν εὐθεῖαι, ἡ περιεχομένη γωνία ὑπὸ τῶν ἐπιζευχθεισῶν εὐθειῶν."", ""GreekTextWordCount"" -> 37, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.9"", ""Text"" -> ""And, when the straight lines containing the angle cut off a circumference, the angle is said to stand upon that circumference."", ""TextWordCount"" -> 21, ""GreekText"" -> ""ὅταν δὲ αἱ περιέχουσαι τὴν γωνίαν εὐθεῖαι ἀπολαμβάνωσί τινα περιφέρειαν, ἐπ᾽ ἐκείνης λέγεται βεβηκέναι ἡ γωνία."", ""GreekTextWordCount"" -> 17, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.10"", ""Text"" -> ""A sector of a circle is the figure which, when an angle is constructed at the centre of the circle, is contained by the straight lines containing the angle and the circumference cut off by them."", ""TextWordCount"" -> 36, ""GreekText"" -> ""Τομεὺς δὲ κύκλου ἐστίν, ὅταν πρὸς τῷ κέντρῳ τοῦ κύκλου συσταθῇ γωνία, τὸ περιεχόμενον σχῆμα ὑπό τε τῶν τὴν γωνίαν περιεχουσῶν εὐθειῶν καὶ τῆς ἀπολαμβανομένης ὑπ᾽ αὐτῶν περιφερείας."", ""GreekTextWordCount"" -> 29, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D3.11"", ""Text"" -> ""Similar segments of circles are those which admit equal angles, or in which the angles are equal to one another."", ""TextWordCount"" -> 20, ""GreekText"" -> ""ὅμοια τμήματα κύκλων ἐστὶ τὰ δεχόμενα γωνίας ἴσας, ἢ ἐν οἷς αἱ γωνίαι ἴσαι ἀλλήλαις εἰσίν."", ""GreekTextWordCount"" -> 16, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D4.1"", ""Text"" -> ""A rectilineal figure is said to be inscribed in a rectilineal figure when the respective angles of the inscribed figure lie on the respective sides of that in which it is inscribed."", ""TextWordCount"" -> 32, ""GreekText"" -> ""σχῆμα εὐθύγραμμον εἰς σχῆμα εὐθύγραμμον ἐγγράφεσθαι λέγεται, ὅταν ἑκάστη τῶν τοῦ ἐγγραφομένου σχήματος γωνιῶν ἑκάστης πλευρᾶς τοῦ, εἰς ὃ ἐγγράφεται, ἅπτηται."", ""GreekTextWordCount"" -> 21, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D4.2"", ""Text"" -> ""Similarly a figure is said to be circumscribed about a figure when the respective sides of the circumscribed figure pass through the respective angles of that about which it is circumscribed."", ""TextWordCount"" -> 31, ""GreekText"" -> ""σχῆμα δὲ ὁμοίως περὶ σχῆμα περιγράφεσθαι λέγεται, ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου ἑκάστης γωνίας τοῦ, περὶ ὃ περιγράφεται, ἅπτηται."", ""GreekTextWordCount"" -> 19, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D4.3"", ""Text"" -> ""A rectilineal figure is said to be inscribed in a circle when each angle of the inscribed figure lies on the circumference of the circle."", ""TextWordCount"" -> 25, ""GreekText"" -> ""σχῆμα εὐθύγραμμον εἰς κύκλον ἐγγράφεσθαι λέγεται, ὅταν ἑκάστη γωνία τοῦ ἐγγραφομένου ἅπτηται τῆς τοῦ κύκλου περιφερείας."", ""GreekTextWordCount"" -> 16, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D4.4"", ""Text"" -> ""A rectilineal figure is said to be circumscribed about a circle, when each side of the circumscribed figure touches the circumference of the circle."", ""TextWordCount"" -> 24, ""GreekText"" -> ""σχῆμα δὲ εὐθύγραμμον περὶ κύκλον περιγράφεσθαι λέγεται, ὅταν ἑκάστη πλευρὰ τοῦ περιγραφομένου ἐφάπτηται τῆς τοῦ κύκλου περιφερείας."", ""GreekTextWordCount"" -> 17, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D4.5"", ""Text"" -> ""Similarly a circle is said to be inscribed in a figure when the circumference of the circle touches each side of the figure in which it is inscribed."", ""TextWordCount"" -> 28, ""GreekText"" -> ""κύκλος δὲ εἰς σχῆμα ὁμοίως ἐγγράφεσθαι λέγεται, ὅταν ἡ τοῦ κύκλου περιφέρεια ἑκάστης πλευρᾶς τοῦ, εἰς ὃ ἐγγράφεται, ἅπτηται."", ""GreekTextWordCount"" -> 19, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D4.6"", ""Text"" -> ""A circle is said to be circumscribed about a figure when the circumference of the circle passes through each angle of the figure about which it is circumscribed."", ""TextWordCount"" -> 28, ""GreekText"" -> ""κύκλος δὲ περὶ σχῆμα περιγράφεσθαι λέγεται, ὅταν ἡ τοῦ κύκλου περιφέρεια ἑκάστης γωνίας τοῦ, περὶ ὃ περιγράφεται, ἅπτηται."", ""GreekTextWordCount"" -> 18, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D4.7"", ""Text"" -> ""A straight line is said to be fitted into a circle when its extremities are on the circumference of the circle."", ""TextWordCount"" -> 21, ""GreekText"" -> ""εὐθεῖα εἰς κύκλον ἐναρμόζεσθαι λέγεται, ὅταν τὰ πέρατα αὐτῆς ἐπὶ τῆς περιφερείας ᾖ τοῦ κύκλου."", ""GreekTextWordCount"" -> 15, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.1"", ""Text"" -> ""A magnitude is a part of a magnitude, the less of the greater, when it measures the greater."", ""TextWordCount"" -> 18, ""GreekText"" -> ""μέρος ἐστὶ μέγεθος μεγέθους τὸ ἔλασσον τοῦ μείζονος, ὅταν καταμετρῇ τὸ μεῖζον."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.2"", ""Text"" -> ""The greater is a multiple of the less when it is measured by the less."", ""TextWordCount"" -> 15, ""GreekText"" -> ""πολλαπλάσιον δὲ τὸ μεῖζον τοῦ ἐλάττονος, ὅταν καταμετρῆται ὑπὸ τοῦ ἐλάττονος."", ""GreekTextWordCount"" -> 11, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.3"", ""Text"" -> ""A ratio is a sort of relation in respect of size between two magnitudes of the same kind."", ""TextWordCount"" -> 18, ""GreekText"" -> ""λόγος ἐστὶ δύο μεγεθῶν ὁμογενῶν ἡ κατὰ πηλικότητά ποια σχέσις."", ""GreekTextWordCount"" -> 10, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.4"", ""Text"" -> ""Magnitudes are said to have a ratio to one another which are capable, when multiplied, of exceeding one another."", ""TextWordCount"" -> 19, ""GreekText"" -> ""λόγον ἔχειν πρὸς ἄλληλα μεγέθη λέγεται, ἃ δύναται πολλαπλασιαζόμενα ἀλλήλων ὑπερέχειν."", ""GreekTextWordCount"" -> 11, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.5"", ""Text"" -> ""Magnitudes are said to be in the same ratio, the first to the second and the third to the fourth, when, if any equimultiples whatever be taken of the first and third, and any equimultiples whatever of the second and fourth, the former equimultiples alike exceed, are alike equal to, or alike fall short of, the latter equimultiples respectively taken in corresponding order."", ""TextWordCount"" -> 63, ""GreekText"" -> ""ἐν τῷ αὐτῷ λόγῳ μεγέθη λέγεται εἶναι πρῶτον πρὸς δεύτερον καὶ τρίτον πρὸς τέταρτον, ὅταν τὰ τοῦ πρώτου καὶ τρίτου ἰσάκις πολλαπλάσια τῶν τοῦ δευτέρου καὶ τετάρτου ἰσάκις πολλαπλασίων καθ᾽ ὁποιονοῦν πολλαπλασιασμὸν ἑκάτερον ἑκατέρου ἢ ἅμα ὑπερέχῃ ἢ ἅμα ἴσα ᾖ ἢ ἅμα ἐλλείπῃ ληφθέντα κατάλληλα."", ""GreekTextWordCount"" -> 47, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.6"", ""Text"" -> ""Let magnitudes which have the same ratio be called proportional."", ""TextWordCount"" -> 10, ""GreekText"" -> ""τὰ δὲ τὸν αὐτὸν ἔχοντα λόγον μεγέθη ἀνάλογον καλείσθω."", ""GreekTextWordCount"" -> 9, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.7"", ""Text"" -> ""When, of the equimultiples, the multiple of the first magnitude exceeds the multiple of the second, but the multiple of the third does not exceed the multiple of the fourth, then the first is said to have a greater ratio to the second than the third has to the fourth."", ""TextWordCount"" -> 50, ""GreekText"" -> ""ὅταν δὲ τῶν ἰσάκις πολλαπλασίων τὸ μὲν τοῦ πρώτου πολλαπλάσιον ὑπερέχῃ τοῦ τοῦ δευτέρου πολλαπλασίου, τὸ δὲ τοῦ τρίτου πολλαπλάσιον μὴ ὑπερέχῃ τοῦ τοῦ τετάρτου πολλαπλασίου, τότε τὸ πρῶτον πρὸς τὸ δεύτερον μείζονα λόγον ἔχειν λέγεται, ἤπερ τὸ τρίτον πρὸς τὸ τέταρτον."", ""GreekTextWordCount"" -> 42, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.8"", ""Text"" -> ""A proportion in three terms is the least possible."", ""TextWordCount"" -> 9, ""GreekText"" -> ""ἀναλογία δὲ ἐν τρισὶν ὅροις ἐλαχίστη ἐστίν."", ""GreekTextWordCount"" -> 7, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.9"", ""Text"" -> ""When three magnitudes are proportional, the first is said to have to the third the duplicate ratio of that which it has to the second."", ""TextWordCount"" -> 25, ""GreekText"" -> ""ὅταν δὲ τρία μεγέθη ἀνάλογον ᾖ, τὸ πρῶτον πρὸς τὸ τρίτον διπλασίονα λόγον ἔχειν λέγεται ἤπερ πρὸς τὸ δεύτερον."", ""GreekTextWordCount"" -> 19, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.10"", ""Text"" -> ""When four magnitudes are continuously proportional, the first is said to have to the fourth the triplicate ratio of that which it has to the second, and so on continually, whatever be the proportion."", ""TextWordCount"" -> 34, ""GreekText"" -> ""ὅταν δὲ τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ πρῶτον πρὸς τὸ τέταρτον τριπλασίονα λόγον ἔχειν λέγεται ἤπερ πρὸς τὸ δεύτερον, καὶ ἀεὶ ἑξῆς ὁμοίως, ὡς ἂν ἡ ἀναλογία ὑπάρχῃ."", ""GreekTextWordCount"" -> 28, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.11"", ""Text"" -> ""The term corresponding magnitudes is used of antecedents in relation to antecedents, and of consequents in relation to consequents."", ""TextWordCount"" -> 19, ""GreekText"" -> ""ὁμόλογα μεγέθη λέγεται τὰ μὲν ἡγούμενα τοῖς ἡγουμένοις τὰ δὲ ἑπόμενα τοῖς ἑπομένοις."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.12"", ""Text"" -> ""Alternate ratio means taking the antecedent in relation to the antecedent and the consequent in relation to the consequent."", ""TextWordCount"" -> 19, ""GreekText"" -> ""ἐναλλὰξ λόγος ἐστὶ λῆψις τοῦ ἡγουμένου πρὸς τὸ ἡγούμενον καὶ τοῦ ἑπομένου πρὸς τὸ ἑπόμενον."", ""GreekTextWordCount"" -> 15, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.13"", ""Text"" -> ""Inverse ratio means taking the consequent as antecedent in relation to the antecedent as consequent."", ""TextWordCount"" -> 15, ""GreekText"" -> ""ἀνάπαλιν λόγος ἐστὶ λῆψις τοῦ ἑπομένου ὡς ἡγουμένου πρὸς τὸ ἡγούμενον ὡς ἑπόμενον."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.14"", ""Text"" -> ""Composition of a ratio means taking the antecedent together with the consequent as one in relation to the consequent by itself."", ""TextWordCount"" -> 21, ""GreekText"" -> ""σύνθεσις λόγου ἐστὶ λῆψις τοῦ ἡγουμένου μετὰ τοῦ ἑπομένου ὡς ἑνὸς πρὸς αὐτὸ τὸ ἑπόμενον."", ""GreekTextWordCount"" -> 15, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.15"", ""Text"" -> ""Separation of a ratio means taking the excess by which the antecedent exceeds the consequent in relation to the consequent by itself."", ""TextWordCount"" -> 22, ""GreekText"" -> ""διαίρεσις λόγου ἐστὶ λῆψις τῆς ὑπεροχῆς, ᾗ ὑπερέχει τὸ ἡγούμενον τοῦ ἑπομένου, πρὸς αὐτὸ τὸ ἑπόμενον."", ""GreekTextWordCount"" -> 16, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.16"", ""Text"" -> ""Conversion of a ratio means taking the antecedent in relation to the excess by which the antecedent exceeds the consequent."", ""TextWordCount"" -> 20, ""GreekText"" -> ""ἀναστροφὴ λόγου ἐστὶ λῆψις τοῦ ἡγουμένου πρὸς τὴν ὑπεροχήν, ᾗ ὑπερέχει τὸ ἡγούμενον τοῦ ἑπομένου."", ""GreekTextWordCount"" -> 15, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.17"", ""Text"" -> ""A ratio ex aequali arises when, there being several magnitudes and another set equal to them in multitude which taken two and two are in the same proportion, as the first is to the last among the first magnitudes, so is the first to the last among the second magnitudes; Or, in other words, it means taking the extreme terms by virtue of the removal of the intermediate terms."", ""TextWordCount"" -> 69, ""GreekText"" -> ""δι᾽ ἴσου λόγος ἐστὶ πλειόνων ὄντων μεγεθῶν καὶ ἄλλων αὐτοῖς ἴσων τὸ πλῆθος σύνδυο λαμβανομένων καὶ ἐν τῷ αὐτῷ λόγῳ, ὅταν ᾖ ὡς ἐν τοῖς πρώτοις μεγέθεσι τὸ πρῶτον πρὸς τὸ ἔσχατον, οὕτως ἐν τοῖς δευτέροις μεγέθεσι τὸ πρῶτον πρὸς τὸ ἔσχατον: ἢ ἄλλως: λῆψις τῶν ἄκρων καθ᾽ ὑπεξαίρεσιν τῶν μέσων."", ""GreekTextWordCount"" -> 53, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D5.18"", ""Text"" -> ""A perturbed proportion arises when, there being three magnitudes and another set equal to them in multitude, as antecedent is to consequent among the first magnitudes, so is antecedent to consequent among the second magnitudes, while, as the consequent is to a third among the first magnitudes, so is a third to the antecedent among the second magnitudes."", ""TextWordCount"" -> 58, ""GreekText"" -> ""τεταραγμένη δὲ ἀναλογία ἐστίν, ὅταν τριῶν ὄντων μεγεθῶν καὶ ἄλλων αὐτοῖς ἴσων τὸ πλῆθος γίνηται ὡς μὲν ἐν τοῖς πρώτοις μεγέθεσιν ἡγούμενον πρὸς ἑπόμενον, οὕτως ἐν τοῖς δευτέροις μεγέθεσιν ἡγούμενον πρὸς ἑπόμενον, ὡς δὲ ἐν τοῖς πρώτοις μεγέθεσιν ἑπόμενον πρὸς ἄλλο τι, οὕτως ἐν τοῖς δευτέροις ἄλλο τι πρὸς ἡγούμενον."", ""GreekTextWordCount"" -> 50, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D6.1"", ""Text"" -> ""Similar rectilineal figures are such as have their angles severally equal and the sides about the equal angles proportional."", ""TextWordCount"" -> 19, ""GreekText"" -> ""ὅμοια σχήματα εὐθύγραμμά ἐστιν, ὅσα τάς τε γωνίας ἴσας ἔχει κατὰ μίαν καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον."", ""GreekTextWordCount"" -> 20, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D6.2"", ""Text"" -> ""When two sides of one figure together with two sides of another figure form antecedents and consequents in a proportion, the figures are reciprocally related."", ""TextWordCount"" -> 25, ""GreekText"" -> ""Ἀντιπεπονθότα δὲ σχήματά ἐστιν, ὅταν ἐν ἑκατέρῳ τῶν σχημάτων ἡγούμενοί τε καὶ ἑπόμενοι λόγοι ὦσιν."", ""GreekTextWordCount"" -> 15, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D6.3"", ""Text"" -> ""A straight line is said to have been cut in extreme and mean ratio when, as the whole line is to the greater segment, so is the greater to the less."", ""TextWordCount"" -> 31, ""GreekText"" -> ""ἄκρον καὶ μέσον λόγον εὐθεῖα τετμῆσθαι λέγεται, ὅταν ᾖ ὡς ἡ ὅλη πρὸς τὸ μεῖζον τμῆμα, οὕτως τὸ μεῖζον πρὸς τὸ ἔλαττον."", ""GreekTextWordCount"" -> 22, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D6.4"", ""Text"" -> ""The height of any figure is the perpendicular drawn from the vertex to the base."", ""TextWordCount"" -> 15, ""GreekText"" -> ""ὕψος ἐστὶ παντὸς σχήματος ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν βάσιν κάθετος ἀγομένη."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D6.5"", ""Text"" -> ""A ratio is said to be compounded of ratios when the sizes of the ratios multiplied together make som (?ratio, or size). A parallelogram not \""filling\"" a line is a figure lesser than the line. If it has a surplus and the line is not enough, it is a figure greater than the line."", ""TextWordCount"" -> 54, ""GreekText"" -> ""Λόγος ἐκ λόγων συγκεῖσθαι λέγεται, ὅταν αἱ τῶν λόγων πηλικότητες ἐφ᾽ ἑαυτὰς πολλαπλασιασθεῖσαι ποιῶσί τινα."", ""GreekTextWordCount"" -> 16, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.1"", ""Text"" -> ""An unit is that by virtue of which each of the things that exist is called one."", ""TextWordCount"" -> 17, ""GreekText"" -> ""μονάς ἐστιν, καθ᾽ ἣν ἕκαστον τῶν ὄντων ἓν λέγεται."", ""GreekTextWordCount"" -> 10, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.2"", ""Text"" -> ""A number is a multitude composed of units."", ""TextWordCount"" -> 8, ""GreekText"" -> ""ἀριθμὸς δὲ τὸ ἐκ μονάδων συγκείμενον πλῆθος."", ""GreekTextWordCount"" -> 7, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.3"", ""Text"" -> ""A number is a part of a number, the less of the greater, when it measures the greater;"", ""TextWordCount"" -> 18, ""GreekText"" -> ""μέρος ἐστὶν ἀριθμὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ μείζονος, ὅταν καταμετρῇ τὸν μείζονα."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.4"", ""Text"" -> ""but parts when it does not measure it."", ""TextWordCount"" -> 8, ""GreekText"" -> ""μέρη δέ, ὅταν μὴ καταμετρῇ."", ""GreekTextWordCount"" -> 5, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.5"", ""Text"" -> ""The greater number is a multiple of the less when it is measured by the less."", ""TextWordCount"" -> 16, ""GreekText"" -> ""πολλαπλάσιος δὲ ὁ μείζων τοῦ ἐλάσσονος, ὅταν καταμετρῆται ὑπὸ τοῦ ἐλάσσονος."", ""GreekTextWordCount"" -> 11, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.6"", ""Text"" -> ""An even number is that which is divisible into two equal parts."", ""TextWordCount"" -> 12, ""GreekText"" -> ""ἄρτιος ἀριθμός ἐστιν ὁ δίχα διαιρούμενος."", ""GreekTextWordCount"" -> 6, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.7"", ""Text"" -> ""An odd number is that which is not divisible into two equal parts, or that which differs by an unit from an even number."", ""TextWordCount"" -> 24, ""GreekText"" -> ""περισσὸς δὲ ὁ μὴ διαιρούμενος δίχα ἢ ὁ μονάδι διαφέρων ἀρτίου ἀριθμοῦ."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.8"", ""Text"" -> ""An even-times even number is that which is measured by an even number according to an even number."", ""TextWordCount"" -> 18, ""GreekText"" -> ""ἀρτιάκις ἄρτιος ἀριθμός ἐστιν ὁ ὑπὸ ἀρτίου ἀριθμοῦ μετρούμενος κατὰ ἄρτιον ἀριθμόν."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.9"", ""Text"" -> ""An even-times odd number is that which is measured by an even number according to an odd number."", ""TextWordCount"" -> 18, ""GreekText"" -> ""ἀρτιάκις δὲ περισσός ἐστιν ὁ ὑπὸ ἀρτίου ἀριθμοῦ μετρούμενος κατὰ περισσὸν ἀριθμόν. Περισσάκις ἀρτιός ἐστιν ὁ ὑπὸ περισσοῦ ἀριθμοῦ μετρούμενος κατὰ ἄρτιον ἀριθμόν."", ""GreekTextWordCount"" -> 23, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.10"", ""Text"" -> ""An odd-times odd number is that which is measured by an odd number according to an odd number."", ""TextWordCount"" -> 18, ""GreekText"" -> ""περισσάκις δὲ περισσὸς ἀριθμός ἐστιν ὁ ὑπὸ περισσοῦ ἀριθμοῦ μετρούμενος κατὰ περισσὸν ἀριθμόν."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.11"", ""Text"" -> ""A prime number is that which is measured by an unit alone."", ""TextWordCount"" -> 12, ""GreekText"" -> ""πρῶτος ἀριθμός ἐστιν ὁ μονάδι μόνῃ μετρούμενος."", ""GreekTextWordCount"" -> 7, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.12"", ""Text"" -> ""Numbers prime to one another are those which are measured by an unit alone as a common measure."", ""TextWordCount"" -> 18, ""GreekText"" -> ""πρῶτοι πρὸς ἀλλήλους ἀριθμοί εἰσιν οἱ μονάδι μόνῃ μετρούμενοι κοινῷ μέτρῳ."", ""GreekTextWordCount"" -> 11, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.13"", ""Text"" -> ""A composite number is that which is measured by some number."", ""TextWordCount"" -> 11, ""GreekText"" -> ""σύνθετος ἀριθμός ἐστιν ὁ ἀριθμῷ τινι μετρούμενος."", ""GreekTextWordCount"" -> 7, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.14"", ""Text"" -> ""Numbers composite to one another are those which are measured by some number as a common measure."", ""TextWordCount"" -> 17, ""GreekText"" -> ""σύνθετοι δὲ πρὸς ἀλλήλους ἀριθμοί εἰσιν οἱ ἀριθμῷ τινι μετρούμενοι κοινῷ μέτρῳ."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.15"", ""Text"" -> ""A number is said to multiply a number when that which is multiplied is added to itself as many times as there are units in the other, and thus some number is produced."", ""TextWordCount"" -> 33, ""GreekText"" -> ""ἀριθμὸς ἀριθμὸν πολλαπλασιάζειν λέγεται, ὅταν, ὅσαι εἰσὶν ἐν αὐτῷ μονάδες, τοσαυτάκις συντεθῇ ὁ πολλαπλασιαζόμενος, καὶ γένηταί τις."", ""GreekTextWordCount"" -> 17, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.16"", ""Text"" -> ""And, when two numbers having multiplied one another make some number, the number so produced is called plane, and its sides are the numbers which have multiplied one another."", ""TextWordCount"" -> 29, ""GreekText"" -> ""ὅταν δὲ δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, ὁ γενόμενος ἐπίπεδος καλεῖται, πλευραὶ δὲ αὐτοῦ οἱ πολλαπλασιάσαντες ἀλλήλους ἀριθμοί."", ""GreekTextWordCount"" -> 19, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.17"", ""Text"" -> ""And, when three numbers having multiplied one another make some number, the number so produced is solid, and its sides are the numbers which have multiplied one another."", ""TextWordCount"" -> 28, ""GreekText"" -> ""ὅταν δὲ τρεῖς ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, ὁ γενόμενος στερεός ἐστιν, πλευραὶ δὲ αὐτοῦ οἱ πολλαπλασιάσαντες ἀλλήλους ἀριθμοί."", ""GreekTextWordCount"" -> 19, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.18"", ""Text"" -> ""A square number is equal multiplied by equal, or a number which is contained by two equal numbers."", ""TextWordCount"" -> 18, ""GreekText"" -> ""τετράγωνος ἀριθμός ἐστιν ὁ ἰσάκις ἴσος ἢ ὁ ὑπὸ δύο ἴσων ἀριθμῶν περιεχόμενος."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.19"", ""Text"" -> ""And a cube is equal multiplied by equal and again by equal, or a number which is contained by three equal numbers."", ""TextWordCount"" -> 22, ""GreekText"" -> ""κύβος δὲ ὁ ἰσάκις ἴσος ἰσάκις ἢ ὁ ὑπὸ τριῶν ἴσων ἀριθμῶν περιεχόμενος."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.20"", ""Text"" -> ""Numbers are proportional when the first is the same multiple, or the same part, or the same parts, of the second that the third is of the fourth."", ""TextWordCount"" -> 28, ""GreekText"" -> ""ἀριθμοὶ ἀνάλογόν εἰσιν, ὅταν ὁ πρῶτος τοῦ δευτέρου καὶ ὁ τρίτος τοῦ τετάρτου ἰσάκις ᾖ πολλαπλάσιος ἢ τὸ αὐτὸ μέρος ἢ τὰ αὐτὰ μέρη ὦσιν."", ""GreekTextWordCount"" -> 25, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.21"", ""Text"" -> ""Similar plane and solid numbers are those which have their sides proportional."", ""TextWordCount"" -> 12, ""GreekText"" -> ""ὅμοιοι ἐπίπεδοι καὶ στερεοὶ ἀριθμοί εἰσιν οἱ ἀνάλογον ἔχοντες τὰς πλευράς."", ""GreekTextWordCount"" -> 11, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D7.22"", ""Text"" -> ""A perfect number is that which is equal to its own parts."", ""TextWordCount"" -> 12, ""GreekText"" -> ""τέλειος ἀριθμός ἐστιν ὁ τοῖς ἑαυτοῦ μέρεσιν ἴσος ὤν."", ""GreekTextWordCount"" -> 9, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.1.1"", ""Text"" -> ""Those magnitudes are said to be commensurable which are measured by the same measure, and those incommensurable which cannot have any common measure."", ""TextWordCount"" -> 23, ""GreekText"" -> ""σύμμετρα μεγέθη λέγεται τὰ τῷ αὐτῷ μέτρῳ μετρούμενα, ἀσύμμετρα δέ, ὧν μηδὲν ἐνδέχεται κοινὸν μέτρον γενέσθαι."", ""GreekTextWordCount"" -> 16, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.1.2"", ""Text"" -> ""Straight lines are commensurable in square when the squares on them are measured by the same area, and incommensurable in square when the squares on them cannot possibly have any area as a common measure."", ""TextWordCount"" -> 35, ""GreekText"" -> ""εὐθεῖαι δυνάμει σύμμετροί εἰσιν, ὅταν τὰ ἀπ᾽ αὐτῶν τετράγωνα τῷ αὐτῷ χωρίῳ μετρῆται, ἀσύμμετροι δέ, ὅταν τοῖς ἀπ᾽ αὐτῶν τετραγώνοις μηδὲν ἐνδέχηται χωρίον κοινὸν μέτρον γενέσθαι."", ""GreekTextWordCount"" -> 28, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.1.3"", ""Text"" -> ""With these hypotheses, it is proved that there exist straight lines infinite in multitude which are commensurable and incommensurable respectively, some in length only, and others in square also, with an assigned straight line. Let then the assigned straight line be called rational, and those straight lines which are commensurable with it, whether in length and in square or in square only, rational, but those which are incommensurable with it irrational."", ""TextWordCount"" -> 71, ""GreekText"" -> ""τούτων ὑποκειμένων δείκνυται, ὅτι τῇ προτεθείσῃ εὐθείᾳ ὑπάρχουσιν εὐθεῖαι πλήθει ἄπειροι σύμμετροί τε καὶ ἀσύμμετροι αἱ μὲν μήκει μόνον, αἱ δὲ καὶ δυνάμει. καλείσθω οὖν ἡ μὲν προτεθεῖσα εὐθεῖα ῥητή, καὶ αἱ ταύτῃ σύμμετροι εἴτε μήκει καὶ δυνάμει εἴτε δυνάμει μόνον ῥηταί, αἱ δὲ ταύτῃ ἀσύμμετροι ἄλογοι καλείσθωσαν."", ""GreekTextWordCount"" -> 48, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.1.4"", ""Text"" -> ""And let the square on the assigned straight line be called rational and those areas which are commensurable with it rational, but those which are incommensurable with it irrational, and the straight lines which produce them irrational, that is, in case the areas are squares, the sides themselves, but in case they are any other rectilineal figures, the straight lines on which are described squares equal to them."", ""TextWordCount"" -> 68, ""GreekText"" -> ""καὶ τὸ μὲν ἀπὸ τῆς προτεθείσης εὐθείας τετράγωνον ῥητόν, καὶ τὰ τούτῳ σύμμετρα ῥητά, τὰ δὲ τούτῳ ἀσύμμετρα ἄλογα καλείσθω, καὶ αἱ δυνάμεναι αὐτὰ ἄλογοι, εἰ μὲν τετράγωνα εἴη, αὐταὶ αἱ πλευραί, εἰ δὲ ἕτερά τινα εὐθύγραμμα, αἱ ἴσα αὐτοῖς τετράγωνα ἀναγράφουσαι."", ""GreekTextWordCount"" -> 42, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.2.1"", ""Text"" -> ""Given a rational straight line and a binomial, divided into its terms, such that the square on the greater term is greater than the square on the lesser by the square on a straight line commensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a first binomial straight line;"", ""TextWordCount"" -> 67, ""GreekText"" -> ""ὑποκειμένης ῥητῆς καὶ τῆς ἐκ δύο ὀνομάτων διῃρημένης εἰς τὰ ὀνόματα, ἧς τὸ μεῖζον ὄνομα τοῦ ἐλάσσονος μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει, ἐὰν μὲν τὸ μεῖζον ὄνομα σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἡ ὅλη ἐκ δύο ὀνομάτων πρώτη."", ""GreekTextWordCount"" -> 42, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.2.2"", ""Text"" -> ""but if the lesser term be commensurable in length with the rational straight line set out, let the whole be called a second binomial;"", ""TextWordCount"" -> 24, ""GreekText"" -> ""ἐὰν δὲ τὸ ἔλασσον ὄνομα σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἐκ δύο ὀνομάτων δευτέρα."", ""GreekTextWordCount"" -> 16, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.2.3"", ""Text"" -> ""and if neither of the terms be commensurable in length with the rational straight line set out, let the whole be called a third binomial."", ""TextWordCount"" -> 25, ""GreekText"" -> ""ἐὰν δὲ μηδέτερον τῶν ὀνομάτων σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἐκ δύο ὀνομάτων τρίτη."", ""GreekTextWordCount"" -> 16, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.2.4"", ""Text"" -> ""Again, if the square on the greater term be greater than the square on the lesser by the square on a straight line incommensurable in length with the greater, then, if the greater term be commensurable in length with the rational straight line set out, let the whole be called a fourth binomial;"", ""TextWordCount"" -> 53, ""GreekText"" -> ""πάλιν δὴ ἐὰν τὸ μεῖζον ὄνομα τοῦ ἐλάσσονος μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, ἐὰν μὲν τὸ μεῖζον ὄνομα σύμμετρον ᾖ μήκει τῇ ἐκκειμένῃ ῥητῇ, καλείσθω ἐκ δύο ὀνομάτων τετάρτη."", ""GreekTextWordCount"" -> 31, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.2.5"", ""Text"" -> ""if the lesser, a fifth binomial;"", ""TextWordCount"" -> 6, ""GreekText"" -> ""ἐὰν δὲ τὸ ἔλασσον, πέμπτη."", ""GreekTextWordCount"" -> 5, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.2.6"", ""Text"" -> ""and if neither, a sixth binomial."", ""TextWordCount"" -> 6, ""GreekText"" -> ""ἐὰν δὲ μηδέτερον, ἕκτη."", ""GreekTextWordCount"" -> 4, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.3.1"", ""Text"" -> ""Given a rational straight line and an apotome, if the square on the whole be greater than the square on the annex by the square on a straight line commensurable in length with the whole, and the whole be commensurable in length with the rational straight line set out, let the apotome be called a first apotome."", ""TextWordCount"" -> 57, ""GreekText"" -> ""ὑποκειμένης ῥητῆς καὶ ἀποτομῆς, ἐὰν μὲν ἡ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει, καὶ ἡ ὅλη σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, καλείσθω ἀποτομὴ πρώτη."", ""GreekTextWordCount"" -> 29, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.3.2"", ""Text"" -> ""But if the annex be commensurable in length with the rational straight line set out, and the square on the whole be greater than that on the annex by the square on a straight line commensurable with the whole, let the apotome be called a second apotome."", ""TextWordCount"" -> 47, ""GreekText"" -> ""ἐὰν δὲ ἡ προσαρμόζουσα σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, καὶ ἡ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καλείσθω ἀποτομὴ δευτέρα."", ""GreekTextWordCount"" -> 24, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.3.3"", ""Text"" -> ""But if neither be commensurable in length with the rational straight line set out, and the square on the whole be greater than the square on the annex by the square on a straight line commensurable with the whole, let the apotome be called a third apotome."", ""TextWordCount"" -> 47, ""GreekText"" -> ""ἐὰν δὲ μηδετέρα σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, ἡ δὲ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ συμμέτρου ἑαυτῇ, καλείσθω ἀποτομὴ τρίτη."", ""GreekTextWordCount"" -> 23, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.3.4"", ""Text"" -> ""Again, if the square on the whole be greater than the square on the annex by the square on a straight line incommensurable with the whole, then, if the whole be commensurable in length with the rational straight line set out, let the apotome be called a fourth apotome;"", ""TextWordCount"" -> 49, ""GreekText"" -> ""πάλιν, ἐὰν ἡ ὅλη τῆς προσαρμοζούσης μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ μήκει, ἐὰν μὲν ἡ ὅλη σύμμετρος ᾖ τῇ ἐκκειμένῃ ῥητῇ μήκει, καλείσθω ἀποτομὴ τετάρτη."", ""GreekTextWordCount"" -> 26, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.3.5"", ""Text"" -> ""if the annex be so commensurable, a fifth;"", ""TextWordCount"" -> 8, ""GreekText"" -> ""ἐὰν δὲ ἡ προσαρμόζουσα, πέμπτη."", ""GreekTextWordCount"" -> 5, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D10.3.6"", ""Text"" -> ""and, if neither, a sixth."", ""TextWordCount"" -> 5, ""GreekText"" -> ""ἐὰν δὲ μηδετέρα, ἕκτη."", ""GreekTextWordCount"" -> 4, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.1"", ""Text"" -> ""A solid is that which has length, breadth, and depth."", ""TextWordCount"" -> 10, ""GreekText"" -> ""στερεόν ἐστι τὸ μῆκος καὶ πλάτος καὶ βάθος ἔχον."", ""GreekTextWordCount"" -> 9, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.2"", ""Text"" -> ""An extremity of a solid is a surface."", ""TextWordCount"" -> 8, ""GreekText"" -> ""στερεοῦ δὲ πέρας ἐπιφάνεια."", ""GreekTextWordCount"" -> 4, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.3"", ""Text"" -> ""A straight line is at right angles to a plane, when it makes right angles with all the straight lines which meet it and are in the plane."", ""TextWordCount"" -> 28, ""GreekText"" -> ""εὐθεῖα πρὸς ἐπίπεδον ὀρθή ἐστιν, ὅταν πρὸς πάσας τὰς ἁπτομένας αὐτῆς εὐθείας καὶ οὔσας ἐν τῷ ὑποκειμένῳ ἐπιπέδῳ ὀρθὰς ποιῇ γωνίας."", ""GreekTextWordCount"" -> 21, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.4"", ""Text"" -> ""A plane is at right angles to a plane when the straight lines drawn, in one of the planes, at right angles to the common section of the planes are at right angles to the remaining plane."", ""TextWordCount"" -> 37, ""GreekText"" -> ""ἐπίπεδον πρὸς ἐπίπεδον ὀρθόν ἐστιν, ὅταν αἱ τῇ κοινῇ τομῇ τῶν ἐπιπέδων πρὸς ὀρθὰς ἀγόμεναι εὐθεῖαι ἐν ἑνὶ τῶν ἐπιπέδων τῷ λοιπῷ ἐπιπέδῳ πρὸς ὀρθὰς ὦσιν."", ""GreekTextWordCount"" -> 26, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.5"", ""Text"" -> ""The inclination of a straight line to a plane is, assuming a perpendicular drawn from the extremity of the straight line which is elevated above the plane to the plane, and a straight line joined from the point thus arising to the extremity of the straight line which is in the plane, the angle contained by the straight line so drawn and the straight line standing up."", ""TextWordCount"" -> 67, ""GreekText"" -> ""εὐθείας πρὸς ἐπίπεδον κλίσις ἐστίν, ὅταν ἀπὸ τοῦ μετεώρου πέρατος τῆς εὐθείας ἐπὶ τὸ ἐπίπεδον κάθετος ἀχθῇ, καὶ ἀπὸ τοῦ γενομένου σημείου ἐπὶ τὸ ἐν τῷ ἐπιπέδῳ πέρας τῆς εὐθείας εὐθεῖα ἐπιζευχθῇ, ἡ περιεχομένη γωνία ὑπὸ τῆς ἀχθείσης καὶ τῆς ἐφεστώσης."", ""GreekTextWordCount"" -> 41, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.6"", ""Text"" -> ""The inclination of a plane to a plane is the acute angle contained by the straight lines drawn at right angles to the common section at the same point, one in each of the planes."", ""TextWordCount"" -> 35, ""GreekText"" -> ""ἐπιπέδου πρὸς ἐπίπεδον κλίσις ἐστὶν ἡ περιεχομένη ὀξεῖα γωνία ὑπὸ τῶν πρὸς ὀρθὰς τῇ κοινῇ τομῇ ἀγομένων πρὸς τῷ αὐτῷ σημείῳ ἐν ἑκατέρῳ τῶν ἐπιπέδων."", ""GreekTextWordCount"" -> 25, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.7"", ""Text"" -> ""A plane is said to be similarly inclined to a plane as another is to another when the said angles of the inclinations are equal to one another."", ""TextWordCount"" -> 28, ""GreekText"" -> ""ἐπίπεδον πρὸς ἐπίπεδον ὁμοίως κεκλίσθαι λέγεται καὶ ἕτερον πρὸς ἕτερον, ὅταν αἱ εἰρημέναι τῶν κλίσεων γωνίαι ἴσαι ἀλλήλαις ὦσιν."", ""GreekTextWordCount"" -> 19, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.8"", ""Text"" -> ""Parallel planes are those which do not meet."", ""TextWordCount"" -> 8, ""GreekText"" -> ""παράλληλα ἐπίπεδά ἐστι τὰ ἀσύμπτωτα."", ""GreekTextWordCount"" -> 5, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.9"", ""Text"" -> ""Similar solid figures are those contained by similar planes equal in multitude."", ""TextWordCount"" -> 12, ""GreekText"" -> ""ὅμοια στερεὰ σχήματά ἐστι τὰ ὑπὸ ὁμοίων ἐπιπέδων περιεχόμενα ἴσων τὸ πλῆθος."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.10"", ""Text"" -> ""Equal and similar solid figures are those contained by similar planes equal in multitude and in magnitude."", ""TextWordCount"" -> 17, ""GreekText"" -> ""ἴσα δὲ καὶ ὅμοια στερεὰ σχήματά ἐστι τὰ ὑπὸ ὁμοίων ἐπιπέδων περιεχόμενα ἴσων τῷ πλήθει καὶ τῷ μεγέθει."", ""GreekTextWordCount"" -> 18, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.11"", ""Text"" -> ""A solid angle is the inclination constituted by more than two lines which meet one another and are not in the same surface, towards all the lines. Otherwise: A solid angle is that which is contained by more than two plane angles which are not in the same plane and are constructed to one point."", ""TextWordCount"" -> 55, ""GreekText"" -> ""στερεὰ γωνία ἐστὶν ἡ ὑπὸ πλειόνων ἢ δύο γραμμῶν ἁπτομένων ἀλλήλων καὶ μὴ ἐν τῇ αὐτῇ ἐπιφανείᾳ οὐσῶν πρὸς πάσαις ταῖς γραμμαῖς κλίσις. ἄλλως: στερεὰ γωνία ἐστὶν ἡ ὑπὸ πλειόνων ἢ δύο γωνιῶν ἐπιπέδων περιεχομένη μὴ οὐσῶν ἐν τῷ αὐτῷ ἐπιπέδῳ πρὸς ἑνὶ σημείῳ συνισταμένων."", ""GreekTextWordCount"" -> 45, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.12"", ""Text"" -> ""A pyramid is a solid figure, contained by planes, which is constructed from one plane to one point."", ""TextWordCount"" -> 18, ""GreekText"" -> ""πυραμίς ἐστι σχῆμα στερεὸν ἐπιπέδοις περιεχόμενον ἀπὸ ἑνὸς ἐπιπέδου πρὸς ἑνὶ σημείῳ συνεστώς."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.13"", ""Text"" -> ""A prism is a solid figure contained by planes two of which, namely those which are opposite, are equal, similar and parallel, while the rest are parallelograms."", ""TextWordCount"" -> 27, ""GreekText"" -> ""πρίσμα ἐστὶ σχῆμα στερεὸν ἐπιπέδοις περιεχόμενον, ὧν δύο τὰ ἀπεναντίον ἴσα τε καὶ ὅμοιά ἐστι καὶ παράλληλα, τὰ δὲ λοιπὰ παραλληλόγραμμα."", ""GreekTextWordCount"" -> 21, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.14"", ""Text"" -> ""When, the diameter of a semicircle remaining fixed, the semicircle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a sphere."", ""TextWordCount"" -> 34, ""GreekText"" -> ""σφαῖρά ἐστιν, ὅταν ἡμικυκλίου μενούσης τῆς διαμέτρου περιενεχθὲν τὸ ἡμικύκλιον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, τὸ περιληφθὲν σχῆμα."", ""GreekTextWordCount"" -> 21, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.15"", ""Text"" -> ""The axis of the sphere is the straight line which remains fixed and about which the semicircle is turned."", ""TextWordCount"" -> 19, ""GreekText"" -> ""ἄξων δὲ τῆς σφαίρας ἐστὶν ἡ μένουσα εὐθεῖα, περὶ ἣν τὸ ἡμικύκλιον στρέφεται."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.16"", ""Text"" -> ""The centre of the sphere is the same as that of the semicircle."", ""TextWordCount"" -> 13, ""GreekText"" -> ""κέντρον δὲ τῆς σφαίρας ἐστὶ τὸ αὐτό, ὃ καὶ τοῦ ἡμικυκλίου."", ""GreekTextWordCount"" -> 11, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.17"", ""Text"" -> ""A diameter of the sphere is any straight line drawn through the centre and terminated in both directions by the surface of the sphere."", ""TextWordCount"" -> 24, ""GreekText"" -> ""διάμετρος δὲ τῆς σφαίρας ἐστὶν εὐθεῖά τις διὰ τοῦ κέντρου ἠγμένη καὶ περατουμένη ἐφ᾽ ἑκάτερα τὰ μέρη ὑπὸ τῆς ἐπιφανείας τῆς σφαίρας."", ""GreekTextWordCount"" -> 23, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.18"", ""Text"" -> ""When, one side of those about the right angle in a right-angled triangle remaining fixed, the triangle is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cone. And, if the straight line which remains fixed be equal to the remaining side about the right angle which is carried round, the cone will be right-angled; if less, obtuse-angled; and if greater, acute-angled."", ""TextWordCount"" -> 75, ""GreekText"" -> ""κῶνός ἐστιν, ὅταν ὀρθογωνίου τριγώνου μενούσης μιᾶς πλευρᾶς τῶν περὶ τὴν ὀρθὴν γωνίαν περιενεχθὲν τὸ τρίγωνον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, τὸ περιληφθὲν σχῆμα. κἂν μὲν ἡ μένουσα εὐθεῖα ἴση ᾖ τῇ λοιπῇ τῇ περὶ τὴν ὀρθὴν περιφερομένῃ, ὀρθογώνιος ἔσται ὁ κῶνος, ἐὰν δὲ ἐλάττων, ἀμβλυγώνιος, ἐὰν δὲ μείζων, ὀξυγώνιος."", ""GreekTextWordCount"" -> 53, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.19"", ""Text"" -> ""The axis of the cone is the straight line which remains fixed and about which the triangle is turned."", ""TextWordCount"" -> 19, ""GreekText"" -> ""ἄξων δὲ τοῦ κώνου ἐστὶν ἡ μένουσα εὐθεῖα, περὶ ἣν τὸ τρίγωνον στρέφεται."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.20"", ""Text"" -> ""And the base is the circle described by the straight line which is carried round."", ""TextWordCount"" -> 15, ""GreekText"" -> ""βάσις δὲ ὁ κύκλος ὁ ὑπὸ τῆς περιφερομένης εὐθείας γραφόμενος."", ""GreekTextWordCount"" -> 10, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.21"", ""Text"" -> ""When, one side of those about the right angle in a rectangular parallelogram remaining fixed, the parallelogram is carried round and restored again to the same position from which it began to be moved, the figure so comprehended is a cylinder."", ""TextWordCount"" -> 41, ""GreekText"" -> ""κύλινδρός ἐστιν, ὅταν ὀρθογωνίου παραλληλογράμμου μενούσης μιᾶς πλευρᾶς τῶν περὶ τὴν ὀρθὴν γωνίαν περιενεχθὲν τὸ παραλληλόγραμμον εἰς τὸ αὐτὸ πάλιν ἀποκατασταθῇ, ὅθεν ἤρξατο φέρεσθαι, τὸ περιληφθὲν σχῆμα."", ""GreekTextWordCount"" -> 27, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.22"", ""Text"" -> ""The axis of the cylinder is the straight line which remains fixed and about which the parallelogram is turned."", ""TextWordCount"" -> 19, ""GreekText"" -> ""ἄξων δὲ τοῦ κυλίνδρου ἐστὶν ἡ μένουσα εὐθεῖα, περὶ ἣν τὸ παραλληλόγραμμον στρέφεται."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.23"", ""Text"" -> ""And the bases are the circles described by the two sides opposite to one another which are carried round."", ""TextWordCount"" -> 19, ""GreekText"" -> ""βάσεις δὲ οἱ κύκλοι οἱ ὑπὸ τῶν ἀπεναντίον περιαγομένων δύο πλευρῶν γραφόμενοι."", ""GreekTextWordCount"" -> 12, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.24"", ""Text"" -> ""Similar cones and cylinders are those in which the axes and the diameters of the bases are proportional."", ""TextWordCount"" -> 18, ""GreekText"" -> ""ὅμοιοι κῶνοι καὶ κύλινδροί εἰσιν, ὧν οἵ τε ἄξονες καὶ αἱ διάμετροι τῶν βάσεων ἀνάλογόν εἰσιν."", ""GreekTextWordCount"" -> 16, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.25"", ""Text"" -> ""A cube is a solid figure contained by six equal squares."", ""TextWordCount"" -> 11, ""GreekText"" -> ""κύβος ἐστὶ σχῆμα στερεὸν ὑπὸ ἓξ τετραγώνων ἴσων περιεχόμενον."", ""GreekTextWordCount"" -> 9, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.26"", ""Text"" -> ""An octahedron is a solid figure contained by eight equal and equilateral triangles."", ""TextWordCount"" -> 13, ""GreekText"" -> ""ὀκτάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ ὀκτὼ τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον."", ""GreekTextWordCount"" -> 11, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.27"", ""Text"" -> ""An icosahedron is a solid figure contained by twenty equal and equilateral triangles."", ""TextWordCount"" -> 13, ""GreekText"" -> ""εἰκοσάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ εἴκοσι τριγώνων ἴσων καὶ ἰσοπλεύρων περιεχόμενον."", ""GreekTextWordCount"" -> 11, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""D11.28"", ""Text"" -> ""A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons."", ""TextWordCount"" -> 14, ""GreekText"" -> ""δωδεκάεδρόν ἐστι σχῆμα στερεὸν ὑπὸ δώδεκα πενταγώνων ἴσων καὶ ἰσοπλεύρων καὶ ἰσογωνίων περιεχόμενον."", ""GreekTextWordCount"" -> 13, ""References"" -> Missing[""NotApplicable""], ""Proof"" -> Missing[""NotApplicable""], ""ProofWordCount"" -> Missing[""NotApplicable""], ""GreekProof"" -> Missing[""NotApplicable""], ""GreekProofWordCount"" -> Missing[""NotApplicable""]|>","<|""VertexLabel"" -> ""1.1"", ""Text"" -> ""On a given finite straight line to construct an equilateral triangle."", ""TextWordCount"" -> 11, ""GreekText"" -> ""ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τρίγωνον ἰσόπλευρον συστήσασθαι."", ""GreekTextWordCount"" -> 8, ""References"" -> {{""Common Notion"" -> 1}, {""Postulate"" -> 1}, {""Postulate"" -> 3}, {""Book"" -> 1, ""Definition"" -> 15}}, ""Proof"" -> ""Let AB be the given finite straight line. Thus it is required to constructan equilateral triangle on the straight line AB. With centre A and distance AB let the circle BCD be described; [Post. 3]again, with centre B and distance BA let the circle ACE be described; [Post. 3] and from the point C, in which the circles cut one another, to the points A, B let the straight lines CA, CB be joined. [Post. 1] Now, since the point A is the centre of the circle CDB, AC is equal to AB. [Def. 15] Again, since the point B is the centre of the circle CAE, BC is equal to BA. [Def. 15] But CA was also proved equal to AB;therefore each of the straight lines CA, CB is equal to AB. And things which are equal to the same thing are also equal to one another; [C. N. 1] therefore CA is also equal to CB. Therefore the three straight lines CA, AB, BC areequal to one another. Therefore the triangle ABC is equilateral; and it has been constructed on the given finite straight line AB."", ""ProofWordCount"" -> 190, ""GreekProof"" -> ""ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ. δεῖ δὴ ἐπὶ τῆς ΑΒ εὐθείας τρίγωνον ἰσόπλευρον συστήσασθαι. κέντρῳ μὲν τῷ Α διαστήματι δὲ τῷ ΑΒ κύκλος γεγράφθω ὁ ΒΓΔ, καὶ πάλιν κέντρῳ μὲν τῷ Β διαστήματι δὲ τῷ ΒΑ κύκλος γεγράφθω ὁ ΑΓΕ, καὶ ἀπὸ τοῦ Γ σημείου, καθ᾽ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι, ἐπὶ τὰ Α, Β σημεῖα ἐπεζεύχθωσαν εὐθεῖαι αἱ ΓΑ, ΓΒ. καὶ ἐπεὶ τὸ Α σημεῖον κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου, ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ: πάλιν, ἐπεὶ τὸ Β σημεῖον κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου, ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ. ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση: ἑκατέρα ἄρα τῶν ΓΑ, ΓΒ τῇ ΑΒ ἐστὶν ἴση. τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα: καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστὶν ἴση: αἱ τρεῖς ἄρα αἱ ΓΑ, ΑΒ, ΒΓ ἴσαι ἀλλήλαις εἰσίν. ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον, καὶ συνέσταται ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τῆς ΑΒ. Ἐπὶ τῆς δοθείσης ἄρα εὐθείας πεπερασμένης τρίγωνον ἰσόπλευρον συνέσταται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 167|>","<|""VertexLabel"" -> ""1.2"", ""Text"" -> ""To place at a given point (as an extremity) a straight line equal to a given straight line."", ""TextWordCount"" -> 18, ""GreekText"" -> ""πρὸς τῷ δοθέντι σημείῳ τῇ δοθείσῃ εὐθείᾳ ἴσην εὐθεῖαν θέσθαι."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Common Notion"" -> 1}, {""Common Notion"" -> 3}, {""Postulate"" -> 1}, {""Postulate"" -> 2}, {""Postulate"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 1}}, ""Proof"" -> ""Let A be the given point, and BC the given straight line. Thus it is required to place at the point A (as an extremity)a straight line equal to the given straight line BC. From the point A to the point B let the straight line AB be joined; [Post. 1] and on it let the equilateral triangleDAB be constructed. [I. 1] Let the straight lines AE, BF be produced in a straight line with DA, DB; [Post. 2] with centre B and distance BC let thecircle CGH be described; [Post. 3] and again, with centre D and distance DG let the circle GKL be described. [Post. 3] Then, since the point B is the centre of the circle CGH, BC is equal to BG. Again, since the point D is the centre of the circle GKL, DL is equal to DG. And in these DA is equal to DB; therefore the remainder AL is equal to the remainder BG. [C. N. 3] But BC was also proved equal to BG; therefore each of the straight lines AL, BC is equal to BG. And things which are equal to the same thing are also equal to one another; [C. N. 1]therefore AL is also equal to BC. Therefore at the given point A the straight line AL is placed equal to the given straight line BC."", ""ProofWordCount"" -> 228, ""GreekProof"" -> ""ἔστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ: δεῖ δὴ πρὸς τῷ Α σημείῳ τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴσην εὐθεῖαν θέσθαι. ἐπεζεύχθω γὰρ ἀπὸ τοῦ Α σημείου ἐπὶ τὸ Β σημεῖον εὐθεῖα ἡ ΑΒ, καὶ συνεστάτω ἐπ᾽ αὐτῆς τρίγωνον ἰσόπλευρον τὸ ΔΑΒ, καὶ ἐκβεβλήσθωσαν ἐπ᾽ εὐθείας ταῖς ΔΑ, ΔΒ εὐθεῖαι αἱ ΑΕ, ΒΖ, καὶ κέντρῳ μὲν τῷ Β διαστήματι δὲ τῷ ΒΓ κύκλος γεγράφθω ὁ ΓΗΘ, καὶ πάλιν κέντρῳ τῷ Δ καὶ διαστήματι τῷ ΔΗ κύκλος γεγράφθω ὁ ΗΚΛ. ἐπεὶ οὖν τὸ Β σημεῖον κέντρον ἐστὶ τοῦ ΓΗΘ κύκλου, ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ. πάλιν, ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ τοῦ ΚΛΗ κύκλου, ἴση ἐστὶν ἡ ΔΛ τῇ ΔΗ, ὧν ἡ ΔΑ τῇ ΔΒ ἴση ἐστίν. λοιπὴ ἄρα ἡ ΑΛ λοιπῇ τῇ ΒΗ ἐστὶν ἴση. ἐδείχθη δὲ καὶ ἡ ΒΓ τῇ ΒΗ ἴση: ἑκατέρα ἄρα τῶν ΑΛ, ΒΓ τῇ ΒΗ ἐστὶν ἴση. τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα: καὶ ἡ ΑΛ ἄρα τῇ ΒΓ ἐστὶν ἴση. πρὸς ἄρα τῷ δοθέντι σημείῳ τῷ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ ἴση εὐθεῖα κεῖται ἡ ΑΛ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 189|>","<|""VertexLabel"" -> ""1.3"", ""Text"" -> ""Given two unequal straight lines, to cut off from the greater a straight line equal to the less."", ""TextWordCount"" -> 18, ""GreekText"" -> ""δύο δοθεισῶν εὐθειῶν ἀνίσων ἀπὸ τῆς μείζονος τῇ ἐλάσσονι ἴσην εὐθεῖαν ἀφελεῖν."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Common Notion"" -> 1}, {""Postulate"" -> 3}, {""Book"" -> 1, ""Definition"" -> 15}, {""Book"" -> 1, ""Theorem"" -> 2}}, ""Proof"" -> ""Let AB, C be the two given unequal straight lines, and let AB be the greater of them. Thus it is required to cut off from AB the greater a straight line equal to C the less. At the point A let AD be placed equal to the straight line C; [I. 2] and with centre A and distance AD let the circle DEF be described. [Post. 3] Now, since the point A is the centre of the circle DEF, AE is equal to AD. [Def. 15]But C is also equal to AD. Therefore each of the straight lines AE, C is equal to AD; so that AE is also equal to C. [C. N. 1] Therefore, given the two straight lines AB, C, from AB the greater AE has been cut off equal to C the less."", ""ProofWordCount"" -> 139, ""GreekProof"" -> ""ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι ἄνισοι αἱ ΑΒ, Γ, ὧν μείζων ἔστω ἡ ΑΒ: δεῖ δὴ ἀπὸ τῆς μείζονος τῆς ΑΒ τῇ ἐλάσσονι τῇ Γ ἴσην εὐθεῖαν ἀφελεῖν. κείσθω πρὸς τῷ Α σημείῳ τῇ Γ εὐθείᾳ ἴση ἡ ΑΔ: καὶ κέντρῳ μὲν τῷ Α διαστήματι δὲ τῷ ΑΔ κύκλος γεγράφθω ὁ ΔΕΖ. καὶ ἐπεὶ τὸ Α σημεῖον κέντρον ἐστὶ τοῦ ΔΕΖ κύκλου, ἴση ἐστὶν ἡ ΑΕ τῇ ΑΔ: ἀλλὰ καὶ ἡ Γ τῇ ΑΔ ἐστιν ἴση. ἑκατέρα ἄρα τῶν ΑΕ, Γ τῇ ΑΔ ἐστιν ἴση: ὥστε καὶ ἡ ΑΕ τῇ Γ ἐστιν ἴση. δύο ἄρα δοθεισῶν εὐθειῶν ἀνίσων τῶν ΑΒ, Γ ἀπὸ τῆς μείζονος τῆς ΑΒ τῇ ἐλάσσονι τῇ Γ ἴση ἀφῄρηται ἡ ΑΕ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 117|>","<|""VertexLabel"" -> ""1.4"", ""Text"" -> ""If two triangles have the two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend."", ""TextWordCount"" -> 60, ""GreekText"" -> ""ἐὰν δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δυσὶ πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν βάσιν τῇ βάσει ἴσην ἕξει, καὶ τὸ τρίγωνον τῷ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν."", ""GreekTextWordCount"" -> 58, ""References"" -> {{""Common Notion"" -> 4}}, ""Proof"" -> ""Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF, and the angle BAC equal to theangle EDF. I say that the base BC is also equal to the base EF, the triangle ABC will be equal to the triangle DEF, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, thatis, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. For, if the triangle ABC be applied to the triangle DEF, and if the point A be placed on the point D and the straight line AB on DE, then the point B will also coincide with E, because AB is equal to DE. Again, AB coinciding with DE, the straight line AC will also coincide with DF, because the angle BAC is equal to the angle EDF; hence the point C will also coincide with the point F, because AC is again equal to DF. But B also coincided with E; hence the base BC will coincide with the base EF. [For if, when B coincides with E and C with F, the base BC does not coincide with the base EF, two straight lines will enclose a space: which is impossible. Therefore the base BC will coincide with EF]and will be equal to it. [C. N. 4] Thus the whole triangle ABC will coincide with the whole triangle DEF, and will be equal to it. And the remaining angles will also coincide with the remaining angles and will be equal to them, the angle ABC to the angle DEF, and the angle ACB to the angle DFE."", ""ProofWordCount"" -> 294, ""GreekProof"" -> ""ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς τὰς ΑΒ, ΑΓ ταῖς δυσὶ πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ καὶ γωνίαν τὴν ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴσην. λέγω, ὅτι καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ, ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ. Ἐφαρμοζομένου γὰρ τοῦ ΑΒΓ τριγώνου ἐπὶ τὸ ΔΕΖ τρίγωνον καὶ τιθεμένου τοῦ μὲν Α σημείου ἐπὶ τὸ Δ σημεῖον τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΔΕ, ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Ε διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΔΕ: ἐφαρμοσάσης δὴ τῆς ΑΒ ἐπὶ τὴν ΔΕ ἐφαρμόσει καὶ ἡ ΑΓ εὐθεῖα ἐπὶ τὴν ΔΖ διὰ τὸ ἴσην εἶναι τὴν ὑπὸ ΒΑΓ γωνίαν τῇ ὑπὸ ΕΔΖ: ὥστε καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ σημεῖον ἐφαρμόσει διὰ τὸ ἴσην πάλιν εἶναι τὴν ΑΓ τῇ ΔΖ. ἀλλὰ μὴν καὶ τὸ Β ἐπὶ τὸ Ε ἐφηρμόκει: ὥστε βάσις ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ἐφαρμόσει. εἰ γὰρ τοῦ μὲν Β ἐπὶ τὸ Ε ἐφαρμόσαντος τοῦ δὲ Γ ἐπὶ τὸ Ζ ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ οὐκ ἐφαρμόσει, δύο εὐθεῖαι χωρίον περιέξουσιν: ὅπερ ἐστὶν ἀδύνατον. ἐφαρμόσει ἄρα ἡ ΒΓ βάσις ἐπὶ τὴν ΕΖ καὶ ἴση αὐτῇ ἔσται: ὥστε καὶ ὅλον τὸ ΑΒΓ τρίγωνον ἐπὶ ὅλον τὸ ΔΕΖ τρίγωνον ἐφαρμόσει καὶ ἴσον αὐτῷ ἔσται, καὶ αἱ λοιπαὶ γωνίαι ἐπὶ τὰς λοιπὰς γωνίας ἐφαρμόσουσι καὶ ἴσαι αὐταῖς ἔσονται, ἡ μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ. ἐὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δύο πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν βάσιν τῇ βάσει ἴσην ἕξει, καὶ τὸ τρίγωνον τῷ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 346|>","<|""VertexLabel"" -> ""1.5"", ""Text"" -> ""In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another."", ""TextWordCount"" -> 33, ""GreekText"" -> ""τῶν ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ προσεκβληθεισῶν τῶν ἴσων εὐθειῶν αἱ ὑπὸ τὴν βάσιν γωνίαι ἴσαι ἀλλήλαις ἔσονται."", ""GreekTextWordCount"" -> 24, ""References"" -> {{""Postulate"" -> 1}, {""Postulate"" -> 2}, {""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 4}}, ""Proof"" -> ""Let ABC be an isosceles triangle having the side AB equal to the side AC; and let the straight lines BD, CE be produced further in a straight line with AB, AC. [Post. 2] I say that the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE. Let a point F be taken at random on BD; from AE the greater let AG be cut off equal to AF the less; [I. 3] and let the straight lines FC, GB be joined. [Post. 1] Then, since AF is equal to AG and AB to AC, the two sides FA, AC are equal to the two sides GA, AB, respectively; and they contain a common angle, the angle FAG.Therefore the base FC is equal to the base GB, and the triangle AFC is equal to the triangle AGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4] And, since the whole AF is equal to the whole AG, and in these AB is equal to AC, the remainder BF is equal to the remainder CG. But FC was also proved equal to GB;therefore the two sides BF, FC are equal to the two sides CG, GB respectively; and the angle BFC is equal to the angle CGB, while the base BC is common to them; therefore the triangle BFC is also equal to the triangle CGB,and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. Accordingly, since the whole angle ABG was proved equal to the angle ACF, and in these the angle CBG is equal to the angle BCF, the remaining angle ABC is equal to the remaining angle ACB;and they are at the base of the triangle ABC. But the angle FBC was also proved equal to the angle GCB; and they are under the base."", ""ProofWordCount"" -> 363, ""GreekProof"" -> ""ἔστω τρίγωνον ἰσοσκελὲς τὸ ΑΒΓ ἴσην ἔχον τὴν ΑΒ πλευρὰν τῇ ΑΓ πλευρᾷ, καὶ προσεκβεβλήσθωσαν ἐπ᾽ εὐθείας ταῖς ΑΒ, ΑΓ εὐθεῖαι αἱ ΒΔ, ΓΕ: λέγω, ὅτι ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΑΓΒ ἴση ἐστίν, ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΒΓΕ. εἰλήφθω γὰρ ἐπὶ τῆς ΒΔ τυχὸν σημεῖον τὸ Ζ, καὶ ἀφῃρήσθω ἀπὸ τῆς μείζονος τῆς ΑΕ τῇ ἐλάσσονι τῇ ΑΖ ἴση ἡ ΑΗ, καὶ ἐπεζεύχθωσαν αἱ ΖΓ, ΗΒ εὐθεῖαι. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΖ τῇ ΑΗ ἡ δὲ ΑΒ τῇ ΑΓ, δύο δὴ αἱ ΖΑ, ΑΓ δυσὶ ταῖς ΗΑ, ΑΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνίαν κοινὴν περιέχουσι τὴν ὑπὸ ΖΑΗ: βάσις ἄρα ἡ ΖΓ βάσει τῇ ΗΒ ἴση ἐστίν, καὶ τὸ ΑΖΓ τρίγωνον τῷ ΑΗΒ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν ὑπὸ ΑΓΖ τῇ ὑπὸ ΑΒΗ, ἡ δὲ ὑπὸ ΑΖΓ τῇ ὑπὸ ΑΗΒ. καὶ ἐπεὶ ὅλη ἡ ΑΖ ὅλῃ τῇ ΑΗ ἐστιν ἴση, ὧν ἡ ΑΒ τῇ ΑΓ ἐστιν ἴση, λοιπὴ ἄρα ἡ ΒΖ λοιπῇ τῇ ΓΗ ἐστιν ἴση. ἐδείχθη δὲ καὶ ἡ ΖΓ τῇ ΗΒ ἴση: δύο δὴ αἱ ΒΖ, ΖΓ δυσὶ ταῖς ΓΗ, ΗΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΒΖΓ γωνίᾳ τῇ ὑπὸ ΓΗΒ ἴση, καὶ βάσις αὐτῶν κοινὴ ἡ ΒΓ: καὶ τὸ ΒΖΓ ἄρα τρίγωνον τῷ ΓΗΒ τριγώνῳ ἴσον ἔσται, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἡ δὲ ὑπὸ ΒΓΖ τῇ ὑπὸ ΓΒΗ. ἐπεὶ οὖν ὅλη ἡ ὑπὸ ΑΒΗ γωνία ὅλῃ τῇ ὑπὸ ΑΓΖ γωνίᾳ ἐδείχθη ἴση, ὧν ἡ ὑπὸ ΓΒΗ τῇ ὑπὸ ΒΓΖ ἴση, λοιπὴ ἄρα ἡ ὑπὸ ΑΒΓ λοιπῇ τῇ ὑπὸ ΑΓΒ ἐστιν ἴση: καί εἰσι πρὸς τῇ βάσει τοῦ ΑΒΓ τριγώνου. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΖΒΓ τῇ ὑπὸ ΗΓΒ ἴση: καί εἰσιν ὑπὸ τὴν βάσιν. τῶν ἄρα ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ προσεκβληθεισῶν τῶν ἴσων εὐθειῶν αἱ ὑπὸ τὴν βάσιν γωνίαι ἴσαι ἀλλήλαις ἔσονται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 349|>","<|""VertexLabel"" -> ""1.6"", ""Text"" -> ""If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another."", ""TextWordCount"" -> 25, ""GreekText"" -> ""ἐὰν τριγώνου αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν, καὶ αἱ ὑπὸ τὰς ἴσας γωνίαις ὑποτείνουσαι πλευραὶ ἴσαι ἀλλήλαις ἔσονται."", ""GreekTextWordCount"" -> 19, ""References"" -> {}, ""Proof"" -> ""Let ABC be a triangle having the angle ABC equal to the angle ACB; I say that the side AB is also equal to the side AC. For, if AB is unequal to AC, one of them is greater. Let AB be greater; and from AB the greater let DB be cut off equal to AC the less; let DC be joined. Then, since DB is equal to AC, and BC is common, the two sides DB, BC are equal to the two sides AC, CB respectively; and the angle DBC is equal to the angle ACB; therefore the base DC is equal to the base AB, and the triangle DBC will be equal to the triangle ACB, the less to the greater: which is absurd. Therefore AB is not unequal to AC; it is therefore equal to it."", ""ProofWordCount"" -> 139, ""GreekProof"" -> ""ἔστω τρίγωνον τὸ ΑΒΓ ἴσην ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν τῇ ὑπὸ ΑΓΒ γωνίᾳ: λέγω, ὅτι καὶ πλευρὰ ἡ ΑΒ πλευρᾷ τῇ ΑΓ ἐστιν ἴση. εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ, ἡ ἑτέρα αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ΑΒ, καὶ ἀφῃρήσθω ἀπὸ τῆς μείζονος τῆς ΑΒ τῇ ἐλάττονι τῇ ΑΓ ἴση ἡ ΔΒ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΒ τῇ ΑΓ κοινὴ δὲ ἡ ΒΓ, δύο δὴ αἱ ΔΒ, ΒΓ δύο ταῖς ΑΓ, ΓΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνία ἡ ὑπὸ ΔΒΓ γωνίᾳ τῇ ὑπὸ ΑΓΒ ἐστιν ἴση: βάσις ἄρα ἡ ΔΓ βάσει τῇ ΑΒ ἴση ἐστίν, καὶ τὸ ΔΒΓ τρίγωνον τῷ ΑΓΒ τριγώνῳ ἴσον ἔσται, τὸ ἔλασσον τῷ μείζονι: ὅπερ ἄτοπον: οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΑΓ: ἴση ἄρα. ἐὰν ἄρα τριγώνου αἱ δύο γωνίαι ἴσαι ἀλλήλαις ὦσιν, καὶ αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι πλευραὶ ἴσαι ἀλλήλαις ἔσονται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 153|>","<|""VertexLabel"" -> ""1.7"", ""Text"" -> ""Given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there cannot be constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it."", ""TextWordCount"" -> 62, ""GreekText"" -> ""ἐπὶ τῆς αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις ἄλλαι δύο εὐθεῖαι ἴσαι ἑκατέρα ἑκατέρᾳ οὐ συσταθήσονται πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι ταῖς ἐξ ἀρχῆς εὐθείαις."", ""GreekTextWordCount"" -> 33, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}}, ""Proof"" -> ""For, if possible, given two straight lines AC, CB constructed on the straight line AB and meeting at the point C, let two other straight linesAD, DB be constructed on the same straight line AB, on the same side of it, meeting in another point D and equal to the former two respectively, namely each to that which has the same extremity with it, so that CA isequal to DA which has the same extremity A with it, and CB to DB which has the same extremity B with it; and let CD be joined. Then, since AC is equal to AD, the angle ACD is also equal to the angle ADC; [I. 5]therefore the angle ADC is greater than the angle DCB; therefore the angle CDB is much greater than the angle DCB. Again, since CB is equal to DB, the angle CDB is also equal to the angle DCB. But it was also proved much greater than it: which is impossible."", ""ProofWordCount"" -> 164, ""GreekProof"" -> ""εἰ γὰρ δυνατόν, ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ δύο ταῖς αὐταῖς εὐθείαις ταῖς ΑΓ, ΓΒ ἄλλαι δύο εὐθεῖαι αἱ ΑΔ, ΔΒ ἴσαι ἑκατέρα ἑκατέρᾳ συνεστάτωσαν πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ τῷ τε Γ καὶ Δ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι, ὥστε ἴσην εἶναι τὴν μὲν ΓΑ τῇ ΔΑ τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Α, τὴν δὲ ΓΒ τῇ ΔΒ τὸ αὐτὸ πέρας ἔχουσαν αὐτῇ τὸ Β, καὶ ἐπεζεύχθω ἡ ΓΔ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΑΔ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΑΔΓ: μείζων ἄρα ἡ ὑπὸ ΑΔΓ τῆς ὑπὸ ΔΓΒ: πολλῷ ἄρα ἡ ὑπὸ ΓΔΒ μείζων ἐστὶ τῆς ὑπὸ ΔΓΒ. πάλιν ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΔΒ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΓΔΒ γωνίᾳ τῇ ὑπὸ ΔΓΒ. ἐδείχθη δὲ αὐτῆς καὶ πολλῷ μείζων: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐπὶ τῆς αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις ἄλλαι δύο εὐθεῖαι ἴσαι ἑκατέρα ἑκατέρᾳ συσταθήσονται πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι ταῖς ἐξ ἀρχῆς εὐθείαις: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 176|>","<|""VertexLabel"" -> ""1.8"", ""Text"" -> ""If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines."", ""TextWordCount"" -> 36, ""GreekText"" -> ""ἐὰν δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δύο πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρα, ἔχῃ δὲ καὶ τὴν βάσιν τῇ βάσει ἴσην, καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἕξει τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην."", ""GreekTextWordCount"" -> 34, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 7}}, ""Proof"" -> ""Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF; and let them have the base BC equalto the base EF; I say that the angle BAC is also equal to the angle EDF. For, if the triangle ABC be applied to the triangle DEF, and if the point B be placed onthe point E and the straight line BC on EF, the point C will also coincide with F, because BC is equal to EF. Then, BC coinciding with EF, BA, AC will also coincide with ED, DF; for, if the base BC coincides with the base EF, and the sides BA, AC do not coincide with ED, DF but fall beside them as EG, GF, then, given two straight lines constructed on a straight line (from its extremities) and meeting in a point, there will have been constructed on the same straight line (from its extremities), and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each to that which has the same extremity with it. But they cannot be so constructed. [I. 7] Therefore it is not possible that, if the base BC be applied to the base EF, the sides BA, AC should not coincide with ED, DF; they will therefore coincide, so that the angle BAC will also coincide with the angle EDF, and will be equal to it."", ""ProofWordCount"" -> 255, ""GreekProof"" -> ""ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ: ἐχέτω δὲ καὶ βάσιν τὴν ΒΓ βάσει τῇ ΕΖ ἴσην: λέγω, ὅτι καὶ γωνία ἡ ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. Ἐφαρμοζομένου γὰρ τοῦ ΑΒΓ τριγώνου ἐπὶ τὸ ΔΕΖ τρίγωνον καὶ τιθεμένου τοῦ μὲν Β σημείου ἐπὶ τὸ Ε σημεῖον τῆς δὲ ΒΓ εὐθείας ἐπὶ τὴν ΕΖ ἐφαρμόσει καὶ τὸ Γ σημεῖον ἐπὶ τὸ Ζ διὰ τὸ ἴσην εἶναι τὴν ΒΓ τῇ ΕΖ: ἐφαρμοσάσης δὴ τῆς ΒΓ ἐπὶ τὴν ΕΖ ἐφαρμόσουσι καὶ αἱ ΒΑ, ΓΑ ἐπὶ τὰς ΕΔ, ΔΖ. εἰ γὰρ βάσις μὲν ἡ ΒΓ ἐπὶ βάσιν τὴν ΕΖ ἐφαρμόσει, αἱ δὲ ΒΑ, ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ, ΔΖ οὐκ ἐφαρμόσουσιν ἀλλὰ παραλλάξουσιν ὡς αἱ ΕΗ, ΗΖ, συσταθήσονται ἐπὶ τῆς αὐτῆς εὐθείας δύο ταῖς αὐταῖς εὐθείαις ἄλλαι δύο εὐθεῖαι ἴσαι ἑκατέρα ἑκατέρᾳ πρὸς ἄλλῳ καὶ ἄλλῳ σημείῳ ἐπὶ τὰ αὐτὰ μέρη τὰ αὐτὰ πέρατα ἔχουσαι. οὐ συνίστανται δέ: οὐκ ἄρα ἐφαρμοζομένης τῆς ΒΓ βάσεως ἐπὶ τὴν ΕΖ βάσιν οὐκ ἐφαρμόσουσι καὶ αἱ ΒΑ, ΑΓ πλευραὶ ἐπὶ τὰς ΕΔ, ΔΖ. ἐφαρμόσουσιν ἄρα: ὥστε καὶ γωνία ἡ ὑπὸ ΒΑΓ ἐπὶ γωνίαν τὴν ὑπὸ ΕΔΖ ἐφαρμόσει καὶ ἴση αὐτῇ ἔσται. ἐὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δύο πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ τὴν βάσιν τῇ βάσει ἴσην ἔχῃ, καὶ τὴν γωνίαν τῇ γωνίᾳ ἴσην ἕξει τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 248|>","<|""VertexLabel"" -> ""1.9"", ""Text"" -> ""To bisect a given rectilineal angle."", ""TextWordCount"" -> 6, ""GreekText"" -> ""τὴν δοθεῖσαν γωνίαν εὐθύγραμμον δίχα τεμεῖν."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 8}}, ""Proof"" -> ""Let the angle BAC be the given rectilineal angle. Thus it is required to bisect it. Let a point D be taken at random on AB; let AE be cut off from AC equal to AD; [I. 3] let DE be joined, and on DE let the equilateral triangle DEF be constructed; let AF be joined. I say that the angle BAC has been bisected by the straight line AF. For, since AD is equal to AE, and AF is common, the two sides DA, AF are equal to the two sides EA, AF respectively. And the base DF is equal to the base EF; therefore the angle DAF is equal to the angle EAF. [I. 8] Therefore the given rectilineal angle BAC has been bisected by the straight line AF."", ""ProofWordCount"" -> 131, ""GreekProof"" -> ""ἔστω ἡ δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΒΑΓ. δεῖ δὴ αὐτὴν δίχα τεμεῖν. εἰλήφθω ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ, καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ τῇ ΑΔ ἴση ἡ ΑΕ, καὶ ἐπεζεύχθω ἡ ΔΕ, καὶ συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ, καὶ ἐπεζεύχθω ἡ ΑΖ: λέγω, ὅτι ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ὑπὸ τῆς ΑΖ εὐθείας. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ, κοινὴ δὲ ἡ ΑΖ, δύο δὴ αἱ ΔΑ, ΑΖ δυσὶ ταῖς ΕΑ, ΑΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ. καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ἴση ἐστίν: γωνία ἄρα ἡ ὑπὸ ΔΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖ ἴση ἐστίν. ἡ ἄρα δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΒΑΓ δίχα τέτμηται ὑπὸ τῆς ΑΖ εὐθείας: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 122|>","<|""VertexLabel"" -> ""1.10"", ""Text"" -> ""To bisect a given finite straight line."", ""TextWordCount"" -> 7, ""GreekText"" -> ""τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην δίχα τεμεῖν."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 1}, {""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 9}}, ""Proof"" -> ""Let AB be the given finite straight line. Thus it is required to bisect the finite straight line AB. Let the equilateral triangle ABC be constructed on it, [I. 1] and let the angle ACB be bisected by the straight line CD; [I. 9] I say that the straight line AB has been bisected at the point D. For, since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD respectively; and the angle ACD is equal to the angle BCD; therefore the base AD is equal to the base BD. [I. 4] Therefore the given finite straight line AB has been bisected at D."", ""ProofWordCount"" -> 117, ""GreekProof"" -> ""ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ: δεῖ δὴ τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα τεμεῖν. συνεστάτω ἐπ᾽ αὐτῆς τρίγωνον ἰσόπλευρον τὸ ΑΒΓ, καὶ τετμήσθω ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ: λέγω, ὅτι ἡ ΑΒ εὐθεῖα δίχα τέτμηται κατὰ τὸ Δ σημεῖον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, κοινὴ δὲ ἡ ΓΔ, δύο δὴ αἱ ΑΓ, ΓΔ δύο ταῖς ΒΓ, ΓΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση ἐστίν: βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔ ἴση ἐστίν. ἡ ἄρα δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ δίχα τέτμηται κατὰ τὸ Δ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 104|>","<|""VertexLabel"" -> ""1.11"", ""Text"" -> ""To draw a straight line at right angles to a given straight line from a given point on it."", ""TextWordCount"" -> 19, ""GreekText"" -> ""τῇ δοθείσῃ εὐθείᾳ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν."", ""GreekTextWordCount"" -> 15, ""References"" -> {{""Book"" -> 1, ""Definition"" -> 10}, {""Book"" -> 1, ""Theorem"" -> 1}, {""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 8}}, ""Proof"" -> ""Let AB be the given straight line, and C the given point on it. Thus it is required to draw from the point C a straight line at right angles to the straight line AB. Let a point D be taken at random on AC;let CE be made equal to CD; [I. 3] on DE let the equilateral triangle FDE be constructed, [I. 1] and let FC be joined; I say that the straight line FC has been drawn at rightangles to the given straight line AB from C the given point on it. For, since DC is equal to CE, and CF is common, the two sides DC, CF are equal to the two sides EC, CF respectively; and the base DF is equal to the base FE; therefore the angle DCF is equal to the angle ECF; [I. 8]and they are adjacent angles. But, when a straight line set up on a straight line makesthe adjacent angles equal to one another, each of the equal angles is right; [Def. 10] therefore each of the angles DCF, FCE is right. Therefore the straight line CF has been drawn at right angles to the given straight line AB from the given pointC on it."", ""ProofWordCount"" -> 206, ""GreekProof"" -> ""ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ τὸ δὲ δοθὲν σημεῖον ἐπ᾽ αὐτῆς τὸ Γ: δεῖ δὴ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν. εἰλήφθω ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ, καὶ κείσθω τῇ ΓΔ ἴση ἡ ΓΕ, καὶ συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον τὸ ΖΔΕ, καὶ ἐπεζεύχθω ἡ ΖΓ: λέγω, ὅτι τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου τοῦ Γ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ, κοινὴ δὲ ἡ ΓΖ, δύο δὴ αἱ ΔΓ, ΓΖ δυσὶ ταῖς ΕΓ, ΓΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ ἴση ἐστίν: γωνία ἄρα ἡ ὑπὸ ΔΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ ἴση ἐστίν: καί εἰσιν ἐφεξῆς. ὅταν δὲ εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν: ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΔΓΖ, ΖΓΕ. τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου τοῦ Γ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 182|>","<|""VertexLabel"" -> ""1.12"", ""Text"" -> ""To a given infinite straight line, from a given point which is not on it, to draw a perpendicular straight line."", ""TextWordCount"" -> 21, ""GreekText"" -> ""ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον ἀπὸ τοῦ δοθέντος σημείου, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν."", ""GreekTextWordCount"" -> 19, ""References"" -> {{""Postulate"" -> 1}, {""Postulate"" -> 3}, {""Book"" -> 1, ""Definition"" -> 10}, {""Book"" -> 1, ""Theorem"" -> 8}, {""Book"" -> 1, ""Theorem"" -> 10}}, ""Proof"" -> ""Let AB be the given infinite straight line, and C the given point which is not on it;thus it is required to draw to the given infinite straight line AB, from the given point C which is not on it, a perpendicular straight line. For let a point D be takenat random on the other side of the straight line AB, and with centre C and distance CD let the circle EFG be described; [Post. 3] let the straight line EG be bisected at H, [I. 10]and let the straight lines CG, CH, CE be joined. [Post. 1] I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it. For, since GH is equal to HE, and HC is common, the two sides GH, HC are equal to the two sides EH, HC respectively; and the base CG is equal to the base CE;therefore the angle CHG is equal to the angle EHC. [I. 8]And they are adjacent angles. But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other iscalled a perpendicular to that on which it stands. [Def. 10] Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it."", ""ProofWordCount"" -> 245, ""GreekProof"" -> ""ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ τὸ δὲ δοθὲν σημεῖον, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, τὸ Γ: δεῖ δὴ ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν. εἰλήφθω γὰρ ἐπὶ τὰ ἕτερα μέρη τῆς ΑΒ εὐθείας τυχὸν σημεῖον τὸ Δ, καὶ κέντρῳ μὲν τῷ Γ διαστήματι δὲ τῷ ΓΔ κύκλος γεγράφθω ὁ ΕΖΗ, καὶ τετμήσθω ἡ ΕΗ εὐθεῖα δίχα κατὰ τὸ Θ, καὶ ἐπεζεύχθωσαν αἱ ΓΗ, ΓΘ, ΓΕ εὐθεῖαι: λέγω, ὅτι ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, κάθετος ἦκται ἡ ΓΘ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ, κοινὴ δὲ ἡ ΘΓ, δύο δὴ αἱ ΗΘ, ΘΓ δύο ταῖς ΕΘ, ΘΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ἐστιν ἴση: γωνία ἄρα ἡ ὑπὸ ΓΘΗ γωνίᾳ τῇ ὑπὸ ΕΘΓ ἐστιν ἴση. καί εἰσιν ἐφεξῆς. ὅταν δὲ εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν, καὶ ἡ ἐφεστηκυῖα εὐθεῖα κάθετος καλεῖται ἐφ᾽ ἣν ἐφέστηκεν. ἐπὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπ᾽ αὐτῆς, κάθετος ἦκται ἡ ΓΘ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 216|>","<|""VertexLabel"" -> ""1.13"", ""Text"" -> ""If a straight line set up on a straight line make angles, it will make either two right angles or angles equal to two right angles."", ""TextWordCount"" -> 26, ""GreekText"" -> ""ἐὰν εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα γωνίας ποιῇ, ἤτοι δύο ὀρθὰς ἢ δυσὶν ὀρθαῖς ἴσας ποιήσει."", ""GreekTextWordCount"" -> 16, ""References"" -> {{""Common Notion"" -> 1}, {""Common Notion"" -> 2}, {""Book"" -> 1, ""Definition"" -> 10}, {""Book"" -> 1, ""Theorem"" -> 11}}, ""Proof"" -> ""For let any straight line AB set up on the straight lineCD make the angles CBA, ABD; I say that the angles CBA, ABD are either two right angles or equal to two right angles. Now, if the angle CBA is equal tothe angle ABD, they are two right angles. [Def. 10] But, if not, let BE be drawn from the point B at right angles to CD; [I. 11] therefore the angles CBE, EBD are two right angles. Then, since the angle CBE is equal to the two angles CBA, ABE, let the angle EBD be added to each; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [C. N. 2] Again, since the angle DBA is equal to the two angles DBE, EBA, let the angle ABC be added to each; therefore the angles DBA. ABC are equal to the three angles DBE, EBA, ABC. [C. N. 2] But the angles CBE, EBD were also proved equal to the same three angles; and things which are equal to the same thing are also equal to one another; [C. N. 1]therefore the angles CBE, EBD are also equal to the angles DBA, ABC. But the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are also equal to two right angles."", ""ProofWordCount"" -> 221, ""GreekProof"" -> ""εὐθεῖα γάρ τις ἡ ΑΒ ἐπ᾽ εὐθεῖαν τὴν ΓΔ σταθεῖσα γωνίας ποιείτω τὰς ὑπὸ ΓΒΑ, ΑΒΔ: λέγω, ὅτι αἱ ὑπὸ ΓΒΑ, ΑΒΔ γωνίαι ἤτοι δύο ὀρθαί εἰσιν ἢ δυσὶν ὀρθαῖς ἴσαι. εἰ μὲν οὖν ἴση ἐστὶν ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ, δύο ὀρθαί εἰσιν. εἰ δὲ οὔ, ἤχθω ἀπὸ τοῦ Β σημείου τῇ ΓΔ εὐθείᾳ πρὸς ὀρθὰς ἡ ΒΕ: αἱ ἄρα ὑπὸ ΓΒΕ, ΕΒΔ δύο ὀρθαί εἰσιν: καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ δυσὶ ταῖς ὑπὸ ΓΒΑ, ΑΒΕ ἴση ἐστίν, κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ: αἱ ἄρα ὑπὸ ΓΒΕ, ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ, ΑΒΕ, ΕΒΔ ἴσαι εἰσίν. πάλιν, ἐπεὶ ἡ ὑπὸ ΔΒΑ δυσὶ ταῖς ὑπὸ ΔΒΕ, ΕΒΑ ἴση ἐστίν, κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ: αἱ ἄρα ὑπὸ ΔΒΑ, ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ, ΕΒΑ, ΑΒΓ ἴσαι εἰσίν. ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ, ΕΒΔ τρισὶ ταῖς αὐταῖς ἴσαι: τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα: καὶ αἱ ὑπὸ ΓΒΕ, ΕΒΔ ἄρα ταῖς ὑπὸ ΔΒΑ, ΑΒΓ ἴσαι εἰσίν: ἀλλὰ αἱ ὑπὸ ΓΒΕ, ΕΒΔ δύο ὀρθαί εἰσιν: καὶ αἱ ὑπὸ ΔΒΑ, ΑΒΓ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἐὰν ἄρα εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα γωνίας ποιῇ, ἤτοι δύο ὀρθὰς ἢ δυσὶν ὀρθαῖς ἴσας ποιήσει: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 199|>","<|""VertexLabel"" -> ""1.14"", ""Text"" -> ""If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another."", ""TextWordCount"" -> 42, ""GreekText"" -> ""ἐὰν πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ δύο εὐθεῖαι μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δυσὶν ὀρθαῖς ἴσας ποιῶσιν, ἐπ᾽ εὐθείας ἔσονται ἀλλήλαις αἱ εὐθεῖαι."", ""GreekTextWordCount"" -> 31, ""References"" -> {{""Common Notion"" -> 1}, {""Common Notion"" -> 3}, {""Postulate"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 13}}, ""Proof"" -> ""For with any straight line AB, and at the point B on it, let the two straight lines BC, BD not lying on the same side make the adjacent angles ABC, ABD equal to two right angles; I say that BD is in a straight line with CB. For, if BD is not in a straight line with BC, let BE be in a straight line with CB. Then, since the straight line AB stands on the straight line CBE,the angles ABC, ABE are equal to two right angles. [I. 13]But the angles ABC, ABD are also equal to two right angles; therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C. N. 1] Let the angle CBA be subtracted from each;therefore the remaining angle ABE is equal to the remaining angle ABD, [C. N. 3] the less to the greater: which is impossible. Therefore BE is not in a straight line with CB. Similarly we can prove that neither is any other straightline except BD. Therefore CB is in a straight line with BD."", ""ProofWordCount"" -> 184, ""GreekProof"" -> ""πρὸς γάρ τινι εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β δύο εὐθεῖαι αἱ ΒΓ, ΒΔ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΑΒΓ, ΑΒΔ δύο ὀρθαῖς ἴσας ποιείτωσαν: λέγω, ὅτι ἐπ᾽ εὐθείας ἐστὶ τῇ ΓΒ ἡ ΒΔ. εἰ γὰρ μή ἐστι τῇ ΒΓ ἐπ᾽ εὐθείας ἡ ΒΔ, ἔστω τῇ ΓΒ ἐπ᾽ εὐθείας ἡ ΒΕ. ἐπεὶ οὖν εὐθεῖα ἡ ΑΒ ἐπ᾽ εὐθεῖαν τὴν ΓΒΕ ἐφέστηκεν, αἱ ἄρα ὑπὸ ΑΒΓ, ΑΒΕ γωνίαι δύο ὀρθαῖς ἴσαι εἰσίν: εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΒΓ, ΑΒΔ δύο ὀρθαῖς ἴσαι: αἱ ἄρα ὑπὸ ΓΒΑ, ΑΒΕ ταῖς ὑπὸ ΓΒΑ, ΑΒΔ ἴσαι εἰσίν. κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΓΒΑ: λοιπὴ ἄρα ἡ ὑπὸ ΑΒΕ λοιπῇ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση, ἡ ἐλάσσων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐπ᾽ εὐθείας ἐστὶν ἡ ΒΕ τῇ ΓΒ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἄλλη τις πλὴν τῆς ΒΔ: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΓΒ τῇ ΒΔ. ἐὰν ἄρα πρός τινι εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ δύο εὐθεῖαι μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δυσὶν ὀρθαῖς ἴσας ποιῶσιν, ἐπ᾽ εὐθείας ἔσονται ἀλλήλαις αἱ εὐθεῖαι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 193|>","<|""VertexLabel"" -> ""1.15"", ""Text"" -> ""If two straight lines cut one another, they make the vertical angles equal to one another."", ""TextWordCount"" -> 16, ""GreekText"" -> ""ἐὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς κατὰ κορυφὴν γωνίας ἴσας ἀλλήλαις ποιοῦσιν."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Common Notion"" -> 1}, {""Common Notion"" -> 3}, {""Postulate"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 13}}, ""Proof"" -> ""For let the straight lines AB, CD cut one another at the point E; I say that the angle AEC is equal to the angle DEB, and the angle CEB to the angle AED. For, since the straight line AE standson the straight line CD, making the angles CEA, AED, the angles CEA, AED are equal to two right angles. [I. 13] Again, since the straight line DE stands on the straight line AB, making the angles AED, DEB, the angles AED, DEB are equal to two right angles. [I. 13] But the angles CEA, AED were also proved equal to two right angles; therefore the angles CEA, AED are equal to the angles AED DEB. [Post. 4 and C. N. 1]Let the angle AED be subtracted from each; therefore the remaining angle CEA is equal to the remaining angle BED. [C. N. 3] Similarly it can be proved that the angles CEB, DEA are also equal."", ""ProofWordCount"" -> 158, ""GreekProof"" -> ""δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΓΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε σημεῖον: λέγω, ὅτι ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΕΓ γωνία τῇ ὑπὸ ΔΕΒ, ἡ δὲ ὑπὸ ΓΕΒ τῇ ὑπὸ ΑΕΔ. ἐπεὶ γὰρ εὐθεῖα ἡ ΑΕ ἐπ᾽ εὐθεῖαν τὴν ΓΔ ἐφέστηκε γωνίας ποιοῦσα τὰς ὑπὸ ΓΕΑ, ΑΕΔ, αἱ ἄρα ὑπὸ ΓΕΑ, ΑΕΔ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. πάλιν, ἐπεὶ εὐθεῖα ἡ ΔΕ ἐπ᾽ εὐθεῖαν τὴν ΑΒ ἐφέστηκε γωνίας ποιοῦσα τὰς ὑπὸ ΑΕΔ, ΔΕΒ, αἱ ἄρα ὑπὸ ΑΕΔ, ΔΕΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΕΑ, ΑΕΔ δυσὶν ὀρθαῖς ἴσαι: αἱ ἄρα ὑπὸ ΓΕΑ, ΑΕΔ ταῖς ὑπὸ ΑΕΔ, ΔΕΒ ἴσαι εἰσίν. κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΑΕΔ: λοιπὴ ἄρα ἡ ὑπὸ ΓΕΑ λοιπῇ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν: ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ ὑπὸ ΓΕΒ, ΔΕΑ ἴσαι εἰσίν. ἐὰν ἄρα δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς κατὰ κορυφὴν γωνίας ἴσας ἀλλήλαις ποιοῦσιν: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερὸν ὅτι, ἐὰν δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὰς πρὸς τῇ τομῇ γωνίας τέτρασιν ὀρθαῖς ἴσας ποιήσουσιν."", ""GreekProofWordCount"" -> 169|>","<|""VertexLabel"" -> ""1.16"", ""Text"" -> ""In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles."", ""TextWordCount"" -> 23, ""GreekText"" -> ""παντὸς τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν μείζων ἐστίν."", ""GreekTextWordCount"" -> 17, ""References"" -> {{""Common Notion"" -> 5}, {""Postulate"" -> 1}, {""Postulate"" -> 2}, {""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 10}, {""Book"" -> 1, ""Theorem"" -> 15}}, ""Proof"" -> ""Let ABC be a triangle, and let one side of it BC be produced to D; I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC. Let AC be bisected at E [I. 10], and let BE be joined and producedin a straight line to F; let EF be made equal to BE [I. 3], let FC be joined [Post. 1], and let AC be drawn through to G [Post. 2]. Then, since AE is equal to EC,and BE to EF, the two sides AE, EB are equal to the two sides CE, EF respectively; and the angle AEB is equal to the angle FEC, for they are vertical angles. [I. 15]Therefore the base AB is equal to the base FC, and the triangle ABE is equal to the triangle CFE, and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]therefore the angle BAE is equal to the angle ECF. But the angle ECD is greater than the angle ECF; [C. N. 5]therefore the angle ACD is greater than the angle BAE. Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15], can be proved greater than the angle ABC as well."", ""ProofWordCount"" -> 221, ""GreekProof"" -> ""ἔστω τρίγωνον τὸ ΑΒΓ, καὶ προσεκβεβλήσθω αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ: λέγω, ὅτι ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ μείζων ἐστὶν ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον τῶν ὑπὸ ΓΒΑ, ΒΑΓ γωνιῶν. τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε, καὶ ἐπιζευχθεῖσα ἡ ΒΕ ἐκβεβλήσθω ἐπ᾽ εὐθείας ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΒΕ ἴση ἡ ΕΖ, καὶ ἐπεζεύχθω ἡ ΖΓ, καὶ διήχθω ἡ ΑΓ ἐπὶ τὸ Η. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΓ, ἡ δὲ ΒΕ τῇ ΕΖ, δύο δὴ αἱ ΑΕ, ΕΒ δυσὶ ταῖς ΓΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΕΒ γωνίᾳ τῇ ὑπὸ ΖΕΓ ἴση ἐστίν: κατὰ κορυφὴν γάρ: βάσις ἄρα ἡ ΑΒ βάσει τῇ ΖΓ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΕΓ τριγώνῳ ἐστὶν ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΕ τῇ ὑπὸ ΕΓΖ. μείζων δέ ἐστιν ἡ ὑπὸ ΕΓΔ τῆς ὑπὸ ΕΓΖ: μείζων ἄρα ἡ ὑπὸ ΑΓΔ τῆς ὑπὸ ΒΑΕ. ὁμοίως δὴ τῆς ΒΓ τετμημένης δίχα δειχθήσεται καὶ ἡ ὑπὸ ΒΓΗ, τουτέστιν ἡ ὑπὸ ΑΓΔ, μείζων καὶ τῆς ὑπὸ ΑΒΓ. παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία ἑκατέρας τῶν ἐντὸς καὶ ἀπεναντίον γωνιῶν μείζων ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 214|>","<|""VertexLabel"" -> ""1.17"", ""Text"" -> ""In any triangle two angles taken together in any manner are less than two right angles."", ""TextWordCount"" -> 16, ""GreekText"" -> ""παντὸς τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσσονές εἰσι πάντῃ μεταλαμβανόμεναι."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Postulate"" -> 2}, {""Book"" -> 1, ""Theorem"" -> 13}, {""Book"" -> 1, ""Theorem"" -> 16}}, ""Proof"" -> ""Let ABC be a triangle; I say that two angles of the triangle ABC taken together in any manner are less than two right angles. For let BC be produced to D. [Post. 2] Then, since the angle ACD is an exterior angle of the triangle ABC, it is greater than the interior and opposite angle ABC. [I. 16] Let the angle ACB be added to each; therefore the angles ACD, ACB are greater than the angles ABC, BCA. But the angles ACD, ACB are equal to two right angles. [I. 13] Therefore the angles ABC, BCA are less than two right angles. Similarly we can prove that the angles BAC, ACB are also less than two right angles, and so are the angles CAB, ABC as well."", ""ProofWordCount"" -> 128, ""GreekProof"" -> ""ἔστω τρίγωνον τὸ ΑΒΓ: λέγω, ὅτι τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάττονές εἰσι πάντῃ μεταλαμβανόμεναι. Ἐκβεβλήσθω γὰρ ἡ ΒΓ ἐπὶ τὸ Δ. καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ. κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ: αἱ ἄρα ὑπὸ ΑΓΔ, ΑΓΒ τῶν ὑπὸ ΑΒΓ, ΒΓΑ μείζονές εἰσιν. ἀλλ᾽ αἱ ὑπὸ ΑΓΔ, ΑΓΒ δύο ὀρθαῖς ἴσαι εἰσίν: αἱ ἄρα ὑπὸ ΑΒΓ, ΒΓΑ δύο ὀρθῶν ἐλάσσονές εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ὑπὸ ΒΑΓ, ΑΓΒ δύο ὀρθῶν ἐλάσσονές εἰσι καὶ ἔτι αἱ ὑπὸ ΓΑΒ, ΑΒΓ. παντὸς ἄρα τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσσονές εἰσι πάντῃ μεταλαμβανόμεναι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 114|>","<|""VertexLabel"" -> ""1.18"", ""Text"" -> ""In any triangle the greater side subtends the greater angle."", ""TextWordCount"" -> 10, ""GreekText"" -> ""παντὸς τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα γωνίαν ὑποτείνει."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 16}}, ""Proof"" -> ""For let ABC be a triangle having the side AC greater than AB; I say that the angle ABC is also greater than the angle BCA. For, since AC is greater than AB, let AD be made equal to AB [I. 3], and let BD bejoined. Then, since the angle ADB is an exterior angle of the triangle BCD, it is greater than the interior and opposite angle DCB. [I. 16] But the angle ADB is equal to the angle ABD, since the side AB is equal to AD; therefore the angle ABD is also greater than the angle ACB; therefore the angle ABC is much greater than the angle ACB."", ""ProofWordCount"" -> 111, ""GreekProof"" -> ""ἔστω γὰρ τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ΑΓ πλευρὰν τῆς ΑΒ: λέγω, ὅτι καὶ γωνία ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ τῆς ὑπὸ ΒΓΑ. ἐπεὶ γὰρ μείζων ἐστὶν ἡ ΑΓ τῆς ΑΒ, κείσθω τῇ ΑΒ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΒΔ. καὶ ἐπεὶ τριγώνου τοῦ ΒΓΔ ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΔΒ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΔΓΒ: ἴση δὲ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ, ἐπεὶ καὶ πλευρὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση: μείζων ἄρα καὶ ἡ ὑπὸ ΑΒΔ τῆς ὑπὸ ΑΓΒ: πολλῷ ἄρα ἡ ὑπὸ ΑΒΓ μείζων ἐστὶ τῆς ὑπὸ ΑΓΒ. παντὸς ἄρα τριγώνου ἡ μείζων πλευρὰ τὴν μείζονα γωνίαν ὑποτείνει: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 111|>","<|""VertexLabel"" -> ""1.19"", ""Text"" -> ""In any triangle the greater angle is subtended by the greater side."", ""TextWordCount"" -> 12, ""GreekText"" -> ""παντὸς τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 18}}, ""Proof"" -> ""Let ABC be a triangle having the angle ABC greater than the angle BCA; I say that the side AC is also greater than the side AB. For, if not, AC is either equal to AB or less. Now AC is not equal to AB; for then the angle ABC would also have been equal to the angle ACB; [I. 5] but it is not; therefore AC is not equal to AB. Neither is AC less than AB, for then the angle ABC would also have been less than the angle ACB; [I. 18] but it is not; therefore AC is not less than AB. And it was proved that it is not equal either. Therefore AC is greater than AB."", ""ProofWordCount"" -> 121, ""GreekProof"" -> ""ἔστω τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν τῆς ὑπὸ ΒΓΑ: λέγω, ὅτι καὶ πλευρὰ ἡ ΑΓ πλευρᾶς τῆς ΑΒ μείζων ἐστίν. εἰ γὰρ μή, ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ἢ ἐλάσσων: ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ: ἴση γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ: οὐκ ἔστι δέ: οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ. οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ: ἐλάσσων γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῆς ὑπὸ ΑΓΒ: οὐκ ἔστι δέ: οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ. ἐδείχθη δέ, ὅτι οὐδὲ ἴση ἐστίν. μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ. παντὸς ἄρα τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 126|>","<|""VertexLabel"" -> ""1.20"", ""Text"" -> ""In any triangle two sides taken together in any manner are greater than the remaining one."", ""TextWordCount"" -> 16, ""GreekText"" -> ""παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Common Notion"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 19}}, ""Proof"" -> ""For let ABC be a triangle; I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one, namely BA, AC greater than BC, AB, BC greater than AC, BC, CA greater than AB. For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined. Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD; [I. 5] therefore the angle BCD is greater than the angle ADC. [C. N. 5] And, since DCB is a triangle having the angle BCD greater than the angle BDC, and the greater angle is subtended by the greater side, [I. 19]therefore DB is greater than BC. But DA is equal to AC; therefore BA, AC are greater than BC. Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB."", ""ProofWordCount"" -> 159, ""GreekProof"" -> ""ἔστω γὰρ τρίγωνον τὸ ΑΒΓ: λέγω, ὅτι τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, αἱ μὲν ΒΑ, ΑΓ τῆς ΒΓ, αἱ δὲ ΑΒ, ΒΓ τῆς ΑΓ, αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. διήχθω γὰρ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον, καὶ κείσθω τῇ ΓΑ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ: μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ: καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν τῆς ὑπὸ ΒΔΓ, ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει, ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων. ἴση δὲ ἡ ΔΑ τῇ ΑΓ: μείζονες ἄρα αἱ ΒΑ, ΑΓ τῆς ΒΓ: ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ μὲν ΑΒ, ΒΓ τῆς ΓΑ μείζονές εἰσιν, αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 160|>","<|""VertexLabel"" -> ""1.21"", ""Text"" -> ""If on one of the sides of a triangle, from its extremities, there be constructed two straight lines meeting within the triangle, the straight lines so constructed will be less than the remaining two sides of the triangle, but will contain a greaterangle."", ""TextWordCount"" -> 43, ""GreekText"" -> ""ἐὰν τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν περάτων δύο εὐθεῖαι ἐντὸς συσταθῶσιν, αἱ συσταθεῖσαι τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ἐλάττονες μὲν ἔσονται, μείζονα δὲ γωνίαν περιέξουσιν."", ""GreekTextWordCount"" -> 28, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 16}, {""Book"" -> 1, ""Theorem"" -> 20}}, ""Proof"" -> ""On BC, one of the sides of the triangle ABC, from its extremities B, C, let the two straight lines BD, DC be constructed meeting within the triangle; I say that BD, DC are less than the remaining two sidesof the triangle BA, AC, but contain an angle BDC greater than the angle BAC. For let BD be drawn through to E. Then, since in any triangle two sides are greater than the remainingone, [I. 20] therefore, in the triangle ABE, the two sides AB, AE are greater than BE. Let EC be added to each; therefore BA, AC are greater than BE, EC. Again, since, in the triangle CED, the two sides CE, ED are greater than CD, let DB be added to each; therefore CE, EB are greater than CD, DB. But BA, AC were proved greater than BE, EC;therefore BA, AC are much greater than BD, DC. Again, since in any triangle the exterior angle is greater than the interior and opposite angle, [I. 16] therefore, in the triangle CDE, the exterior angle BDC is greater than the angle CED. For the same reason, moreover, in the triangle ABE also, the exterior angle CEB is greater than the angle BAC. But the angle BDC was proved greater than the angle CEB; therefore the angle BDC is much greater than the angle BAC."", ""ProofWordCount"" -> 226, ""GreekProof"" -> ""τριγώνου γὰρ τοῦ ΑΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ ἀπὸ τῶν περάτων τῶν Β, Γ δύο εὐθεῖαι ἐντὸς συνεστάτωσαν αἱ ΒΔ, ΔΓ: λέγω, ὅτι αἱ ΒΔ, ΔΓ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν τῶν ΒΑ, ΑΓ ἐλάσσονες μέν εἰσιν, μείζονα δὲ γωνίαν περιέχουσι τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ. διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε. καὶ ἐπεὶ παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσιν, τοῦ ΑΒΕ ἄρα τριγώνου αἱ δύο πλευραὶ αἱ ΑΒ, ΑΕ τῆς ΒΕ μείζονές εἰσιν: κοινὴ προσκείσθω ἡ ΕΓ: αἱ ἄρα ΒΑ, ΑΓ τῶν ΒΕ, ΕΓ μείζονές εἰσιν. πάλιν, ἐπεὶ τοῦ ΓΕΔ τριγώνου αἱ δύο πλευραὶ αἱ ΓΕ, ΕΔ τῆς ΓΔ μείζονές εἰσιν, κοινὴ προσκείσθω ἡ ΔΒ: αἱ ΓΕ, ΕΒ ἄρα τῶν ΓΔ, ΔΒ μείζονές εἰσιν. ἀλλὰ τῶν ΒΕ, ΕΓ μείζονες ἐδείχθησαν αἱ ΒΑ, ΑΓ: πολλῷ ἄρα αἱ ΒΑ, ΑΓ τῶν ΒΔ, ΔΓ μείζονές εἰσιν. πάλιν, ἐπεὶ παντὸς τριγώνου ἡ ἐκτὸς γωνία τῆς ἐντὸς καὶ ἀπεναντίον μείζων ἐστίν, τοῦ ΓΔΕ ἄρα τριγώνου ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ. διὰ ταὐτὰ τοίνυν καὶ τοῦ ΑΒΕ τριγώνου ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ. ἀλλὰ τῆς ὑπὸ ΓΕΒ μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ: πολλῷ ἄρα ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ. ἐὰν ἄρα τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν περάτων δύο εὐθεῖαι ἐντὸς συσταθῶσιν, αἱ συσταθεῖσαι τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ἐλάττονες μέν εἰσιν, μείζονα δὲ γωνίαν περιέχουσιν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 239|>","<|""VertexLabel"" -> ""1.22"", ""Text"" -> ""Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one."", ""TextWordCount"" -> 39, ""GreekText"" -> ""ἐκ τριῶν εὐθειῶν, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις εὐθείαις, τρίγωνον συστήσασθαι: δεῖ δὲ τὰς δύο τῆς λοιπῆς μείζονας εἶναι πάντῃ μεταλαμβανομένας διὰ τὸ καὶ παντὸς τριγώνου τὰς δύο πλευρὰς τῆς λοιπῆς μείζονας εἶναι πάντῃ μεταλαμβανομένας."", ""GreekTextWordCount"" -> 36, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 20}}, ""Proof"" -> ""Let the three given straight lines be A, B, C, and of these let two taken together in any manner be greater than the remaining one, namely A, B greater than C, A, C greater than B, and B, C greater than A; thus it is required to construct a triangle out of straight lines equal to A, B, C. Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C. [I. 3] With centre F and distance FD let the circle DKL be described; again, with centre G and distance GH let the circle KLH be described; and let KF, KG be joined; I say that the triangle KFG has been constructed out of three straight lines equal to A, B, C. For, since the point F is the centre of the circle DKL, FD is equal to FK. But FD is equal to A; therefore KF is also equal to A. Again, since the point G is the centre of the circle LKH, GH is equal to GK. But GH is equal to C; therefore KG is also equal to C. And FG is also equal to B; therefore the three straight lines KF, FG, GK are equal to the three straight lines A, B, C. Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constructed."", ""ProofWordCount"" -> 263, ""GreekProof"" -> ""ἔστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ, ὧν αἱ δύο τῆς λοιπῆς μείζονες ἔστωσαν πάντῃ μεταλαμβανόμεναι, αἱ μὲν Α, Β τῆς Γ, αἱ δὲ Α, Γ τῆς Β, καὶ ἔτι αἱ Β, Γ τῆς Α: δεῖ δὴ ἐκ τῶν ἴσων ταῖς Α, Β, Γ τρίγωνον συστήσασθαι. Ἐκκείσθω τις εὐθεῖα ἡ ΔΕ πεπερασμένη μὲν κατὰ τὸ Δ ἄπειρος δὲ κατὰ τὸ Ε, καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΖ, τῇ δὲ Β ἴση ἡ ΖΗ, τῇ δὲ Γ ἴση ἡ ΗΘ: καὶ κέντρῳ μὲν τῷ Ζ, διαστήματι δὲ τῷ ΖΔ κύκλος γεγράφθω ὁ ΔΚΛ: πάλιν κέντρῳ μὲν τῷ Η, διαστήματι δὲ τῷ ΗΘ κύκλος γεγράφθω ὁ ΚΛΘ, καὶ ἐπεζεύχθωσαν αἱ ΚΖ, ΚΗ: λέγω, ὅτι ἐκ τριῶν εὐθειῶν τῶν ἴσων ταῖς Α, Β, Γ τρίγωνον συνέσταται τὸ ΚΖΗ. ἐπεὶ γὰρ τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΔΚΛ κύκλου, ἴση ἐστὶν ἡ ΖΔ τῇ ΖΚ: ἀλλὰ ἡ ΖΔ τῇ Α ἐστιν ἴση. καὶ ἡ ΚΖ ἄρα τῇ Α ἐστιν ἴση. πάλιν, ἐπεὶ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΛΚΘ κύκλου, ἴση ἐστὶν ἡ ΗΘ τῇ ΗΚ: ἀλλὰ ἡ ΗΘ τῇ Γ ἐστιν ἴση: καὶ ἡ ΚΗ ἄρα τῇ Γ ἐστιν ἴση. ἐστὶ δὲ καὶ ἡ ΖΗ τῇ Β ἴση: αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΚΖ, ΖΗ, ΗΚ τρισὶ ταῖς Α, Β, Γ ἴσαι εἰσίν. ἐκ τριῶν ἄρα εὐθειῶν τῶν ΚΖ, ΖΗ, ΗΚ, αἵ εἰσιν ἴσαι τρισὶ ταῖς δοθείσαις εὐθείαις ταῖς Α, Β, Γ, τρίγωνον συνέσταται τὸ ΚΖΗ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 240|>","<|""VertexLabel"" -> ""1.23"", ""Text"" -> ""On a given straight line and at a point on it to construct a rectilineal angle equal to a given rectilineal angle."", ""TextWordCount"" -> 22, ""GreekText"" -> ""πρὸς τῇ δοθείσῃ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ ἴσην γωνίαν εὐθύγραμμον συστήσασθαι."", ""GreekTextWordCount"" -> 17, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 8}}, ""Proof"" -> ""Let AB be the given straight line, A the point on it, and the angle DCE the given rectilineal angle; thus it is required to construct on the given straight line AB, and at the point A on it, a rectilineal angle equal to the given rectilineal angle DCE. On the straight lines CD, CE respectively let the points D, E be taken at random; let DE be joined, and out of three straight lines which are equal to the three straight lines CD, DE, CE let the triangle AFG be constructed in such a way that CD is equal to AF, CE to AG, and further DE to FG. Then, since the two sides DC, CE are equal to the two sides FA, AG respectively, and the base DE is equal to the base FG, the angle DCE is equal to the angle FAG. [I. 8] Therefore on the given straight line AB, and at the point A on it, the rectilineal angle FAG has been constructed equal to the given rectilineal angle DCE."", ""ProofWordCount"" -> 175, ""GreekProof"" -> ""ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ πρὸς αὐτῇ σημεῖον τὸ Α, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΔΓΕ: δεῖ δὴ πρὸς τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ ΔΓΕ ἴσην γωνίαν εὐθύγραμμον συστήσασθαι. εἰλήφθω ἐφ᾽ ἑκατέρας τῶν ΓΔ, ΓΕ τυχόντα σημεῖα τὰ Δ, Ε, καὶ ἐπεζεύχθω ἡ ΔΕ: καὶ ἐκ τριῶν εὐθειῶν, αἵ εἰσιν ἴσαι τρισὶ ταῖς ΓΔ, ΔΕ, ΓΕ, τρίγωνον συνεστάτω τὸ ΑΖΗ, ὥστε ἴσην εἶναι τὴν μὲν ΓΔ τῇ ΑΖ, τὴν δὲ ΓΕ τῇ ΑΗ, καὶ ἔτι τὴν ΔΕ τῇ ΖΗ. ἐπεὶ οὖν δύο αἱ ΔΓ, ΓΕ δύο ταῖς ΖΑ, ΑΗ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ βάσις ἡ ΔΕ βάσει τῇ ΖΗ ἴση, γωνία ἄρα ἡ ὑπὸ ΔΓΕ γωνίᾳ τῇ ὑπὸ ΖΑΗ ἐστιν ἴση. πρὸς ἄρα τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ ὑπὸ ΔΓΕ ἴση γωνία εὐθύγραμμος συνέσταται ἡ ὑπὸ ΖΑΗ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 163|>","<|""VertexLabel"" -> ""1.24"", ""Text"" -> ""If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base."", ""TextWordCount"" -> 39, ""GreekText"" -> ""ἐὰν δύο τρίγωνα τὰς δύο πλευρὰς ταῖς δύο πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας μείζονα ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν βάσιν τῆς βάσεως μείζονα ἕξει."", ""GreekTextWordCount"" -> 33, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 19}, {""Book"" -> 1, ""Theorem"" -> 23}}, ""Proof"" -> ""Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF, and let the angle at A be greater than the angle at D; I say that the base BC is also greater than the base EF. For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG equal to the angle BAC; [I. 23] let DG be made equal to either of the two straight lines AC, DF, and let EG, FG be joined. Then, since AB is equal to DE, and AC to DG,the two sides BA, AC are equal to the two sides ED, DG, respectively; and the angle BAC is equal to the angle EDG; therefore the base BC is equal to the base EG. [I. 4] Again, since DF is equal to DG,the angle DGF is also equal to the angle DFG; [I. 5]therefore the angle DFG is greater than the angle EGF. Therefore the angle EFG is much greater than the angle EGF. And, since EFG is a triangle having the angle EFG greater than the angle EGF, and the greater angle is subtended by the greater side, [I. 19]the side EG is also greater than EF. But EG is equal to BC. Therefore BC is also greater than EF."", ""ProofWordCount"" -> 246, ""GreekProof"" -> ""ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ, ἡ δὲ πρὸς τῷ Α γωνία τῆς πρὸς τῷ Δ γωνίας μείζων ἔστω: λέγω, ὅτι καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ μείζων ἐστίν. ἐπεὶ γὰρ μείζων ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ γωνίας, συνεστάτω πρὸς τῇ ΔΕ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Δ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ἡ ὑπὸ ΕΔΗ, καὶ κείσθω ὁποτέρᾳ τῶν ΑΓ, ΔΖ ἴση ἡ ΔΗ, καὶ ἐπεζεύχθωσαν αἱ ΕΗ, ΖΗ. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΑΓ τῇ ΔΗ, δύο δὴ αἱ ΒΑ, ΑΓ δυσὶ ταῖς ΕΔ, ΔΗ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση: βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΗ ἐστιν ἴση. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΔΖ τῇ ΔΗ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ ΔΖΗ: μείζων ἄρα ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ: πολλῷ ἄρα μείζων ἐστὶν ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ. καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΕΖΗ μείζονα ἔχον τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ ΕΗΖ, ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει, μείζων ἄρα καὶ πλευρὰ ἡ ΕΗ τῆς ΕΖ. ἴση δὲ ἡ ΕΗ τῇ ΒΓ: μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ. ἐὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας μείζονα ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ τὴν βάσιν τῆς βάσεως μείζονα ἕξει: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 264|>","<|""VertexLabel"" -> ""1.25"", ""Text"" -> ""If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other."", ""TextWordCount"" -> 39, ""GreekText"" -> ""ἐὰν δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ βάσιν τῆς βάσεως μείζονα ἔχῃ, καὶ τὴν γωνίαν τῆς γωνίας μείζονα ἕξει τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην."", ""GreekTextWordCount"" -> 32, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 24}}, ""Proof"" -> ""Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF; and let the base BC be greater than the base EF; I say that the angle BAC is also greater than the angle EDF. For, if not, it is either equal to it or less. Now the angle BAC is not equal to the angle EDF; for then the base BC would also have been equal to the base EF, [I. 4] but it is not; therefore the angle BAC is not equal to the angle EDF. Neither again is the angle BAC less than the angle EDF; for then the base BC would also have been less than the base EF, [I. 24] but it is not; therefore the angle BAC is not less than the angle EDF. But it was proved that it is not equal either; therefore the angle BAC is greater than the angle EDF."", ""ProofWordCount"" -> 167, ""GreekProof"" -> ""ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ, τὴν δὲ ΑΓ τῇ ΔΖ: βάσις δὲ ἡ ΒΓ βάσεως τῆς ΕΖ μείζων ἔστω: λέγω, ὅτι καὶ γωνία ἡ ὑπὸ ΒΑΓ γωνίας τῆς ὑπὸ ΕΔΖ μείζων ἐστίν: εἰ γὰρ μή, ἤτοι ἴση ἐστὶν αὐτῇ ἢ ἐλάσσων: ἴση μὲν οὖν οὐκ ἔστιν ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ: ἴση γὰρ ἂν ἦν καὶ βάσις ἡ ΒΓ βάσει τῇ ΕΖ: οὐκ ἔστι δέ. οὐκ ἄρα ἴση ἐστὶ γωνία ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ: οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ: ἐλάσσων γὰρ ἂν ἦν καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ: οὐκ ἔστι δέ: οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ. ἐδείχθη δὲ ὅτι οὐδὲ ἴση: μείζων ἄρα ἐστὶν ἡ ὑπὸ ΒΑΓ τῆς ὑπὸ ΕΔΖ. ἐὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς ἴσας ἔχῃ ἑκατέραν ἑκάτερᾳ, τὴν δὲ βάσιν τῆς βάσεως μείζονα ἔχῃ, καὶ τὴν γωνίαν τῆς γωνίας μείζονα ἕξει τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 184|>","<|""VertexLabel"" -> ""1.26"", ""Text"" -> ""If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal tothe remaining sides and the remaining angle to the remaining angle."", ""TextWordCount"" -> 54, ""GreekText"" -> ""ἐὰν δύο τρίγωνα τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις ἢ τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν, καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ."", ""GreekTextWordCount"" -> 50, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 16}}, ""Proof"" -> ""Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively, namely the angle ABC to the angle DEF, and the angleBCA to the angle EFD; and let them also have one side equal to one side, first that adjoining the equal angles, namely BC to EF; I say that they will also have the remaining sides equal to the remaining sides respectively, namely AB to DE andAC to DF, and the remaining angle to the remaining angle, namely the angle BAC to the angle EDF. For, if AB is unequal to DE, one of them is greater. Let AB be greater, and let BG be made equal to DE; and let GC be joined. Then, since BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF respectively; and the angle GBC is equal to the angle DEF; therefore the base GC is equal to the base DF, and the triangle GBC is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4]therefore the angle GCB is equal to the angle DFE. But the angle DFE is by hypothesis equal to the angle BCA;therefore the angle BCG is equal to the angle BCA, the less to the greater: which is impossible. Therefore AB is not unequal to DE, and is therefore equal to it. But BC is also equal to EF;therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and the angle ABC is equal to the angle DEF; therefore the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF. [I. 4] Again, let sides subtending equal angles be equal, as AB to DE; I say again that the remaining sides will be equal to the remaining sides, namely AC to DF and BC to EF, andfurther the remaining angle BAC is equal to the remaining angle EDF. For, if BC is unequal to EF, one of them is greater. Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined. Then, since BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles; therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DEF,and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I. 4] therefore the angle BHA is equal to the angle EFD. But the angle EFD is equal to the angle BCA; therefore, in the triangle AHC, the exterior angle BHA isequal to the interior and opposite angle BCA: which is impossible. [I. 16] Therefore BC is not unequal to EF, and is therefore equal to it. But AB is also equal to DE;therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles; therefore the base AC is equal to the base DF, the triangle ABC equal to the triangle DEF, and the remaining angle BAC equal to the remaining angleEDF. [I. 4]"", ""ProofWordCount"" -> 566, ""GreekProof"" -> ""ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ, ΒΓΑ δυσὶ ταῖς ὑπὸ ΔΕΖ, ΕΖΔ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ, τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ, τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ: ἐχέτω δὲ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην, πρότερον τὴν πρὸς ταῖς ἴσαις γωνίαις τὴν ΒΓ τῇ ΕΖ: λέγω, ὅτι καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ, καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ, τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ. εἰ γὰρ ἄνισός ἐστιν ἡ ΑΒ τῇ ΔΕ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ΑΒ, καὶ κείσθω τῇ ΔΕ ἴση ἡ ΒΗ, καὶ ἐπεζεύχθω ἡ ΗΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ μὲν ΒΗ τῇ ΔΕ, ἡ δὲ ΒΓ τῇ ΕΖ, δύο δὴ αἱ ΒΗ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΗΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἴση ἐστίν: βάσις ἄρα ἡ ΗΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΗΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἡ ὑπὸ ΗΓΒ γωνία τῇ ὑπὸ ΔΖΕ. ἀλλὰ ἡ ὑπὸ ΔΖΕ τῇ ὑπὸ ΒΓΑ ὑπόκειται ἴση: καὶ ἡ ὑπὸ ΒΓΗ ἄρα τῇ ὑπὸ ΒΓΑ ἴση ἐστίν, ἡ ἐλάσσων τῇ μείζονι: ὅπερ ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ΑΒ τῇ ΔΕ. ἴση ἄρα. ἔστι δὲ καὶ ἡ ΒΓ τῇ ΕΖ ἴση: δύο δὴ αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση: βάσις ἄρα ἡ ΑΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση ἐστίν. ἀλλὰ δὴ πάλιν ἔστωσαν αἱ ὑπὸ τὰς ἴσας γωνίας πλευραὶ ὑποτείνουσαι ἴσαι, ὡς ἡ ΑΒ τῇ ΔΕ: λέγω πάλιν, ὅτι καὶ αἱ λοιπαὶ πλευραὶ ταῖς λοιπαῖς πλευραῖς ἴσαι ἔσονται, ἡ μὲν ΑΓ τῇ ΔΖ, ἡ δὲ ΒΓ τῇ ΕΖ καὶ ἔτι ἡ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση ἐστίν. εἰ γὰρ ἄνισός ἐστιν ἡ ΒΓ τῇ ΕΖ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων, εἰ δυνατόν, ἡ ΒΓ, καὶ κείσθω τῇ ΕΖ ἴση ἡ ΒΘ, καὶ ἐπεζεύχθω ἡ ΑΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΒΘ τῇ ΕΖ ἡ δὲ ΑΒ τῇ ΔΕ, δύο δὴ αἱ ΑΒ, ΒΘ δυσὶ ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνίας ἴσας περιέχουσιν: βάσις ἄρα ἡ ΑΘ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΑΒΘ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΘΑ γωνία τῇ ὑπὸ ΕΖΔ. ἀλλὰ ἡ ὑπὸ ΕΖΔ τῇ ὑπὸ ΒΓΑ ἐστιν ἴση: τριγώνου δὴ τοῦ ΑΘΓ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΘΑ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΒΓΑ: ὅπερ ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ΒΓ τῇ ΕΖ: ἴση ἄρα. ἐστὶ δὲ καὶ ἡ ΑΒ τῇ ΔΕ ἴση. δύο δὴ αἱ ΑΒ, ΒΓ δύο ταῖς ΔΕ, ΕΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνίας ἴσας περιέχουσι: βάσις ἄρα ἡ ΑΓ βάσει τῇ ΔΖ ἴση ἐστίν, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ ἴσον καὶ λοιπὴ γωνία ἡ ὑπὸ ΒΑΓ τῇ λοιπῇ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴση. ἐὰν ἄρα δύο τρίγωνα τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην ἤτοι τὴν πρὸς ταῖς ἴσαις γωνίαις, ἢ τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν, καὶ τὰς λοιπὰς πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 594|>","<|""VertexLabel"" -> ""1.27"", ""Text"" -> ""If a straight line falling on two straight lines make the alternate angles equal to one another, the straight lines will be parallel to one another."", ""TextWordCount"" -> 26, ""GreekText"" -> ""ἐὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιῇ, παράλληλοι ἔσονται ἀλλήλαις αἱ εὐθεῖαι."", ""GreekTextWordCount"" -> 17, ""References"" -> {{""Book"" -> 1, ""Definition"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 16}}, ""Proof"" -> ""For let the straight line EF falling on the two straightlines AB, CD make the alternate angles AEF, EFD equal to one another; I say that AB is parallel to CD. For, if not, AB, CD when produced will meet either in the directionof B, D or towards A, C. Let them be produced and meet, in the direction of B, D, at G. Then, in the triangle GEF, the exterior angle AEF is equal to the interior and oppositeangle EFG: which is impossible. [I. 16] Therefore AB, CD when produced will not meet in the direction of B, D. Similarly it can be proved that neither will they meettowards A, C. But straight lines which do not meet in either direction are parallel; [Def. 23] therefore AB is parallel to CD."", ""ProofWordCount"" -> 132, ""GreekProof"" -> ""εἰς γὰρ δύο εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΑΕΖ, ΕΖΔ ἴσας ἀλλήλαις ποιείτω: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. εἰ γὰρ μή, ἐκβαλλόμεναι αἱ ΑΒ, ΓΔ συμπεσοῦνται ἤτοι ἐπὶ τὰ Β, Δ μέρη ἢ ἐπὶ τὰ Α, Γ. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν ἐπὶ τὰ Β, Δ μέρη κατὰ τὸ Η. τριγώνου δὴ τοῦ ΗΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΕΖΗ: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα αἱ ΑΒ, ΓΔ ἐκβαλλόμεναι συμπεσοῦνται ἐπὶ τὰ Β, Δ μέρη. ὁμοίως δὴ δειχθήσεται, ὅτι οὐδὲ ἐπὶ τὰ Α, Γ: αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη συμπίπτουσαι παράλληλοί εἰσιν: παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. ἐὰν ἄρα εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιῇ, παράλληλοι ἔσονται αἱ εὐθεῖαι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 138|>","<|""VertexLabel"" -> ""1.28"", ""Text"" -> ""If a straight line falling on two straight lines make the exterior angle equal to the interior and opposite angle on the same side, or the interior angles on the same side equal to two right angles, the straight lines will be parallel to one another."", ""TextWordCount"" -> 46, ""GreekText"" -> ""ἐὰν εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὴν ἐκτὸς γωνίαν τῇ ἐντὸς καὶ ἀπεναντίον καὶ ἐπὶ τὰ αὐτὰ μέρη ἴσην ποιῇ ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας, παράλληλοι ἔσονται ἀλλήλαις αἱ εὐθεῖαι."", ""GreekTextWordCount"" -> 36, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 13}, {""Book"" -> 1, ""Theorem"" -> 15}, {""Book"" -> 1, ""Theorem"" -> 27}}, ""Proof"" -> ""For let the straight line EF falling on the two straight lines AB, CD make the exterior angle EGB equal to the interior and opposite angle GHD, or the interior angles on the same side, namely BGH, GHD, equal to two right angles; I say that AB is parallel to CD. For, since the angle EGB is equal to the angle GHD, while the angle EGB is equal to the angle AGH, [I. 15] the angle AGH is also equal to the angle GHD; and they are alternate; therefore AB is parallel to CD. [I. 27] Again, since the angles BGH, GHD are equal to two right angles, and the angles AGH, BGH are also equal to two right angles, [I. 13] the angles AGH, BGH are equal to the angles BGH, GHD. Let the angle BGH be subtracted from each; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate; therefore AB is parallel to CD. [I. 27]"", ""ProofWordCount"" -> 165, ""GreekProof"" -> ""εἰς γὰρ δύο εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπίπτουσα ἡ ΕΖ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον γωνίᾳ τῇ ὑπὸ ΗΘΔ ἴσην ποιείτω ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη τὰς ὑπὸ ΒΗΘ, ΗΘΔ δυσὶν ὀρθαῖς ἴσας: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΗΘΔ, ἀλλὰ ἡ ὑπὸ ΕΗΒ τῇ ὑπὸ ΑΗΘ ἐστιν ἴση, καὶ ἡ ὑπὸ ΑΗΘ ἄρα τῇ ὑπὸ ΗΘΔ ἐστιν ἴση: καί εἰσιν ἐναλλάξ: παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. πάλιν, ἐπεὶ αἱ ὑπὸ ΒΗΘ, ΗΘΔ δύο ὀρθαῖς ἴσαι εἰσίν, εἰσὶ δὲ καὶ αἱ ὑπὸ ΑΗΘ, ΒΗΘ δυσὶν ὀρθαῖς ἴσαι, αἱ ἄρα ὑπὸ ΑΗΘ, ΒΗΘ ταῖς ὑπὸ ΒΗΘ, ΗΘΔ ἴσαι εἰσίν: κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΒΗΘ: λοιπὴ ἄρα ἡ ὑπὸ ΑΗΘ λοιπῇ τῇ ὑπὸ ΗΘΔ ἐστιν ἴση: καί εἰσιν ἐναλλάξ: παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. ἐὰν ἄρα εἰς δύο εὐθείας εὐθεῖα ἐμπίπτουσα τὴν ἐκτὸς γωνίαν τῇ ἐντὸς καὶ ἀπεναντίον καὶ ἐπὶ τὰ αὐτὰ μέρη ἴσην ποιῇ ἢ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας, παράλληλοι ἔσονται αἱ εὐθεῖαι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 185|>","<|""VertexLabel"" -> ""1.29"", ""Text"" -> ""A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle, and the interior angles on the same side equal to two right angles."", ""TextWordCount"" -> 39, ""GreekText"" -> ""ἡ εἰς τὰς παραλλήλους εὐθείας εὐθεῖα ἐμπίπτουσα τάς τε ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιεῖ καὶ τὴν ἐκτὸς τῇ ἐντὸς καὶ ἀπεναντίον ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας."", ""GreekTextWordCount"" -> 33, ""References"" -> {{""Common Notion"" -> 1}, {""Common Notion"" -> 2}, {""Postulate"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 13}, {""Book"" -> 1, ""Theorem"" -> 15}}, ""Proof"" -> ""For let the straight line EF fall on the parallel straight lines AB, CD; I say that it makes the alternate angles AGH, GHD equal, the exterior angle EGB equal to the interior and opposite angle GHD, and the interior angles on the sameside, namely BGH, GHD, equal to two right angles. For, if the angle AGH is unequal to the angle GHD, one of them is greater. Let the angle AGH be greater. Let the angle BGH be added to each; therefore the angles AGH, BGH are greater than the angles BGH, GHD. But the angles AGH, BGH are equal to two right angles; [I. 13] therefore the angles BGH, GHD are less than two right angles. But straight lines produced indefinitely from angles less than two right angles meet; [Post. 5]therefore AB, CD, if produced indefinitely, will meet; but they do not meet, because they are by hypothesis parallel. Therefore the angle AGH is not unequal to the angle GHD, and is therefore equal to it. Again, the angle AGH is equal to the angle EGB; [I. 15] therefore the angle EGB is also equal to the angle GHD. [C. N. 1] Let the angle BGH be added to each; therefore the angles EGB, BGH are equal to the angles BGH, GHD. [C. N. 2] But the angles EGB, BGH are equal to two right angles; [I. 13] therefore the angles BGH, GHD are also equal to two right angles."", ""ProofWordCount"" -> 243, ""GreekProof"" -> ""εἰς γὰρ παραλλήλους εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπιπτέτω ἡ ΕΖ: λέγω, ὅτι τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΑΗΘ, ΗΘΔ ἴσας ποιεῖ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη τὰς ὑπὸ ΒΗΘ, ΗΘΔ δυσὶν ὀρθαῖς ἴσας. εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΑΗΘ: κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ: αἱ ἄρα ὑπὸ ΑΗΘ, ΒΗΘ τῶν ὑπὸ ΒΗΘ, ΗΘΔ μείζονές εἰσιν. ἀλλὰ αἱ ὑπὸ ΑΗΘ, ΒΗΘ δυσὶν ὀρθαῖς ἴσαι εἰσίν. καὶ αἱ ἄρα ὑπὸ ΒΗΘ, ΗΘΔ δύο ὀρθῶν ἐλάσσονές εἰσιν. αἱ δὲ ἀπ᾽ ἐλασσόνων ἢ δύο ὀρθῶν ἐκβαλλόμεναι εἰς ἄπειρον συμπίπτουσιν: αἱ ἄρα ΑΒ, ΓΔ ἐκβαλλόμεναι εἰς ἄπειρον συμπεσοῦνται: οὐ συμπίπτουσι δὲ διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι: οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ: ἴση ἄρα. ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ ἐστιν ἴση: καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ ἐστιν ἴση. κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ: αἱ ἄρα ὑπὸ ΕΗΒ, ΒΗΘ ταῖς ὑπὸ ΒΗΘ, ΗΘΔ ἴσαι εἰσίν. ἀλλὰ αἱ ὑπὸ ΕΗΒ, ΒΗΘ δύο ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΒΗΘ, ΗΘΔ ἄρα δύο ὀρθαῖς ἴσαι εἰσίν. ἡ ἄρα εἰς τὰς παραλλήλους εὐθείας εὐθεῖα ἐμπίπτουσα τάς τε ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιεῖ καὶ τὴν ἐκτὸς τῇ ἐντὸς καὶ ἀπεναντίον ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας: ὅπερ ἔδει δεῖξαι. αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλληλοι. ἔστω ἑκατέρα τῶν ΑΒ, ΓΔ τῇ ΕΖ παράλληλος: λέγω, ὅτι καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος. ἐμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ. καὶ ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΑΒ, ΕΖ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ. πάλιν, ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΕΖ, ΓΔ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση. καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ ἐστιν ἴση: καί εἰσιν ἐναλλάξ. παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. Αἱ ἄρα τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλληλοι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 353|>","<|""VertexLabel"" -> ""1.30"", ""Text"" -> ""Straight lines parallel to the same straight line are also parallel to one another."", ""TextWordCount"" -> 14, ""GreekText"" -> ""αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλληλοι."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Common Notion"" -> 1}, {""Book"" -> 1, ""Theorem"" -> 29}}, ""Proof"" -> ""Let each of the straight lines AB, CD be parallel to EF; I say that AB is also parallel to CD. For let the straight line GK fall upon them; Then, since the straight line GK has fallen on the parallel straight lines AB, EF,the angle AGK is equal to the angle GHF. [I. 29] Again, since the straight line GK has fallen on the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD. [I. 29] But the angle AGK was also proved equal to the angle GHF; therefore the angle AGK is also equal to the angle GKD; [C. N. 1]and they are alternate. Therefore AB is parallel to CD."", ""ProofWordCount"" -> 118, ""GreekProof"" -> ""῎Εστω ἑκατέρα τῶν ΑΒ, ΓΔ τῇ ΕΖ παράλληλος: λέγω, ὅτι καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος. Ἐμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ. Καὶ ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΑΒ, ΕΖ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ. πάλιν, ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΕΖ, ΓΔ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση. καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ ἐστιν ἴση: καί εἰσιν ἐναλλάξ. παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ. [Αἱ ἄρα τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλληλοι:] ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 108|>","<|""VertexLabel"" -> ""1.31"", ""Text"" -> ""Through a given point to draw a straight line parallel to a given straight line."", ""TextWordCount"" -> 15, ""GreekText"" -> ""διὰ τοῦ δοθέντος σημείου τῇ δοθείσῃ εὐθείᾳ παράλληλον εὐθεῖαν γραμμὴν ἀγαγεῖν."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 27}}, ""Proof"" -> ""Let A be the given point, and BC the given straight line; thus it is required to draw through the point A a straight line parallel to the straight line BC. Let a point D be taken at random on BC, and let AD be joined; on the straight line DA, and at the point A on it, let the angle DAE be constructed equal to the angle ADC [I. 23]; and let the straight line AF be produced in a straight line with EA. Then, since the straight line AD falling on the two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another, therefore EAF is parallel to BC. [I. 27] Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC."", ""ProofWordCount"" -> 138, ""GreekProof"" -> ""ἔστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ: δεῖ δὴ διὰ τοῦ Α σημείου τῇ ΒΓ εὐθείᾳ παράλληλον εὐθεῖαν γραμμὴν ἀγαγεῖν. εἰλήφθω ἐπὶ τῆς ΒΓ τυχὸν σημεῖον τὸ Δ, καὶ ἐπεζεύχθω ἡ ΑΔ: καὶ συνεστάτω πρὸς τῇ ΔΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ἡ ὑπὸ ΔΑΕ: καὶ ἐκβεβλήσθω ἐπ᾽ εὐθείας τῇ ΕΑ εὐθεῖα ἡ ΑΖ. καὶ ἐπεὶ εἰς δύο εὐθείας τὰς ΒΓ, ΕΖ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΕΑΔ, ΑΔΓ ἴσας ἀλλήλαις πεποίηκεν, παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ. διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 119|>","<|""VertexLabel"" -> ""1.32"", ""Text"" -> ""In any triangle, if one of the sides be produced, the exterior angle is equal to the two interior and opposite angles, and the three interior angles of the triangle are equal to two right angles."", ""TextWordCount"" -> 36, ""GreekText"" -> ""παντὸς τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ἴση ἐστίν, καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν."", ""GreekTextWordCount"" -> 27, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 13}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 1, ""Theorem"" -> 31}}, ""Proof"" -> ""Let ABC be a triangle, and let one side of it BC be produced to D; I say that the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle ABC, BCA, CAB are equal to two right angles. For let CE be drawn through the point C parallel to the straight line AB. [I. 31] Then, since AB is parallel to CE, and AC has fallen upon them, the alternate angles BAC, ACE are equal to one another. [I. 29] Again, since AB is parallel to CE, and the straight line BD has fallen upon them, the exterior angle ECD is equal to the interior and opposite angle ABC. [I. 29] But the angle ACE was also proved equal to the angle BAC; therefore the whole angle ACD is equal to the two interior and opposite angles BAC, ABC. Let the angle ACB be added to each; therefore the angles ACD, ACB are equal to the three angles ABC, BCA, CAB. But the angles ACD, ACB are equal to two right angles; [I. 13] therefore the angles ABC, BCA, CAB are also equal to two right angles."", ""ProofWordCount"" -> 200, ""GreekProof"" -> ""ἔστω τρίγωνον τὸ ΑΒΓ, καὶ προσεκβεβλήσθω αὐτοῦ μία πλευρὰ ἡ ΒΓ ἐπὶ τὸ Δ: λέγω, ὅτι ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΓΔ ἴση ἐστὶ δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ταῖς ὑπὸ ΓΑΒ, ΑΒΓ, καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι αἱ ὑπὸ ΑΒΓ, ΒΓΑ, ΓΑΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἤχθω γὰρ διὰ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ παράλληλος ἡ ΓΕ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ, ΑΓΕ ἴσαι ἀλλήλαις εἰσίν. πάλιν, ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ, καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΒΔ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ ἴση: ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ἴση ἐστὶ δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ταῖς ὑπὸ ΒΑΓ, ΑΒΓ. κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ: αἱ ἄρα ὑπὸ ΑΓΔ, ΑΓΒ τρισὶ ταῖς ὑπὸ ΑΒΓ, ΒΓΑ, ΓΑΒ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΑΓΔ, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΑΓΒ, ΓΒΑ, ΓΑΒ ἄρα δυσίν ὀρθαῖς ἴσαι εἰσίν. παντὸς ἄρα τριγώνου μιᾶς τῶν πλευρῶν προσεκβληθείσης ἡ ἐκτὸς γωνία δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ἴση ἐστίν, καὶ αἱ ἐντὸς τοῦ τριγώνου τρεῖς γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 212|>","<|""VertexLabel"" -> ""1.33"", ""Text"" -> ""The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel."", ""TextWordCount"" -> 25, ""GreekText"" -> ""αἱ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι εὐθεῖαι καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν."", ""GreekTextWordCount"" -> 19, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 27}, {""Book"" -> 1, ""Theorem"" -> 29}}, ""Proof"" -> ""Let AB, CD be equal and parallel, and let the straightlines AC, BD join them (at the extremities which are) in the same directions (respectively); I say that AC, BD are also equal and parallel. Let BC be joined. Then, since AB is parallel to CD,and BC has fallen upon them, the alternate angles ABC, BCD are equal to one another. [I. 29] And, since AB is equal to CD, and BC is common, the two sides AB, BC are equal to the two sides DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the griangle ABC is equal to the triangle DCB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4]therefore the angle ACB is equal to the angle CBD. And, since the straight line BC falling on the two straight lines AC, BD has made the alternate angles equal to one another,AC is parallel to BD. [I. 27] And it was also proved equal to it."", ""ProofWordCount"" -> 187, ""GreekProof"" -> ""ἔστωσαν ἴσαι τε καὶ παράλληλοι αἱ ΑΒ, ΓΔ, καὶ ἐπιζευγνύτωσαν αὐτὰς ἐπὶ τὰ αὐτὰ μέρη εὐθεῖαι αἱ ΑΓ, ΒΔ: λέγω, ὅτι καὶ αἱ ΑΓ, ΒΔ ἴσαι τε καὶ παράλληλοί εἰσιν. ἐπεζεύχθω ἡ ΒΓ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ, ΒΓΔ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ κοινὴ δὲ ἡ ΒΓ, δύο δὴ αἱ ΑΒ, ΒΓ δύο ταῖς ΒΓ, ΓΔ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση: βάσις ἄρα ἡ ΑΓ βάσει τῇ ΒΔ ἐστιν ἴση, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΒΓΔ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται ἑκατέρα ἑκατέρᾳ, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΓΒΔ. καὶ ἐπεὶ εἰς δύο εὐθείας τὰς ΑΓ, ΒΔ εὐθεῖα ἐμπίπτουσα ἡ ΒΓ τὰς ἐναλλὰξ γωνίας ἴσας ἀλλήλαις πεποίηκεν, παράλληλος ἄρα ἐστὶν ἡ ΑΓ τῇ ΒΔ. ἐδείχθη δὲ αὐτῇ καὶ ἴση. αἱ ἄρα τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι εὐθεῖαι καὶ αὐταὶ ἴσαι τε καὶ παράλληλοί εἰσιν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 188|>","<|""VertexLabel"" -> ""1.34"", ""Text"" -> ""In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas."", ""TextWordCount"" -> 19, ""GreekText"" -> ""τῶν παραλληλογράμμων χωρίων αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει."", ""GreekTextWordCount"" -> 18, ""References"" -> {{""Common Notion"" -> 2}, {""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 26}, {""Book"" -> 1, ""Theorem"" -> 29}}, ""Proof"" -> ""Let ACDB be a parallelogrammic area, and BC its diameter;I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it. For, since AB is parallel to CD, and the straight line BC has fallenupon them, the alternate angles ABC, BCD are equal to one another. [I. 29] Again, since AC is parallel to BD, and BC has fallen upon them,the alternate angles ACB, CBD are equal to one another. [I. 29] Therefore ABC, DCB are two triangles having the two angles ABC, BCA equal to the two angles DCB, CBD respectively, and one side equal to one side, namely thatadjoining the equal angles and common to both of them, BC; therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [I. 26]therefore the side AB is equal to CD, and AC to BD, and further the angle BAC is equal to the angle CDB. And, since the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD. [C. N. 2] And the angle BAC was also proved equal to the angle CDB. Therefore in parallelogrammic areas the opposite sides and angles are equal to one another. I say, next, that the diameter also bisects the areas. For, since AB is equal to CD,and BC is common, the two sides AB, BC are equal to the two sides DC, CB respectively; and the angle ABC is equal to the angle BCD; therefore the base AC is also equal to DB, and the triangle ABC is equal to the triangle DCB. [I. 4] Therefore the diameter BC bisects the parallelogram ACDB."", ""ProofWordCount"" -> 304, ""GreekProof"" -> ""ἔστω παραλληλόγραμμον χωρίον τὸ ΑΓΔΒ, διάμετρος δὲ αὐτοῦ ἡ ΒΓ: λέγω, ὅτι τοῦ ΑΓΔΒ παραλληλογράμμου αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν, καὶ ἡ ΒΓ διάμετρος αὐτὸ δίχα τέμνει. ἐπεὶ γὰρ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΒΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΒΓ, ΒΓΔ ἴσαι ἀλλήλαις εἰσίν. πάλιν, ἐπεὶ παράλληλός ἐστιν ἡ ΑΓ τῇ ΒΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΑΓΒ, ΓΒΔ ἴσαι ἀλλήλαις εἰσίν. δύο δὴ τρίγωνά ἐστι τὰ ΑΒΓ, ΒΓΔ τὰς δύο γωνίας τὰς ὑπὸ ΑΒΓ, ΒΓΑ δυσὶ ταῖς ὑπὸ ΒΓΔ, ΓΒΔ ἴσας ἔχοντα ἑκατέραν ἑκατέρᾳ καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν πρὸς ταῖς ἴσαις γωνίαις κοινὴν αὐτῶν τὴν ΒΓ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς ἴσας ἕξει ἑκατέραν ἑκατέρᾳ καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ: ἴση ἄρα ἡ μὲν ΑΒ πλευρὰ τῇ ΓΔ, ἡ δὲ ΑΓ τῇ ΒΔ, καὶ ἔτι ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΓΔΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΒΓΔ, ἡ δὲ ὑπὸ ΓΒΔ τῇ ὑπὸ ΑΓΒ, ὅλη ἄρα ἡ ὑπὸ ΑΒΔ ὅλῃ τῇ ὑπὸ ΑΓΔ ἐστιν ἴση. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΒ ἴση. τῶν ἄρα παραλληλογράμμων χωρίων αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν. λέγω δή, ὅτι καὶ ἡ διάμετρος αὐτὰ δίχα τέμνει. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΒ τῇ ΓΔ, κοινὴ δὲ ἡ ΒΓ, δύο δὴ αἱ ΑΒ, ΒΓ δυσὶ ταῖς ΓΔ, ΒΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΒΓ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση. καὶ βάσις ἄρα ἡ ΑΓ τῇ ΔΒ ἴση. καὶ τὸ ΑΒΓ ἄρα τρίγωνον τῷ ΒΓΔ τριγώνῳ ἴσον ἐστίν. ἡ ἄρα ΒΓ διάμετρος δίχα τέμνει τὸ ΑΒΓΔ παραλληλόγραμμον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 287|>","<|""VertexLabel"" -> ""1.35"", ""Text"" -> ""Parallelograms which are on the same base and in the same parallels are equal to one another."", ""TextWordCount"" -> 17, ""GreekText"" -> ""τὰ παραλληλόγραμμα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν."", ""GreekTextWordCount"" -> 16, ""References"" -> {{""Common Notion"" -> 1}, {""Common Notion"" -> 2}, {""Common Notion"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 1, ""Theorem"" -> 34}}, ""Proof"" -> ""Let ABCD, EBCF be parallelograms on the same base BC and in the same parallels AF, BC;I say that ABCD is equal to the parallelogram EBCF. For, since ABCD is a parallelogram, AD is equal to BC. [I. 34] For the same reason also EF is equal to BC, so that AD is also equal to EF; [C. N. 1]and DE is common; therefore the whole AE is equal to the whole DF. [C. N. 2] But AB is also equal to DC; [I. 34] therefore the two sides EA, AB are equal to the two sidesFD, DC respectively, and the angle FDC is equal to the angle EAB, the exterior to the interior; [I. 29]therefore the base EB is equal to the base FC, and the triangle EAB will be equal to the triangle FDC. [I. 4] Let DGE be subtracted from each; therefore the trapezium ABGD which remains is equal to the trapezium EGCF which remains. [C. N. 3] Let the triangle GBC be added to each; therefore the whole parallelogram ABCD is equal to the whole parallelogram EBCF. [C. N. 2]"", ""ProofWordCount"" -> 187, ""GreekProof"" -> ""ἔστω παραλληλόγραμμα τὰ ΑΒΓΔ, ΕΒΓΖ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΖ, ΒΓ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔ τῷ ΕΒΓΖ παραλληλογράμμῳ. ἐπεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ, ἴση ἐστὶν ἡ ΑΔ τῇ ΒΓ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση: ὥστε καὶ ἡ ΑΔ τῇ ΕΖ ἐστιν ἴση: καὶ κοινὴ ἡ ΔΕ: ὅλη ἄρα ἡ ΑΕ ὅλῃ τῇ ΔΖ ἐστιν ἴση. ἔστι δὲ καὶ ἡ ΑΒ τῇ ΔΓ ἴση: δύο δὴ αἱ ΕΑ, ΑΒ δύο ταῖς ΖΔ, ΔΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΖΔΓ γωνίᾳ τῇ ὑπὸ ΕΑΒ ἐστιν ἴση ἡ ἐκτὸς τῇ ἐντός: βάσις ἄρα ἡ ΕΒ βάσει τῇ ΖΓ ἴση ἐστίν, καὶ τὸ ΕΑΒ τρίγωνον τῷ ΔΖΓ τριγώνῳ ἴσον ἔσται: κοινὸν ἀφῃρήσθω τὸ ΔΗΕ: λοιπὸν ἄρα τὸ ΑΒΗΔ τραπέζιον λοιπῷ τῷ ΕΗΓΖ τραπεζίῳ ἐστὶν ἴσον: κοινὸν προσκείσθω τὸ ΗΒΓ τρίγωνον: ὅλον ἄρα τὸ ΑΒΓΔ παραλληλόγραμμον ὅλῳ τῷ ΕΒΓΖ παραλληλογράμμῳ ἴσον ἐστίν. τὰ ἄρα παραλληλόγραμμα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 177|>","<|""VertexLabel"" -> ""1.36"", ""Text"" -> ""Parallelograms which are on equal bases and in the same parallels are equal to one another."", ""TextWordCount"" -> 16, ""GreekText"" -> ""τὰ παραλληλόγραμμα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν."", ""GreekTextWordCount"" -> 15, ""References"" -> {{""Common Notion"" -> 1}, {""Book"" -> 1, ""Theorem"" -> 33}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 1, ""Theorem"" -> 35}}, ""Proof"" -> ""Let ABCD, EFGH be parallelograms which are on equal bases BC, FG and in the same parallels AH, BG; I say that the parallelogram ABCD is equal to EFGH. For let BE, CH be joined. Then, since BC is equal to FG while FG is equal to EH, BC is also equal to EH. [C. N. 1] But they are also parallel. And EB, HC join them; but straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are equal and parallel. [I. 33] Therefore EBCH is a parallelogram. [I. 34] And it is equal to ABCD; for it has the same base BC with it, and is in the same parallels BC, AH with it. [I. 35] For the same reason also EFGH is equal to the same EBCH; [I. 35] so that the parallelogram ABCD is also equal to EFGH. [C. N. 1]"", ""ProofWordCount"" -> 153, ""GreekProof"" -> ""ἔστω παραλληλόγραμμα τὰ ΑΒΓΔ, ΕΖΗΘ ἐπὶ ἴσων βάσεων ὄντα τῶν ΒΓ, ΖΗ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΘ, ΒΗ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΖΗΘ. ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ, ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση, καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση. εἰσὶ δὲ καὶ παράλληλοι. καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ, ΘΓ: αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι ἴσαι τε καὶ παράλληλοί εἰσι: καὶ αἱ ΕΒ, ΘΓ ἄρα ἴσαι τέ εἰσι καὶ παράλληλοι. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ. καί ἐστιν ἴσον τῷ ΑΒΓΔ: βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει τὴν ΒΓ, καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστὶν αὐτῷ ταῖς ΒΓ, ΑΘ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ἐστιν ἴσον: ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΖΗΘ ἐστιν ἴσον. τὰ ἄρα παραλληλόγραμμα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 164|>","<|""VertexLabel"" -> ""1.37"", ""Text"" -> ""Triangles which are on the same base and in the same parallels are equal to one another."", ""TextWordCount"" -> 17, ""GreekText"" -> ""τὰ τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν."", ""GreekTextWordCount"" -> 16, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 1, ""Theorem"" -> 35}}, ""Proof"" -> ""Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC;I say that the triangle ABC is equal to the triangle DBC. Let AD be produced in both directions to E, F; through B let BE be drawn parallel to CA, [I. 31]and through C let CF be drawn parallel to BD. [I. 31] Then each of the figures EBCA, DBCF is a parallelogram; and they are equal, for they are on the same base BC and in the same parallels BC, EF. [I. 35] Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I. 34] And the triangle DBC is half of the parallelogram DBCF;for the diameter DC bisects it. [I. 34] [But the halves of equal things are equal to one another.] Therefore the triangle ABC is equal to the triangle DBC."", ""ProofWordCount"" -> 150, ""GreekProof"" -> ""ἔστω τρίγωνα τὰ ΑΒΓ, ΔΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΔ, ΒΓ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ. Ἐκβεβλήσθω ἡ ΑΔ ἐφ᾽ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε, Ζ, καὶ διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΕ, διὰ δὲ τοῦ Γ τῇ ΒΔ παράλληλος ἤχθω ἡ ΓΖ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΕΒΓΑ, ΔΒΓΖ: καί εἰσιν ἴσα: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΕΖ: καί ἐστι τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ἥμισυ τὸ ΑΒΓ τρίγωνον: ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει: τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ἥμισυ τὸ ΔΒΓ τρίγωνον: ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει. τὰ δὲ τῶν ἴσων ἡμίση ἴσα ἀλλήλοις ἐστίν. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ. τὰ ἄρα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 158|>","<|""VertexLabel"" -> ""1.38"", ""Text"" -> ""Triangles which are on equal bases and in the same parallels are equal to one another."", ""TextWordCount"" -> 16, ""GreekText"" -> ""τὰ τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν."", ""GreekTextWordCount"" -> 15, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 1, ""Theorem"" -> 36}}, ""Proof"" -> ""Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD; I say that the triangle ABC is equal to the triangle DEF. For let AD be produced in both directions to G, H; through B let BG be drawn parallel to CA, [I. 31] and through F let FH be drawn parallel to DE. Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH; for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36] Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34] And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34] [But the halves of equal things are equal to one another.] Therefore the triangle ABC is equal to the triangle DEF."", ""ProofWordCount"" -> 151, ""GreekProof"" -> ""ἔστω τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἐπὶ ἴσων βάσεων τῶν ΒΓ, ΕΖ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΑΔ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. Ἐκβεβλήσθω γὰρ ἡ ΑΔ ἐφ᾽ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η, θ, καὶ διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΗ, διὰ δὲ τοῦ Ζ τῇ ΔΕ παράλληλος ἤχθω ἡ ΖΘ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΗΒΓΑ, ΔΕΖΘ: καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, ΕΖ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΗΘ: καί ἐστι τοῦ μὲν ΗΒΓΑ παραλληλογράμμου ἥμισυ τὸ ΑΒΓ τρίγωνον. ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει: τοῦ δὲ ΔΕΖΘ παραλληλογράμμου ἥμισυ τὸ ΖΕΔ τρίγωνον: ἡ γὰρ ΔΖ διάμετρος αὐτὸ δίχα τέμνει: τὰ δὲ τῶν ἴσων ἡμίση ἴσα ἀλλήλοις ἐστίν. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. τὰ ἄρα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 161|>","<|""VertexLabel"" -> ""1.39"", ""Text"" -> ""Equal triangles which are on the same base and on the same side are also in the same parallels."", ""TextWordCount"" -> 19, ""GreekText"" -> ""τὰ ἴσα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν."", ""GreekTextWordCount"" -> 20, ""References"" -> {{""Common Notion"" -> 1}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 37}}, ""Proof"" -> ""Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it;[I say that they are also in the same parallels.] And [For]let AD be joined; I say that AD is parallel to BC. For, if not, let AE be drawn through the point A parallel to the straight lineBC, [I. 31] and let EC be joined. Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the sameparallels. [I. 37] But ABC is equal to DBC; therefore DBC is also equal to EBC, [C. N. 1]the greater to the less: which is impossible. Therefore AE is not parallel to BC. Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BC."", ""ProofWordCount"" -> 143, ""GreekProof"" -> ""Ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΔΒΓ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη τῆς ΒΓ: λέγω, ὅτι καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. Ἐπεζεύχθω γὰρ ἡ ΑΔ: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΓ. Εἰ γὰρ μή, ἤχθω διὰ τοῦ Α σημείου τῇ ΒΓ εὐθείᾳ παράλληλος ἡ ΑΕ, καὶ ἐπεζεύχθω ἡ ΕΓ. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΕΒΓ τριγώνῳ: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις. ἀλλὰ τὸ ΑΒΓ τῷ ΔΒΓ ἐστιν ἴσον: καὶ τὸ ΔΒΓ ἄρα τῷ ΕΒΓ ἴσον ἐστὶ τὸ μεῖζον τῷ ἐλάσσονι: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα παράλληλός ἐστιν ἡ ΑΕ τῇ ΒΓ. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλη τις πλὴν τῆς ΑΔ: ἡ ΑΔ ἄρα τῇ ΒΓ ἐστι παράλληλος. Τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 152|>","<|""VertexLabel"" -> ""1.40"", ""Text"" -> ""Equal triangles which are on equal bases and on the same side are also in the same parallels."", ""TextWordCount"" -> 18, ""GreekText"" -> ""τὰ ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν."", ""GreekTextWordCount"" -> 19, ""References"" -> {{""Common Notion"" -> 1}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 38}}, ""Proof"" -> ""Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side. I say that they are also in the same parallels. For let AD be joined; I say that AD is parallel to BE. For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined. Therefore the triangle ABC is equal to the triangle FCE; for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38] But the triangle ABC is equal to the triangle DCE; therefore the triangle DCE is also equal to the triangle FCE, [C. N. 1]the greater to the less: which is impossible. Therefore AF is not parallel to BE. Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE."", ""ProofWordCount"" -> 143, ""GreekProof"" -> ""ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΓΔΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ, ΓΕ καὶ ἐπὶ τὰ αὐτὰ μέρη. λέγω, ὅτι καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. ἐπεζεύχθω γὰρ ἡ ΑΔ: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ. εἰ γὰρ μή, ἤχθω διὰ τοῦ Α τῇ ΒΕ παράλληλος ἡ ΑΖ, καὶ ἐπεζεύχθω ἡ ΖΕ. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΖΓΕ τριγώνῳ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, ΓΕ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΕ, ΑΖ. ἀλλὰ τὸ ΑΒΓ τρίγωνον ἴσον ἐστὶ τῷ ΔΓΕ τριγώνῳ: καὶ τὸ ΔΓΕ ἄρα τρίγωνον ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ τὸ μεῖζον τῷ ἐλάσσονι: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλη τις πλὴν τῆς ΑΔ: ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος. τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 153|>","<|""VertexLabel"" -> ""1.41"", ""Text"" -> ""If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle."", ""TextWordCount"" -> 23, ""GreekText"" -> ""ἐὰν παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστι τὸ παραλληλόγραμμον τοῦ τριγώνου."", ""GreekTextWordCount"" -> 20, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 1, ""Theorem"" -> 37}}, ""Proof"" -> ""For let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC, AE; I say that the parallelogram ABCD is double of the triangle BEC. For let AC be joined. Then the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels BC, AE. [I. 37] But the parallelogram ABCD is double of the triangle ABC; for the diameter AC bisects it; [I. 34]so that the parallelogram ABCD is also double of the triangle EBC."", ""ProofWordCount"" -> 100, ""GreekProof"" -> ""παραλληλόγραμμον γὰρ τὸ ΑΒΓΔ τριγώνῳ τῷ ΕΒΓ βάσιν τε ἐχέτω τὴν αὐτὴν τὴν ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἔστω ταῖς ΒΓ, ΑΕ: λέγω, ὅτι διπλάσιόν ἐστι τὸ ΑΒΓΔ παραλληλόγραμμον τοῦ ΒΕΓ τριγώνου. ἐπεζεύχθω γὰρ ἡ ΑΓ. ἴσον δή ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΕΒΓ τριγώνῳ: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς ἐστιν αὐτῷ τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΑΕ. ἀλλὰ τὸ ΑΒΓΔ παραλληλόγραμμον διπλάσιόν ἐστι τοῦ ΑΒΓ τριγώνου: ἡ γὰρ ΑΓ διάμετρος αὐτὸ δίχα τέμνει: ὥστε τὸ ΑΒΓΔ παραλληλόγραμμον καὶ τοῦ ΕΒΓ τριγώνου ἐστὶ διπλάσιον. ἐὰν ἄρα παραλληλόγραμμον τριγώνῳ βάσιν τε ἔχῃ τὴν αὐτὴν καὶ ἐν ταῖς αὐταῖς παραλλήλοις ᾖ, διπλάσιόν ἐστι τὸ παραλληλόγραμμον τοῦ τριγώνου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 114|>","<|""VertexLabel"" -> ""1.42"", ""Text"" -> ""To construct, in a given rectilineal angle, a parallelogram equal to a given triangle."", ""TextWordCount"" -> 14, ""GreekText"" -> ""τῷ δοθέντι τριγώνῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 38}, {""Book"" -> 1, ""Theorem"" -> 41}}, ""Proof"" -> ""Let ABC be the given triangle, and D the given rectilineal angle; thus it is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC. Let BC be bisected at E, and let AE be joined; on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I. 23] through A let AG be drawn parallel to EC, and [I. 31] through C let CG be drawn parallel to EF. Then FECG is a parallelogram. And, since BE is equal to EC, the triangle ABE is also equal to the triangle AEC, for they are on equal bases BE, EC and in the same parallels BC, AG; [I. 38]therefore the triangle ABC is double of the triangle AEC. But the parallelogram FECG is also double of the triangle AEC, for it has the same base with it and is in the same parallels with it; [I. 41] therefore the parallelogram FECG is equal to the triangle ABC. And it has the angle CEF equal to the given angle D. Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D."", ""ProofWordCount"" -> 209, ""GreekProof"" -> ""ἔστω τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ: δεῖ δὴ τῷ ΑΒΓ τριγώνῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ. τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε, καὶ ἐπεζεύχθω ἡ ΑΕ, καὶ συνεστάτω πρὸς τῇ ΕΓ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε τῇ Δ γωνίᾳ ἴση ἡ ὑπὸ ΓΕΖ, καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ἤχθω ἡ ΑΗ, διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος ἤχθω ἡ ΓΗ: παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ, ἴσον ἐστὶ καὶ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΕ, ΕΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΑΗ: διπλάσιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου. ἔστι δὲ καὶ τὸ ΖΕΓΗ παραλληλόγραμμον διπλάσιον τοῦ ΑΕΓ τριγώνου: βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει καὶ ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις: ἴσον ἄρα ἐστὶ τὸ ΖΕΓΗ παραλληλόγραμμον τῷ ΑΒΓ τριγώνῳ. καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ἴσην τῇ δοθείσῃ τῇ Δ. τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ ἴσον παραλληλόγραμμον συνέσταται τὸ ΖΕΓΗ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ, ἥτις ἐστὶν ἴση τῇ Δ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 195|>","<|""VertexLabel"" -> ""1.43"", ""Text"" -> ""In any parallelogram the complements of the parallelograms about the diameter are equal to one another."", ""TextWordCount"" -> 16, ""GreekText"" -> ""παντὸς παραλληλογράμμου τῶν περὶ τὴν διάμετρον παραλληλογράμμων τὰ παραπληρώματα ἴσα ἀλλήλοις ἐστίν."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Common Notion"" -> 2}, {""Common Notion"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 34}}, ""Proof"" -> ""Let ABCD be a parallelogram, and AC its diameter; and about AC let EH, FG be parallelograms, and BK, KD the so-called complements; I say that the complement BK is equal to the complement KD. For, since ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ACD. [I. 34] Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK. For the same reason the triangle KFC is also equal to KGC. Now, since the triangle AEK is equal to the triangle AHK, and KFC to KGC, the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C. N. 2] And the whole triangle ABC is also equal to the whole ADC; therefore the complement BK which remains is equal to thecomplement KD which remains. [C. N. 3]"", ""ProofWordCount"" -> 149, ""GreekProof"" -> ""ἔστω παραλληλόγραμμον τὸ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἡ ΑΓ, περὶ δὲ τὴν ΑΓ παραλληλόγραμμα μὲν ἔστω τὰ ΖΘ, ΖΗ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΒΚ, ΚΔ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα τῷ ΚΔ παραπληρώματι. ἐπεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἡ ΑΓ, ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ. πάλιν, ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΕΘ, διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ, ἴσον ἐστὶ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον. ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον, τὸ δὲ ΚΖΓ τῷ ΚΗΓ, τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ: ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ὅλῳ τῷ ΑΔΓ ἴσον: λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα λοιπῷ τῷ ΚΔ παραπληρώματί ἐστιν ἴσον. παντὸς ἄρα παραλληλογράμμου χωρίου τῶν περὶ τὴν διάμετρον παραλληλογράμμων τὰ παραπληρώματα ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 156|>","<|""VertexLabel"" -> ""1.44"", ""Text"" -> ""To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle."", ""TextWordCount"" -> 19, ""GreekText"" -> ""παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι τριγώνῳ ἴσον παραλληλόγραμμον παραβαλεῖν ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ."", ""GreekTextWordCount"" -> 15, ""References"" -> {{""Common Notion"" -> 1}, {""Postulate"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 15}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 42}, {""Book"" -> 1, ""Theorem"" -> 43}}, ""Proof"" -> ""Let AB be the given straight line, C the given triangle and D the given rectilineal angle;thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C. Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42];let it be placed so that BE is in a straight line with AB; let FG be drawn through to H, and let AH be drawn through A parallel to either BG or EF. [I. 31] Let HB be joined. Then, since the straight line HF falls upon the parallelsAH, EF, the angles AHF, HFE are equal to two right angles. [I. 29]Therefore the angles BHG, GFE are less than two right angles; and straight lines produced indefinitely from angles less thantwo right angles meet; [Post. 5] therefore HB, FE, when produced, will meet. Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, [I. 31] and let HA, GB be produced to the points L, M. Then HLKF is a parallelogram, HK is its diameter, and AG, ME are parallelograms. and LB, BF the so-called complements, about HK; therefore LB is equal to BF. [I. 43] But BF is equal to the triangle C;therefore LB is also equal to C. [C. N. 1] And, since the angle GBE is equal to the angle ABM, [I. 15] while the angle GBE is equal to D, the angle ABM is also equal to the angle D. Therefore the parallelogram LB equal to the given triangleC has been applied to the given straight line AB, in the angle ABM which is equal to D."", ""ProofWordCount"" -> 303, ""GreekProof"" -> ""ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν τρίγωνον τὸ Γ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ: δεῖ δὴ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον παραλληλόγραμμον παραβαλεῖν ἐν ἴσῃ τῇ Δ γωνίᾳ. συνεστάτω τῷ Γ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΒΕΖΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ, ἥ ἐστιν ἴση τῇ Δ: καὶ κείσθω ὥστε ἐπ᾽ εὐθείας εἶναι τὴν ΒΕ τῇ ΑΒ, καὶ διήχθω ἡ ΖΗ ἐπὶ τὸ Θ, καὶ διὰ τοῦ Α ὁποτέρᾳ τῶν ΒΗ, ΕΖ παράλληλος ἤχθω ἡ ΑΘ, καὶ ἐπεζεύχθω ἡ ΘΒ. καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ, ΕΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ, αἱ ἄρα ὑπὸ ΑΘΖ, ΘΖΕ γωνίαι δυσὶν ὀρθαῖς εἰσιν ἴσαι. αἱ ἄρα ὑπὸ ΒΘΗ, ΗΖΕ δύο ὀρθῶν ἐλάσσονές εἰσιν: αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο ὀρθῶν εἰς ἄπειρον ἐκβαλλόμεναι συμπίπτουσιν: αἱ ΘΒ, ΖΕ ἄρα ἐκβαλλόμεναι συμπεσοῦνται. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν κατὰ τὸ Κ, καὶ διὰ τοῦ Κ σημείου ὁποτέρᾳ τῶν ΕΑ, ΖΘ παράλληλος ἤχθω ἡ ΚΛ, καὶ ἐκβεβλήσθωσαν αἱ ΘΑ, ΗΒ ἐπὶ τὰ Λ, Μ σημεῖα. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ, διάμετρος δὲ αὐτοῦ ἡ ΘΚ, περὶ δὲ τὴν ΘΚ παραλληλόγραμμα μὲν τὰ ΑΗ, ΜΕ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΛΒ, ΒΖ: ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ. ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον: καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ, ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση, καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ ἐστὶν ἴση. παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται τὸ ΛΒ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ, ἥ ἐστιν ἴση τῇ Δ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 277|>","<|""VertexLabel"" -> ""1.45"", ""Text"" -> ""To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure."", ""TextWordCount"" -> 15, ""GreekText"" -> ""τῷ δοθέντι εὐθυγράμμῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Common Notion"" -> 1}, {""Common Notion"" -> 2}, {""Book"" -> 1, ""Theorem"" -> 14}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 1, ""Theorem"" -> 30}, {""Book"" -> 1, ""Theorem"" -> 33}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 1, ""Theorem"" -> 42}, {""Book"" -> 1, ""Theorem"" -> 44}}, ""Proof"" -> ""Let ABCD be the given rectilineal figure and E the given rectilineal angle;thus it is required to construct, in the given angle E, a parallelogram equal to the rectilineal figure ABCD. Let DB be joined, and let the parallelogram FH be constructed equal to the triangle ABD, in the angle HKF which is equal to E; [I. 42]let the parallelogram GM equal to the triangle DBC be applied to the straight line GH, in the angle GHM which is equal to E. [I. 44] Then, since the angle E is equal to each of the angles HKF, GHM,the angle HKF is also equal to the angle GHM. [C. N. 1] Let the angle KHG be added to each; therefore the angles FKH, KHG are equal to the angles KHG, GHM. But the angles FKH, KHG are equal to two right angles; [I. 29]therefore the angles KHG, GHM are also equal to two right angles. Thus, with a straight line GH, and at the point H on it, two straight lines KH, HM not lying on the same side make the adjacent angles equal to two right angles;therefore KH is in a straight line with HM. [I. 14] And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another. [I. 29] Let the angle HGL be added to each;therefore the angles MHG, HGL are equal to the angles HGF, HGL. [C. N. 2] But the angles MHG, HGL are equal to two right angles; [I. 29] therefore the angles HGF, HGL are also equal to two right angles. [C. N. 1]Therefore FG is in a straight line with GL. [I. 14] And, since FK is equal and parallel to HG, [I. 34] and HG to ML also, KF is also equal and parallel to ML; [C. N. 1; I. 30] and the straight lines KM, FL join them (at their extremities);therefore KM, FL are also equal and parallel. [I. 33] Therefore KFLM is a parallelogram. And, since the triangle ABD is equal to the parallelogram FH, and DBC to GM, the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. Therefore the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E."", ""ProofWordCount"" -> 396, ""GreekProof"" -> ""ἔστω τὸ μὲν δοθὲν εὐθύγραμμον τὸ ΑΒΓΔ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Ε: δεῖ δὴ τῷ ΑΒΓΔ εὐθυγράμμῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ δοθείσῃ γωνίᾳ τῇ Ε. ἐπεζεύχθω ἡ ΔΒ, καὶ συνεστάτω τῷ ΑΒΔ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΖΘ ἐν τῇ ὑπὸ ΘΚΖ γωνίᾳ, ἥ ἐστιν ἴση τῇ Ε: καὶ παραβεβλήσθω παρὰ τὴν ΗΘ εὐθεῖαν τῷ ΔΒΓ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΗΜ ἐν τῇ ὑπὸ ΗΘΜ γωνίᾳ, ἥ ἐστιν ἴση τῇ Ε. καὶ ἐπεὶ ἡ Ε γωνία ἑκατέρᾳ τῶν ὑπὸ ΘΚΖ, ΗΘΜ ἐστιν ἴση, καὶ ἡ ὑπὸ ΘΚΖ ἄρα τῇ ὑπὸ ΗΘΜ ἐστιν ἴση. κοινὴ προσκείσθω ἡ ὑπὸ ΚΘΗ: αἱ ἄρα ὑπὸ ΖΚΘ, ΚΘΗ ταῖς ὑπὸ ΚΘΗ, ΗΘΜ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΖΚΘ, ΚΘΗ δυσὶν ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΚΘΗ, ΗΘΜ ἄρα δύο ὀρθαῖς ἴσαι εἰσίν. πρὸς δή τινι εὐθείᾳ τῇ ΗΘ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Θ δύο εὐθεῖαι αἱ ΚΘ, ΘΜ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δύο ὀρθαῖς ἴσας ποιοῦσιν: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΚΘ τῇ ΘΜ: καὶ ἐπεὶ εἰς παραλλήλους τὰς ΚΜ, ΖΗ εὐθεῖα ἐνέπεσεν ἡ ΘΗ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΜΘΗ, ΘΗΖ ἴσαι ἀλλήλαις εἰσίν. κοινὴ προσκείσθω ἡ ὑπὸ ΘΗΛ: αἱ ἄρα ὑπὸ ΜΘΗ, ΘΗΛ ταῖς ὑπὸ ΘΗΖ, ΘΗΛ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΜΘΗ, ΘΗΛ δύο ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΘΗΖ, ΘΗΛ ἄρα δύο ὀρθαῖς ἴσαι εἰσίν: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΖΗ τῇ ΗΛ. καὶ ἐπεὶ ἡ ΖΚ τῇ ΘΗ ἴση τε καὶ παράλληλός ἐστιν, ἀλλὰ καὶ ἡ ΘΗ τῇ ΜΛ, καὶ ἡ ΚΖ ἄρα τῇ ΜΛ ἴση τε καὶ παράλληλός ἐστιν: καὶ ἐπιζευγνύουσιν αὐτὰς εὐθεῖαι αἱ ΚΜ, ΖΛ: καὶ αἱ ΚΜ, ΖΛ ἄρα ἴσαι τε καὶ παράλληλοί εἰσιν: παραλληλόγραμμον ἄρα ἐστὶ τὸ ΚΖΛΜ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΒΔ τρίγωνον τῷ ΖΘ παραλληλογράμμῳ, τὸ δὲ ΔΒΓ τῷ ΗΜ, ὅλον ἄρα τὸ ΑΒΓΔ εὐθύγραμμον ὅλῳ τῷ ΚΖΛΜ παραλληλογράμμῳ ἐστὶν ἴσον. τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓΔ ἴσον παραλληλόγραμμον συνέσταται τὸ ΚΖΛΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΚΜ, ἥ ἐστιν ἴση τῇ δοθείσῃ τῇ Ε: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 340|>","<|""VertexLabel"" -> ""1.46"", ""Text"" -> ""On a given straight line to describe a square."", ""TextWordCount"" -> 9, ""GreekText"" -> ""ἀπὸ τῆς δοθείσης εὐθείας τετράγωνον ἀναγράψαι."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 11}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 34}}, ""Proof"" -> ""Let AB be the given straight line; thus it is required to describe a square on the straight line AB. Let AC be drawn at right angles to the straight line AB from the point A on it [I. 11], and let AD be made equal to AB; through the point D let DE be drawnparallel to AB, and through the point B let BE be drawn parallel to AD. [I. 31] Therefore ADEB is a parallelogram; therefore AB is equal to DE, and AD to BE. [I. 34] But AB is equal to AD;therefore the four straight lines BA, AD, DE, EB are equal to one another; therefore the parallelogram ADEB is equilateral. I say next that it is also right-angled. For, since the straight line AD falls upon the parallelsAB, DE, the angles BAD, ADE are equal to two right angles. [I. 29] But the angle BAD is right; therefore the angle ADE is also right. And in parallelogrammic areas the opposite sides andangles are equal to one another; [I. 34] therefore each of the opposite angles ABE, BED is also right. Therefore ADEB is right-angled. And it was also proved equilateral. Therefore it is a square; and it is described on the straight line AB."", ""ProofWordCount"" -> 209, ""GreekProof"" -> ""ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ: δεῖ δὴ ἀπὸ τῆς ΑΒ εὐθείας τετράγωνον ἀναγράψαι. ἤχθω τῇ ΑΒ εὐθείᾳ ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ α πρὸς ὀρθὰς ἡ ΑΓ, καὶ κείσθω τῇ ΑΒ ἴση ἡ ΑΔ: καὶ διὰ μὲν τοῦ Δ σημείου τῇ ΑΒ παράλληλος ἤχθω ἡ ΔΕ, διὰ δὲ τοῦ Β σημείου τῇ ΑΔ παράλληλος ἤχθω ἡ ΒΕ. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒ: ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΑΔ τῇ ΒΕ. ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση: αἱ τέσσαρες ἄρα αἱ ΒΑ, ΑΔ, ΔΕ, ΕΒ ἴσαι ἀλλήλαις εἰσίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ, ΔΕ εὐθεῖα ἐνέπεσεν ἡ ΑΔ, αἱ ἄρα ὑπὸ ΒΑΔ, ΑΔΕ γωνίαι δύο ὀρθαῖς ἴσαι εἰσίν. ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ. τῶν δὲ παραλληλογράμμων χωρίων αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν: ὀρθὴ ἄρα καὶ ἑκατέρα τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ, ΒΕΔ γωνιῶν: ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ. ἐδείχθη δὲ καὶ ἰσόπλευρον. τετράγωνον ἄρα ἐστίν: καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ἀναγεγραμμένον: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 184|>","<|""VertexLabel"" -> ""1.47"", ""Text"" -> ""In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle."", ""TextWordCount"" -> 24, ""GreekText"" -> ""ἐν τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν τετραγώνοις."", ""GreekTextWordCount"" -> 24, ""References"" -> {{""Common Notion"" -> 2}, {""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 14}, {""Book"" -> 1, ""Theorem"" -> 41}, {""Book"" -> 1, ""Theorem"" -> 46}}, ""Proof"" -> ""Let ABC be a right-angled triangle having the angleBAC right; I say that the square on BC is equal to the squares on BA, AC. For let there be described on BC the square BDEC,and on BA, AC the squares GB, HC; [I. 46] through A let AL be drawn parallel to either BD or CE, and let AD, FC be joined. Then, since each of the angles BAC, BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight linesAC, AG not lying on the same side make the adjacent angles equal to two right angles; therefore CA is in a straight line with AG. [I. 14] For the same reason BA is also in a straight line with AH. And, since the angle DBC is equal to the angle FBA: for each is right: let the angle ABC be added to each;therefore the whole angle DBA is equal to the whole angle FBC. [C. N. 2] And, since DB is equal to BC, and FB to BA, the two sides AB, BD are equal to the two sides FB, BC respectively,and the angle ABD is equal to the angle FBC; therefore the base AD is equal to the base FC, and the triangle ABD is equal to the triangle FBC. [I. 4] Now the parallelogram BL is double of the triangle ABD, for they have the same base BD and are in the same parallelsBD, AL. [I. 41] And the square GB is double of the triangle FBC, for they again have the same base FB and are in the same parallels FB, GC. [I. 41] [But the doubles of equals are equal to one another.]Therefore the parallelogram BL is also equal to the square GB. Similarly, if AE, BK be joined, the parallelogram CL can also be proved equal to the square HC;therefore the whole square BDEC is equal to the two squares GB, HC. [C. N. 2] And the square BDEC is described on BC, and the squares GB, HC on BA, AC. Therefore the square on the side BC is equal to thesquares on the sides BA, AC."", ""ProofWordCount"" -> 368, ""GreekProof"" -> ""ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν: λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνοις. Ἀναγεγράφθω γὰρ ἀπὸ μὲν τῆς ΒΓ τετράγωνον τὸ ΒΔΕΓ, ἀπὸ δὲ τῶν ΒΑ, ΑΓ τὰ ΗΒ, ΘΓ, καὶ διὰ τοῦ α ὁποτέρᾳ τῶν ΒΔ, ΓΕ παράλληλος ἤχθω ἡ ΑΛ: καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΖΓ. καὶ ἐπεὶ ὀρθή ἐστιν ἑκατέρα τῶν ὑπὸ ΒΑΓ, ΒΑΗ γωνιῶν, πρὸς δή τινι εὐθείᾳ τῇ ΒΑ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α δύο εὐθεῖαι αἱ ΑΓ, ΑΗ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας δυσὶν ὀρθαῖς ἴσας ποιοῦσιν: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΓΑ τῇ ΑΗ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΒΑ τῇ ΑΘ ἐστιν ἐπ᾽ εὐθείας. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΖΒΑ: ὀρθὴ γὰρ ἑκατέρα: κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ: ὅλη ἄρα ἡ ὑπὸ ΔΒΑ ὅλῃ τῇ ὑπὸ ΖΒΓ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΔΒ τῇ ΒΓ, ἡ δὲ ΖΒ τῇ ΒΑ, δύο δὴ αἱ ΔΒ, ΒΑ δύο ταῖς ΖΒ, ΒΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΔΒΑ γωνίᾳ τῇ ὑπὸ ΖΒΓ ἴση: βάσις ἄρα ἡ ΑΔ βάσει τῇ ΖΓ ἐστιν ἴση, καὶ τὸ ΑΒΔ τρίγωνον τῷ ΖΒΓ τριγώνῳ ἐστὶν ἴσον: καὶ ἐστὶ τοῦ μὲν ΑΒΔ τριγώνου διπλάσιον τὸ ΒΛ παραλληλόγραμμον: βάσιν τε γὰρ τὴν αὐτὴν ἔχουσι τὴν ΒΔ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ταῖς ΒΔ, ΑΛ: τοῦ δὲ ΖΒΓ τριγώνου διπλάσιον τὸ ΗΒ τετράγωνον: βάσιν τε γὰρ πάλιν τὴν αὐτὴν ἔχουσι τὴν ΖΒ καὶ ἐν ταῖς αὐταῖς εἰσι παραλλήλοις ταῖς ΖΒ, ΗΓ. τὰ δὲ τῶν ἴσων διπλάσια ἴσα ἀλλήλοις ἐστίν: ἴσον ἄρα ἐστὶ καὶ τὸ ΒΛ παραλληλόγραμμον τῷ ΗΒ τετραγώνῳ. ὁμοίως δὴ ἐπιζευγνυμένων τῶν ΑΕ, ΒΚ δειχθήσεται καὶ τὸ ΓΛ παραλληλόγραμμον ἴσον τῷ ΘΓ τετραγώνῳ: ὅλον ἄρα τὸ ΒΔΕΓ τετράγωνον δυσὶ τοῖς ΗΒ, ΘΓ τετραγώνοις ἴσον ἐστίν. καί ἐστι τὸ μὲν ΒΔΕΓ τετράγωνον ἀπὸ τῆς ΒΓ ἀναγραφέν, τὰ δὲ ΗΒ, ΘΓ ἀπὸ τῶν ΒΑ, ΑΓ. τὸ ἄρα ἀπὸ τῆς ΒΓ πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ πλευρῶν τετραγώνοις. ἐν ἄρα τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν τετραγώνοις: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 368|>","<|""VertexLabel"" -> ""1.48"", ""Text"" -> ""If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right."", ""TextWordCount"" -> 37, ""GreekText"" -> ""ἐὰν τριγώνου τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον ἴσον ᾖ τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ὀρθή ἐστιν."", ""GreekTextWordCount"" -> 31, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 8}, {""Book"" -> 1, ""Theorem"" -> 47}}, ""Proof"" -> ""For in the triangle ABC let the square on one side BC be equal to the squares on the sides BA, AC; I say that the angle BAC is right. For let AD be drawn from the point A at right angles to the straight line AC, let AD be made equal to BA, and let DC be joined. Since DA is equal to AB, the square on DA is also equal to the square on AB. Let the square on AC be added to each; therefore the squares on DA, AC are equal to the squares on BA, AC. But the square on DC is equal to the squares on DA, AC, for the angle DAC is right; [I. 47] and the square on BC is equal to the squares on BA, AC, for this is the hypothesis; therefore the square on DC is equal to the square on BC, so that the side DC is also equal to BC. And, since DA is equal to AB, and AC is common, the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8]But the angle DAC is right; therefore the angle BAC is also right."", ""ProofWordCount"" -> 220, ""GreekProof"" -> ""τριγώνου γὰρ τοῦ ΑΒΓ τὸ ἀπὸ μιᾶς τῆς ΒΓ πλευρᾶς τετράγωνον ἴσον ἔστω τοῖς ἀπὸ τῶν ΒΑ, ΑΓ πλευρῶν τετραγώνοις: λέγω, ὅτι ὀρθή ἐστιν ἡ ὑπὸ ΒΑΓ γωνία. ἤχθω γὰρ ἀπὸ τοῦ Α σημείου τῇ ΑΓ εὐθείᾳ πρὸς ὀρθὰς ἡ ΑΔ καὶ κείσθω τῇ ΒΑ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΔΑ τετράγωνον τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΑΓ τετράγωνον: τὰ ἄρα ἀπὸ τῶν ΔΑ, ΑΓ τετράγωνα ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνοις. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΔΑ, ΑΓ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΔΓ: ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΔΑΓ γωνία: τοῖς δὲ ἀπὸ τῶν ΒΑ, ΑΓ ἴσον ἐστὶ τὸ ἀπὸ ΒΓ: ὑπόκειται γάρ: τὸ ἄρα ἀπὸ τῆς ΔΓ τετράγωνον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ: ὥστε καὶ πλευρὰ ἡ ΔΓ τῇ ΒΓ ἐστιν ἴση: καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, κοινὴ δὲ ἡ ΑΓ, δύο δὴ αἱ ΔΑ, ΑΓ δύο ταῖς ΒΑ, ΑΓ ἴσαι εἰσίν: καὶ βάσις ἡ ΔΓ βάσει τῇ ΒΓ ἴση: γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΑΓ. ἐὰν ἄρα τριγώνου τὸ ἀπὸ μιᾶς τῶν πλευρῶν τετράγωνον ἴσον ᾖ τοῖς ἀπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν τετραγώνοις, ἡ περιεχομένη γωνία ὑπὸ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ὀρθή ἐστιν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 237|>","<|""VertexLabel"" -> ""2.1"", ""Text"" -> ""If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments."", ""TextWordCount"" -> 42, ""GreekText"" -> ""ἐὰν ὦσι δύο εὐθεῖαι, τμηθῇ δὲ ἡ ἑτέρα αὐτῶν εἰς ὁσαδηποτοῦν τμήματα, τὸ περιεχόμενον ὀρθογώνιον ὑπὸ τῶν δύο εὐθειῶν ἴσον ἐστὶ τοῖς ὑπό τε τῆς ἀτμήτου καὶ ἑκάστου τῶν τμημάτων περιεχομένοις ὀρθογωνίοις."", ""GreekTextWordCount"" -> 32, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 11}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 34}}, ""Proof"" -> ""Let A, BC be two straight lines, and let BC be cut at random at the points D, E; I say that the rectangle contained by A, BC is equal to the rectangle contained by A, BD, that contained by A, DE and that contained by A, EC. For let BF be drawn from B at right angles to BC; [I. 11] let BG be made equal to A, [I. 3] through G let GH be drawn parallel to BC, [I. 31] and through D, E, C let DK, EL, CH be drawn parallel to BG. Then BH is equal to BK, DL, EH. Now BH is the rectangle A, BC, for it is contained by GB, BC, and BG is equal to A; BK is the rectangle A, BD, for it is contained by GB, BD, and BG is equal to A; and DL is the rectangle A, DE, for DK, that is BG [I. 34], is equal to A. Similarly also EH is the rectangle A, EC. Therefore the rectangle A, BC is equal to the rectangle A, BD, the rectangle A, DE and the rectangle A, EC."", ""ProofWordCount"" -> 190, ""GreekProof"" -> ""ἔστωσαν δύο εὐθεῖαι αἱ Α, ΒΓ, καὶ τετμήσθω ἡ ΒΓ, ὡς ἔτυχεν, κατὰ τὰ Δ, Ε σημεῖα: λέγω, ὅτι τὸ ὑπὸ τῶν Α, ΒΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ τε ὑπὸ τῶν Α, ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ὑπὸ τῶν Α, ΔΕ καὶ ἔτι τῷ ὑπὸ τῶν Α, ΕΓ. ἤχθω γὰρ ἀπὸ τοῦ Β τῇ ΒΓ πρὸς ὀρθὰς ἡ ΒΖ, καὶ κείσθω τῇ Α ἴση ἡ ΒΗ, καὶ διὰ μὲν τοῦ Η τῇ ΒΓ παράλληλος ἤχθω ἡ ΗΘ, διὰ δὲ τῶν Δ, Ε, Γ τῇ ΒΗ παράλληλοι ἤχθωσαν αἱ ΔΚ, ΕΛ, ΓΘ. ἴσον δή ἐστι τὸ ΒΘ τοῖς ΒΚ, ΔΛ, ΕΘ. καί ἐστι τὸ μὲν ΒΘ τὸ ὑπὸ τῶν Α, ΒΓ: περιέχεται μὲν γὰρ ὑπὸ τῶν ΗΒ, ΒΓ, ἴση δὲ ἡ ΒΗ τῇ Α: τὸ δὲ ΒΚ τὸ ὑπὸ τῶν Α, ΒΔ: περιέχεται μὲν γὰρ ὑπὸ τῶν ΗΒ, ΒΔ, ἴση δὲ ἡ ΒΗ τῇ Α. τὸ δὲ ΔΛ τὸ ὑπὸ τῶν Α, ΔΕ: ἴση γὰρ ἡ ΔΚ, τουτέστιν ἡ ΒΗ, τῇ Α. καὶ ἔτι ὁμοίως τὸ ΕΘ τὸ ὑπὸ τῶν Α, ΕΓ: τὸ ἄρα ὑπὸ τῶν Α, ΒΓ ἴσον ἐστὶ τῷ τε ὑπὸ Α, ΒΔ καὶ τῷ ὑπὸ Α, ΔΕ καὶ ἔτι τῷ ὑπὸ Α, ΕΓ. ἐὰν ἄρα ὦσι δύο εὐθεῖαι, τμηθῇ δὲ ἡ ἑτέρα αὐτῶν εἰς ὁσαδηποτοῦν τμήματα, τὸ περιεχόμενον ὀρθογώνιον ὑπὸ τῶν δύο εὐθειῶν ἴσον ἐστὶ τοῖς ὑπό τε τῆς ἀτμήτου καὶ ἑκάστου τῶν τμημάτων περιεχομένοις ὀρθογωνίοις: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 232|>","<|""VertexLabel"" -> ""2.2"", ""Text"" -> ""If a straight line be cut at random, the rectangle contained by the whole and both of the segments is equal to the square on the whole."", ""TextWordCount"" -> 27, ""GreekText"" -> ""ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης καὶ ἑκατέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ὅλης τετραγώνῳ."", ""GreekTextWordCount"" -> 23, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 46}}, ""Proof"" -> ""For let the straight line AB be cut at random at the point C; I say that the rectangle contained by AB, BC together with the rectangle contained by BA, AC is equal to the square on AB. For let the square ADEB be described on AB [I. 46], and let CF be drawn through C parallel to either AD or BE. [I. 31] Then AE is equal to AF, CE. Now AE is the square on AB; AF is the rectangle contained by BA, AC, for it is contained by DA, AC, and AD is equal to AB; and CE is the rectangle AB, BC, for BE is equal to AB. Therefore the rectangle BA, AC together with the rectangle AB, BC is equal to the square on AB."", ""ProofWordCount"" -> 130, ""GreekProof"" -> ""εὐθεῖα γὰρ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ σημεῖον: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ὑπὸ ΒΑ, ΑΓ περιεχομένου ὀρθογωνίου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, καὶ ἤχθω διὰ τοῦ Γ ὁποτέρᾳ τῶν ΑΔ, ΒΕ παράλληλος ἡ ΓΖ. ἴσον δή ἐστι τὸ ΑΕ τοῖς ΑΖ, ΓΕ. καί ἐστι τὸ μὲν ΑΕ τὸ ἀπὸ τῆς ΑΒ τετράγωνον, τὸ δὲ ΑΖ τὸ ὑπὸ τῶν ΒΑ, ΑΓ περιεχόμενον ὀρθογώνιον: περιέχεται μὲν γὰρ ὑπὸ τῶν ΔΑ, ΑΓ, ἴση δὲ ἡ ΑΔ τῇ ΑΒ: τὸ δὲ ΓΕ τὸ ὑπὸ τῶν ΑΒ, ΒΓ: ἴση γὰρ ἡ ΒΕ τῇ ΑΒ. τὸ ἄρα ὑπὸ τῶν ΒΑ, ΑΓ μετὰ τοῦ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης καὶ ἑκατέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ὅλης τετραγώνῳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 155|>","<|""VertexLabel"" -> ""2.3"", ""Text"" -> ""If a straight line be cut at random, the rectangle contained by the whole and one of the segments is equal to the rectangle contained by the segments and the square on the aforesaid segment."", ""TextWordCount"" -> 35, ""GreekText"" -> ""ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ τε ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ προειρημένου τμήματος τετραγώνῳ."", ""GreekTextWordCount"" -> 32, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 46}}, ""Proof"" -> ""For let the straight line AB be cut at random at C; I say that the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB together with the square on BC. For let the square CDEB be described on CB; [I. 46] let ED be drawn through to F, and through A let AF be drawn parallel to either CD or BE. [I. 31] Then AE is equal to AD, CE. Now AE is the rectangle contained by AB, BC, for it is contained by AB, BE, and BE is equal to BC; AD is the rectangle AC, CB, for DC is equal to CB; and DB is the square on CB. Therefore the rectangle contained by AB, BC is equal to the rectangle contained by AC, CB together with the square on BC."", ""ProofWordCount"" -> 139, ""GreekProof"" -> ""εὐθεῖα γὰρ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ τε ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΒΓ τετραγώνου. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΓΒ τετράγωνον τὸ ΓΔΕΒ, καὶ διήχθω ἡ ΕΔ ἐπὶ τὸ Ζ, καὶ διὰ τοῦ Α ὁποτέρᾳ τῶν ΓΔ, ΒΕ παράλληλος ἤχθω ἡ ΑΖ. ἴσον δή ἐστι τὸ ΑΕ τοῖς ΑΔ, ΓΕ: καί ἐστι τὸ μὲν ΑΕ τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον: περιέχεται μὲν γὰρ ὑπὸ τῶν ΑΒ, ΒΕ, ἴση δὲ ἡ ΒΕ τῇ ΒΓ: τὸ δὲ ΑΔ τὸ ὑπὸ τῶν ΑΓ, ΓΒ: ἴση γὰρ ἡ ΔΓ τῇ ΓΒ: τὸ δὲ ΔΒ τὸ ἀπὸ τῆς ΓΒ τετράγωνον: τὸ ἄρα ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΒΓ τετραγώνου. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ὑπὸ τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ τε ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ προειρημένου τμήματος τετραγώνῳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 176|>","<|""VertexLabel"" -> ""2.4"", ""Text"" -> ""If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments."", ""TextWordCount"" -> 29, ""GreekText"" -> ""ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν τμημάτων τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ."", ""GreekTextWordCount"" -> 27, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 6}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 1, ""Theorem"" -> 46}}, ""Proof"" -> ""For let the straight line AB be cut at random at C; I say that the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB. For let the square ADEB be described on AB, [I. 46] let BD be joined; through C let CF be drawn parallel to either AD or EB, and through G let HK be drawn parallel to either AB or DE. [I. 31] Then, since CF is parallel to AD, and BD has fallen on them, the exterior angle CGB is equal to the interior and opposite angle ADB. [I. 29] But the angle ADB is equal to the angle ABD, since the side BA is also equal to AD; [I. 5]therefore the angle CGB is also equal to the angle GBC, so that the side BC is also equal to the side CG. [I. 6] But CB is equal to GK, and CG to KB; [I. 34] therefore GK is also equal to KB; therefore CGKB is equilateral. I say next that it is also right-angled. For, since CG is parallel to BK, the angles KBC, GCB are equal to two right angles. [I. 29] But the angle KBC is right; therefore the angle BCG is also right, so that the opposite angles CGK, GKB are also right. [I. 34] Therefore CGKB is right-angled; and it was also proved equilateral; therefore it is a square; and it is described on CB. For the same reason HF is also a square; and it is described on HG, that is AC. [I. 34] Therefore the squares HF, KC are the squares on AC, CB. Now, since AG is equal to GE, and AG is the rectangle AC, CB, for GC is equal to CB, therefore GE is also equal to the rectangle AC, CB. Therefore AG, GE are equal to twice the rectangle AC, CB. But the squares HF, CK are also the squares on AC, CB; therefore the four areas HF, CK, AG, GE are equal to the squares on AC, CB and twice the rectangle contained by AC, CB. But HF, CK, AG, GE are the whole ADEB, which is the square on AB. Therefore the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB."", ""ProofWordCount"" -> 391, ""GreekProof"" -> ""εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ. λέγω, ὅτι τὸ ἀπὸ τῆς ΑΒ τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, καὶ ἐπεζεύχθω ἡ ΒΔ, καὶ διὰ μὲν τοῦ Γ ὁποτέρᾳ τῶν ΑΔ, ΕΒ παράλληλος ἤχθω ἡ ΓΖ, διὰ δὲ τοῦ Η ὁποτέρᾳ τῶν ΑΒ, ΔΕ παράλληλος ἤχθω ἡ ΘΚ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΓΖ τῇ ΑΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΔ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΗΒ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΔΒ. ἀλλ᾽ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΒΑ τῇ ΑΔ ἐστιν ἴση: καὶ ἡ ὑπὸ ΓΗΒ ἄρα γωνία τῇ ὑπὸ ΗΒΓ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΒΓ πλευρᾷ τῇ ΓΗ ἐστιν ἴση: ἀλλ᾽ ἡ μὲν ΓΒ τῇ ΗΚ ἐστιν ἴση, ἡ δὲ ΓΗ τῇ ΚΒ: καὶ ἡ ΗΚ ἄρα τῇ ΚΒ ἐστιν ἴση: ἰσόπλευρον ἄρα ἐστὶ τὸ ΓΗΚΒ. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ παράλληλός ἐστιν ἡ ΓΗ τῇ ΒΚ καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΓΒ, αἱ ἄρα ὑπὸ ΚΒΓ, ΗΓΒ γωνίαι δύο ὀρθαῖς εἰσιν ἴσαι. ὀρθὴ δὲ ἡ ὑπὸ ΚΒΓ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΓΗ: ὥστε καὶ αἱ ἀπεναντίον αἱ ὑπὸ ΓΗΚ, ΗΚΒ ὀρθαί εἰσιν. ὀρθογώνιον ἄρα ἐστὶ τὸ ΓΗΚΒ: ἐδείχθη δὲ καὶ ἰσόπλευρον: τετράγωνον ἄρα ἐστίν: καί ἐστιν ἀπὸ τῆς ΓΒ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΘΖ τετράγωνόν ἐστιν: καί ἐστιν ἀπὸ τῆς ΘΗ, τουτέστιν ἀπὸ τῆς ΑΓ: τὰ ἄρα ΘΖ, ΚΓ τετράγωνα ἀπὸ τῶν ΑΓ, ΓΒ εἰσιν. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, καί ἐστι τὸ ΑΗ τὸ ὑπὸ τῶν ΑΓ, ΓΒ: ἴση γὰρ ἡ ΗΓ τῇ ΓΒ: καὶ τὸ ΗΕ ἄρα ἴσον ἐστὶ τῷ ὑπὸ ΑΓ, ΓΒ: τὰ ἄρα ΑΗ, ΗΕ ἴσα ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. ἔστι δὲ καὶ τὰ ΘΖ, ΓΚ τετράγωνα ἀπὸ τῶν ΑΓ, ΓΒ: τὰ ἄρα τέσσαρα τὰ ΘΖ, ΓΚ, ΑΗ, ΗΕ ἴσα ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ. ἀλλὰ τὰ ΘΖ, ΓΚ, ΑΗ, ΗΕ ὅλον ἐστὶ τὸ ΑΔΕΒ, ὅ ἐστιν ἀπὸ τῆς ΑΒ τετράγωνον: τὸ ἄρα ἀπὸ τῆς ΑΒ τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν τμημάτων τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ: ὅπερ ἔδει δεῖξαι. πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐν τοῖς τετραγώνοις χωρίοις τὰ περὶ τὴν διάμετρον παραλληλόγραμμα τετράγωνά ἐστιν."", ""GreekProofWordCount"" -> 435|>","<|""VertexLabel"" -> ""2.5"", ""Text"" -> ""If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half."", ""TextWordCount"" -> 42, ""GreekText"" -> ""ἐὰν εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὸ ὑπὸ τῶν ἀνίσων τῆς ὅλης τμημάτων περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ἡμισείας τετραγώνῳ."", ""GreekTextWordCount"" -> 32, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 36}, {""Book"" -> 1, ""Theorem"" -> 43}, {""Book"" -> 1, ""Theorem"" -> 46}}, ""Proof"" -> ""For let a straight line AB be cut into equal segments at C and into unequal segments at D; I say that the rectangle contained by AD, DB together with the square on CD is equal to the square on CB. For let the square CEFB be described on CB, [I. 46] and let BE be joined; through D let DG be drawn parallel to either CE or BF, through H again let KM be drawn parallel to either AB or EF, and again through A let AK be drawn parallel to either CL or BM. [I. 31] Then, since the complement CH is equal to the complement HF, [I. 43] let DM be added to each; therefore the whole CM is equal to the whole DF. But CM is equal to AL, since AC is also equal to CB; [I. 36]therefore AL is also equal to DF. Let CH be added to each; therefore the whole AH is equal to the gnomon NOP. But AH is the rectangle AD, DB, for DH is equal to DB, therefore the gnomon NOP is also equal to the rectangle AD, DB. Let LG, which is equal to the square on CD, be added to each; therefore the gnomon NOP and LG are equal to the rectangle contained by AD, DB and the square on CD. But the gnomon NOP and LG are the whole square CEFB, which is described on CB; therefore the rectangle contained by AD, DB together with the square on CD is equal to the square on CB."", ""ProofWordCount"" -> 260, ""GreekProof"" -> ""εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω εἰς μὲν ἴσα κατὰ τὸ Γ, εἰς δὲ ἄνισα κατὰ τὸ Δ: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΔ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΒ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΓΒ τετράγωνον τὸ ΓΕΖΒ, καὶ ἐπεζεύχθω ἡ ΒΕ, καὶ διὰ μὲν τοῦ Δ ὁποτέρᾳ τῶν ΓΕ, ΒΖ παράλληλος ἤχθω ἡ ΔΗ, διὰ δὲ τοῦ Θ ὁποτέρᾳ τῶν ΑΒ, ΕΖ παράλληλος πάλιν ἤχθω ἡ ΚΜ, καὶ πάλιν διὰ τοῦ Α ὁποτέρᾳ τῶν ΓΛ, ΒΜ παράλληλος ἤχθω ἡ ΑΚ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΓΘ παραπλήρωμα τῷ ΘΖ παραπληρώματι, κοινὸν προσκείσθω τὸ ΔΜ: ὅλον ἄρα τὸ ΓΜ ὅλῳ τῷ ΔΖ ἴσον ἐστίν. ἀλλὰ τὸ ΓΜ τῷ ΑΛ ἴσον ἐστίν, ἐπεὶ καὶ ἡ ΑΓ τῇ ΓΒ ἐστιν ἴση: καὶ τὸ ΑΛ ἄρα τῷ ΔΖ ἴσον ἐστίν. κοινὸν προσκείσθω τὸ ΓΘ: ὅλον ἄρα τὸ ΑΘ τῷ ΜΝΞ γνώμονι ἴσον ἐστίν. ἀλλὰ τὸ ΑΘ τὸ ὑπὸ τῶν ΑΔ, ΔΒ ἐστιν: ἴση γὰρ ἡ ΔΘ τῇ ΔΒ: καὶ ὁ ΜΝΞ ἄρα γνώμων ἴσος ἐστὶ τῷ ὑπὸ ΑΔ, ΔΒ. κοινὸν προσκείσθω τὸ ΛΗ, ὅ ἐστιν ἴσον τῷ ἀπὸ τῆς ΓΔ: ὁ ἄρα ΜΝΞ γνώμων καὶ τὸ ΛΗ ἴσα ἐστὶ τῷ ὑπὸ τῶν ΑΔ, ΔΒ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΓΔ τετραγώνῳ. ἀλλὰ ὁ ΜΝΞ γνώμων καὶ τὸ ΛΗ ὅλον ἐστὶ τὸ ΓΕΖΒ τετράγωνον, ὅ ἐστιν ἀπὸ τῆς ΓΒ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΔ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΒ τετραγώνῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὸ ὑπὸ τῶν ἀνίσων τῆς ὅλης τμημάτων περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ἡμισείας τετραγώνῳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 283|>","<|""VertexLabel"" -> ""2.6"", ""Text"" -> ""If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line."", ""TextWordCount"" -> 60, ""GreekText"" -> ""ἐὰν εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας, τὸ ὑπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ καὶ τῆς προσκειμένης περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης τετραγώνῳ."", ""GreekTextWordCount"" -> 45, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 36}, {""Book"" -> 1, ""Theorem"" -> 43}, {""Book"" -> 1, ""Theorem"" -> 46}}, ""Proof"" -> ""For let a straight line AB be bisected at the point C, and let a straight line BD be added to it in a straight line; I say that the rectangle contained by AD, DB together with the square on CB is equal to the square on CD. For let the square CEFD be described on CD, [I. 46] and let DE be joined; through the point B let BG be drawn parallel to either EC or DF, through the point H let KM be drawn parallel to either AB or EF, and further through A let AK be drawn parallel to either CL or DM. [I. 31] Then, since AC is equal to CB, AL is also equal to CH. [I. 36]But CH is equal to HF. [I. 43] Therefore AL is also equal to HF. Let CM be added to each; therefore the whole AM is equal to the gnomon NOP. But AM is the rectangle AD, DB, for DM is equal to DB; therefore the gnomon NOP is also equal to the rectangle AD, DB. Let LG, which is equal to the square on BC, be added to each; therefore the rectangle contained by AD, DB together with the square on CB is equal to the gnomon NOP and LG. But the gnomon NOP and LG are the whole square CEFD, which is described on CD; therefore the rectangle contained by AD, DB together with the square on CB is equal to the square on CD."", ""ProofWordCount"" -> 250, ""GreekProof"" -> ""εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω δίχα κατὰ τὸ Γ σημεῖον, προσκείσθω δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας ἡ ΒΔ: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΒ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΔ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΓΔ τετράγωνον τὸ ΓΕΖΔ, καὶ ἐπεζεύχθω ἡ ΔΕ, καὶ διὰ μὲν τοῦ Β σημείου ὁποτέρᾳ τῶν ΕΓ, ΔΖ παράλληλος ἤχθω ἡ ΒΗ, διὰ δὲ τοῦ Θ σημείου ὁποτέρᾳ τῶν ΑΒ, ΕΖ παράλληλος ἤχθω ἡ ΚΜ, καὶ ἔτι διὰ τοῦ Α ὁποτέρᾳ τῶν ΓΛ, ΔΜ παράλληλος ἤχθω ἡ ΑΚ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, ἴσον ἐστὶ καὶ τὸ ΑΛ τῷ ΓΘ. ἀλλὰ τὸ ΓΘ τῷ ΘΖ ἴσον ἐστίν. καὶ τὸ ΑΛ ἄρα τῷ ΘΖ ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΓΜ: ὅλον ἄρα τὸ ΑΜ τῷ ΝΞΟ γνώμονί ἐστιν ἴσον. ἀλλὰ τὸ ΑΜ ἐστι τὸ ὑπὸ τῶν ΑΔ, ΔΒ: ἴση γάρ ἐστιν ἡ ΔΜ τῇ ΔΒ: καὶ ὁ ΝΞΟ ἄρα γνώμων ἴσος ἐστὶ τῷ ὑπὸ τῶν ΑΔ, ΔΒ περιεχομένῳ ὀρθογωνίῳ. κοινὸν προσκείσθω τὸ ΛΗ, ὅ ἐστιν ἴσον τῷ ἀπὸ τῆς ΒΓ τετραγώνῳ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΒ τετραγώνου ἴσον ἐστὶ τῷ ΝΞΟ γνώμονι καὶ τῷ ΛΗ. ἀλλὰ ὁ ΝΞΟ γνώμων καὶ τὸ ΛΗ ὅλον ἐστὶ τὸ ΓΕΖΔ τετράγωνον, ὅ ἐστιν ἀπὸ τῆς ΓΔ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΒ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΓΒ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΔ τετραγώνῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας, τὸ ὑπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ καὶ τῆς προσκειμένης περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ἡμισείας τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης τετραγώνῳ: ὅπερ ἔδει δεῖξαι. ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ᾽ ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ. εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ σημεῖον: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ: καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, κοινὸν προσκείσθω τὸ ΓΖ: ὅλον ἄρα τὸ ΑΖ ὅλῳ τῷ ΓΕ ἴσον ἐστίν: τὰ ἄρα ΑΖ, ΓΕ διπλάσιά ἐστι τοῦ ΑΖ. ἀλλὰ τὰ ΑΖ, ΓΕ ὁ ΚΛΜ ἐστι γνώμων καὶ τὸ ΓΖ τετράγωνον: ὁ ΚΛΜ ἄρα γνώμων καὶ τὸ ΓΖ διπλάσιά ἐστι τοῦ ΑΖ. ἔστι δὲ τοῦ ΑΖ διπλάσιον καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ἴση γὰρ ἡ ΒΖ τῇ ΒΓ: ὁ ἄρα ΚΛΜ γνώμων καὶ τὸ ΓΖ τετράγωνον ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. κοινὸν προσκείσθω τὸ ΔΗ, ὅ ἐστιν ἀπὸ τῆς ΑΓ τετράγωνον: ὁ ἄρα ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ. ἀλλὰ ὁ ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ὅλον ἐστὶ τὸ ΑΔΕΒ καὶ τὸ ΓΖ, ἅ ἐστιν ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα: τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΑΓ τετραγώνου. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ᾽ ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 593|>","<|""VertexLabel"" -> ""2.7"", ""Text"" -> ""If a straight line be cut at random, the square on the whole and that on one of the segments both together are equal to twice the rectangle contained by the whole and the said segment and the square on the remaining segment."", ""TextWordCount"" -> 43, ""GreekText"" -> ""ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ᾽ ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ."", ""GreekTextWordCount"" -> 41, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 43}, {""Book"" -> 1, ""Theorem"" -> 46}}, ""Proof"" -> ""For let a straight line AB be cut at random at the point C; I say that the squares on AB, BC are equal to twice the rectangle contained by AB, BC and the square on CA. For let the square ADEB be described on AB, [I. 46] and let the figure be drawn. Then, since AG is equal to GE, [I. 43] let CF be added to each; therefore the whole AF is equal to the whole CE. Therefore AF, CE are double of AF. But AF, CE are the gnomon KLM and the square CF; therefore the gnomon KLM and the square CF are double of AF. But twice the rectangle AB, BC is also double of AF; for BF is equal to BC; therefore the gnomon KLM and the square CF are equal to twice the rectangle AB, BC. Let DG, which is the square on AC, be added to each; therefore the gnomon KLM and the squares BG, GD are equal to twice the rectangle contained by AB, BC and the square on AC. But the gnomon KLM and the squares BG, GD are the whole ADEB and CF, which are squares described on AB, BC; therefore the squares on AB, BC are equal to twice the rectangle contained by AB, BC together with the square on AC."", ""ProofWordCount"" -> 222, ""GreekProof"" -> ""Εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ σημεῖον: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ: καὶ καταγεγράφθω τὸ σχῆμα. Ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, κοινὸν προσκείσθω τὸ ΓΖ: ὅλον ἄρα τὸ ΑΖ ὅλῳ τῷ ΓΕ ἴσον ἐστίν: τὰ ἄρα ΑΖ, ΓΕ διπλάσιά ἐστι τοῦ ΑΖ. ἀλλὰ τὰ ΑΖ, ΓΕ ὁ ΚΛΜ ἐστι γνώμων καὶ τὸ ΓΖ τετράγωνον: ὁ ΚΛΜ ἄρα γνώμων καὶ τὸ ΓΖ διπλάσιά ἐστι τοῦ ΑΖ. ἔστι δὲ τοῦ ΑΖ διπλάσιον καὶ τὸ δὶς ὑπὸ τῶν ΑΒ, ΒΓ: ἴση γὰρ ἡ ΒΖ τῇ ΒΓ: ὁ ἄρα ΚΛΜ γνώμων καὶ τὸ ΓΖ τετράγωνον ἴσον ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΒ, ΒΓ. κοινὸν προσκείσθω τὸ ΔΗ, ὅ ἐστιν ἀπὸ τῆς ΑΓ τετράγωνον: ὁ ἄρα ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ. ἀλλὰ ὁ ΚΛΜ γνώμων καὶ τὰ ΒΗ, ΗΔ τετράγωνα ὅλον ἐστὶ τὸ ΑΔΕΒ καὶ τὸ ΓΖ, ἅ ἐστιν ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα: τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα ἴσα ἐστὶ τῷ [τε] δὶς ὑπὸ τῶν ΑΒ, ΒΓ περιεχομένῳ ὀρθογωνίῳ μετὰ τοῦ ἀπὸ τῆς ΑΓ τετραγώνου. Ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης καὶ τὸ ἀφ᾽ ἑνὸς τῶν τμημάτων τὰ συναμφότερα τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 263|>","<|""VertexLabel"" -> ""2.8"", ""Text"" -> ""If a straight line be cut at random, four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square described on the whole and the aforesaid segment as on one straight line."", ""TextWordCount"" -> 47, ""GreekText"" -> ""ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ τετράκις ὑπὸ τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνου ἴσον ἐστὶ τῷ ἀπό τε τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος ὡς ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ."", ""GreekTextWordCount"" -> 40, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 36}, {""Book"" -> 1, ""Theorem"" -> 43}}, ""Proof"" -> ""For let a straight line AB be cut at random at the point C; I say that four times the rectangle contained by AB, BC together with the square on AC is equal to the square described on AB, BC as on one straight line. For let [the straight line]BD be produced in a straight line [with AB], and let BD be made equal to CB; let the square AEFD be described on AD, and let the figure be drawn double. Then, since CB is equal to BD, while CB is equal to GK, and BD to KN, therefore GK is also equal to KN. For the same reason QR is also equal to RP. And, since BC is equal to BD, and GK to KN, therefore CK is also equal to KD, and GR to RN. [I. 36] But CK is equal to RN, for they are complements of the parallelogram CP; [I. 43] therefore KD is also equal to GR; therefore the four areas DK, CK, GR, RN are equal to one another. Therefore the four are quadruple of CK. Again, since CB is equal to BD, while BD is equal to BK, that is CG, and CB is equal to GK, that is GQ, therefore CG is also equal to GQ. And, since CG is equal to GQ, and QR to RP, AG is also equal to MQ, and QL to RF. [I. 36] But MQ is equal to QL, for they are complements of the parallelogram ML; [I. 43] therefore AG is also equal to RF; therefore the four areas AG, MQ, QL, RF are equal to one another. Therefore the four are quadruple of AG. But the four areas CK, KD, GR, RN were proved to be quadruple of CK; therefore the eight areas, which contain the gnomon STU, are quadruple of AK. Now, since AK is the rectangle AB, BD, for BK is equal to BD, therefore four times the rectangle AB, BD is quadruple of AK. But the gnomon STU was also proved to be quadruple of AK; therefore four times the rectangle AB, BD is equal to the gnomon STU. Let OH, which is equal to the square on AC, be added to each; therefore four times the rectangle AB, BD together with the square on AC is equal to the gnomon STU and OH. But the gnomon STU and OH are the whole square AEFD, which is described on AD. therefore four times the rectangle AB, BD together with the square on AC is equal to the square on AD But BD is equal to BC; therefore four times the rectangle contained by AB, BC together with the square on AC is equal to the square on AD, that is to the square described on AB and BC as on one straight line."", ""ProofWordCount"" -> 473, ""GreekProof"" -> ""εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ σημεῖον: λέγω, ὅτι τὸ τετράκις ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΑΓ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ, ΒΓ ὡς ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ. Ἐκβεβλήσθω γὰρ ἐπ᾽ εὐθείας τῇ ΑΒ εὐθεῖα ἡ ΒΔ, καὶ κείσθω τῇ ΓΒ ἴση ἡ ΒΔ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΑΔ τετράγωνον τὸ ΑΕΖΔ, καὶ καταγεγράφθω διπλοῦν τὸ σχῆμα. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΓΒ τῇ ΒΔ, ἀλλὰ ἡ μὲν ΓΒ τῇ ΗΚ ἐστιν ἴση, ἡ δὲ ΒΔ τῇ ΚΝ, καὶ ἡ ΗΚ ἄρα τῇ ΚΝ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΠΡ τῇ ΡΟ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΒΔ, ἡ δὲ ΗΚ τῇ ΚΝ, ἴσον ἄρα ἐστὶ καὶ τὸ μὲν ΓΚ τῷ ΚΔ, τὸ δὲ ΗΡ τῷ ΡΝ. ἀλλὰ τὸ ΓΚ τῷ ΡΝ ἐστιν ἴσον: παραπληρώματα γὰρ τοῦ ΓΟ παραλληλογράμμου: καὶ τὸ ΚΔ ἄρα τῷ ΗΡ ἴσον ἐστίν: τὰ τέσσαρα ἄρα τὰ ΔΚ, ΓΚ, ΗΡ, ΡΝ ἴσα ἀλλήλοις ἐστίν. τὰ τέσσαρα ἄρα τετραπλάσιά ἐστι τοῦ ΓΚ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΓΒ τῇ ΒΔ, ἀλλὰ ἡ μὲν ΒΔ τῇ ΒΚ, τουτέστι τῇ ΓΗ ἴση, ἡ δὲ ΓΒ τῇ ΗΚ, τουτέστι τῇ ΗΠ, ἐστιν ἴση, καὶ ἡ ΓΗ ἄρα τῇ ΗΠ ἴση ἐστίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ΓΗ τῇ ΗΠ, ἡ δὲ ΠΡ τῇ ΡΟ, ἴσον ἐστὶ καὶ τὸ μὲν ΑΗ τῷ ΜΠ, τὸ δὲ ΠΛ τῷ ΡΖ. ἀλλὰ τὸ ΜΠ τῷ ΠΛ ἐστιν ἴσον: παραπληρώματα γὰρ τοῦ ΜΛ παραλληλογράμμου: καὶ τὸ ΑΗ ἄρα τῷ ΡΖ ἴσον ἐστίν: τὰ τέσσαρα ἄρα τὰ ΑΗ, ΜΠ, ΠΛ, ΡΖ ἴσα ἀλλήλοις ἐστίν: τὰ τέσσαρα ἄρα τοῦ ΑΗ ἐστι τετραπλάσια. ἐδείχθη δὲ καὶ τὰ τέσσαρα τὰ ΓΚ, ΚΔ, ΗΡ, ΡΝ τοῦ ΓΚ τετραπλάσια: τὰ ἄρα ὀκτώ, ἃ περιέχει τὸν ΣΤΥ γνώμονα, τετραπλάσιά ἐστι τοῦ ΑΚ. καὶ ἐπεὶ τὸ ΑΚ τὸ ὑπὸ τῶν ΑΒ, ΒΔ ἐστιν: ἴση γὰρ ἡ ΒΚ τῇ ΒΔ: τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΔ τετραπλάσιόν ἐστι τοῦ ΑΚ. ἐδείχθη δὲ τοῦ ΑΚ τετραπλάσιος καὶ ὁ ΣΤΥ γνώμων: τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΔ ἴσον ἐστὶ τῷ ΣΤΥ γνώμονι. κοινὸν προσκείσθω τὸ ΞΘ, ὅ ἐστιν ἴσον τῷ ἀπὸ τῆς ΑΓ τετραγώνῳ: τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΔ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ ΑΓ τετραγώνου ἴσον ἐστὶ τῷ ΣΤΥ γνώμονι καὶ τῷ ΞΘ. ἀλλὰ ὁ ΣΤΥ γνώμων καὶ τὸ ΞΘ ὅλον ἐστὶ τὸ ΑΕΖΔ τετράγωνον, ὅ ἐστιν ἀπὸ τῆς ΑΔ: τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΔ μετὰ τοῦ ἀπὸ ΑΓ ἴσον ἐστὶ τῷ ἀπὸ ΑΔ τετραγώνῳ: ἴση δὲ ἡ ΒΔ τῇ ΒΓ. τὸ ἄρα τετράκις ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ ΑΓ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΔ, τουτέστι τῷ ἀπὸ τῆς ΑΒ καὶ ΒΓ ὡς ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ τετράκις ὑπὸ τῆς ὅλης καὶ ἑνὸς τῶν τμημάτων περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνου ἴσον ἐστὶ τῷ ἀπό τε τῆς ὅλης καὶ τοῦ εἰρημένου τμήματος ὡς ἀπὸ μιᾶς ἀναγραφέντι τετραγώνῳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 499|>","<|""VertexLabel"" -> ""2.9"", ""Text"" -> ""If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section."", ""TextWordCount"" -> 41, ""GreekText"" -> ""ἐὰν εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὰ ἀπὸ τῶν ἀνίσων τῆς ὅλης τμημάτων τετράγωνα διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου."", ""GreekTextWordCount"" -> 31, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 6}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 1, ""Theorem"" -> 47}}, ""Proof"" -> ""For let a straight line AB be cut into equal segments at C, and into unequal segments at D; I say that the squares on AD, DB are double of the squares on AC, CD. For let CE be drawn from C at right angles to AB, and let it be made equal to either AC or CB; let EA, EB be joined, let DF be drawn through D parallel to EC, and FG through F parallel to AB, and let AF be joined. Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC. And, since the angle at C is right, the remaining angles EAC, AEC are equal to one right angle. [I. 32] And they are equal; therefore each of the angles CEA, CAE is half a right angle. For the same reason each of the angles CEB, EBC is also half a right angle; therefore the whole angle AEB is right. And, since the angle GEF is half a right angle, and the angle EGF is right, for it is equal to the interior and opposite angle ECB, [I. 29] the remaining angle EFG is half a right angle; [I. 32] therefore the angle GEF is equal to the angle EFG, so that the side EG is also equal to GF. [I. 6] Again, since the angle at B is half a right angle, and the angle FDB is right, for it is again equal to the interior and opposite angle ECB, [I. 29] the remaining angle BFD is half a right angle; [I. 32]therefore the angle at B is equal to the angle DFB, so that the side FD is also equal to the side DB. [I. 6] Now, since AC is equal to CE, the square on AC is also equal to the square on CE; therefore the squares on AC, CE are double of the square on AC. But the square on EA is equal to the squares on AC, CE, for the angle ACE is right; [I. 47] therefore the square on EA is double of the square on AC. Again, since EG is equal to GF, the square on EG is also equal to the square on GF; therefore the squares on EG, GF are double of the square on GF. But the square on EF is equal to the squares on EG, GF; therefore the square on EF is double of the square on GF. But GF is equal to CD; [I. 34] therefore the square on EF is double of the square on CD. But the square on EA is also double of the square on AC; therefore the squares on AE, EF are double of the squares on AC, CD. And the square on AF is equal to the squares on AE, EF, for the angle AEF is right; [I. 47] therefore the square on AF is double of the squares on AC, CD. But the squares on AD, DF are equal to the square on AF, for the angle at D is right; [I. 47] therefore the squares on AD, DF are double of the squares on AC, CD. And DF is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD."", ""ProofWordCount"" -> 549, ""GreekProof"" -> ""εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω εἰς μὲν ἴσα κατὰ τὸ Γ, εἰς δὲ ἄνισα κατὰ τὸ Δ: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἤχθω γὰρ ἀπὸ τοῦ Γ τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΕ, καὶ κείσθω ἴση ἑκατέρᾳ τῶν ΑΓ, ΓΒ, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, ΕΒ, καὶ διὰ μὲν τοῦ Δ τῇ ΕΓ παράλληλος ἤχθω ἡ ΔΖ, διὰ δὲ τοῦ Ζ τῇ ΑΒ ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΑΖ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΕ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΕΑΓ γωνία τῇ ὑπὸ ΑΕΓ. καὶ ἐπεὶ ὀρθή ἐστιν ἡ πρὸς τῷ Γ, λοιπαὶ ἄρα αἱ ὑπὸ ΕΑΓ, ΑΕΓ μιᾷ ὀρθῇ ἴσαι εἰσίν: καί εἰσιν ἴσαι: ἡμίσεια ἄρα ὀρθῆς ἐστιν ἑκατέρα τῶν ὑπὸ ΓΕΑ, ΓΑΕ. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΓΕΒ, ΕΒΓ ἡμίσειά ἐστιν ὀρθῆς: ὅλη ἄρα ἡ ὑπὸ ΑΕΒ ὀρθή ἐστιν. καὶ ἐπεὶ ἡ ὑπὸ ΗΕΖ ἡμίσειά ἐστιν ὀρθῆς, ὀρθὴ δὲ ἡ ὑπὸ ΕΗΖ: ἴση γάρ ἐστι τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΕΓΒ: λοιπὴ ἄρα ἡ ὑπὸ ΕΖΗ ἡμίσειά ἐστιν ὀρθῆς: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΗΕΖ γωνία τῇ ὑπὸ ΕΖΗ: ὥστε καὶ πλευρὰ ἡ ΕΗ τῇ ΗΖ ἐστιν ἴση. πάλιν ἐπεὶ ἡ πρὸς τῷ β γωνία ἡμίσειά ἐστιν ὀρθῆς, ὀρθὴ δὲ ἡ ὑπὸ ΖΔΒ: ἴση γὰρ πάλιν ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΕΓΒ: λοιπὴ ἄρα ἡ ὑπὸ ΒΖΔ ἡμίσειά ἐστιν ὀρθῆς: ἴση ἄρα ἡ πρὸς τῷ Β γωνία τῇ ὑπὸ ΔΖΒ: ὥστε καὶ πλευρὰ ἡ ΖΔ πλευρᾷ τῇ ΔΒ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΕ, ἴσον ἐστὶ καὶ τὸ ἀπὸ ΑΓ τῷ ἀπὸ ΓΕ: τὰ ἄρα ἀπὸ τῶν ΑΓ, ΓΕ τετράγωνα διπλάσιά ἐστι τοῦ ἀπὸ ΑΓ. τοῖς δὲ ἀπὸ τῶν ΑΓ, ΓΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΑ τετράγωνον: ὀρθὴ γὰρ ἡ ὑπὸ ΑΓΕ γωνία: τὸ ἄρα ἀπὸ τῆς ΕΑ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΑΓ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΕΗ τῇ ΗΖ, ἴσον καὶ τὸ ἀπὸ τῆς ΕΗ τῷ ἀπὸ τῆς ΗΖ: τὰ ἄρα ἀπὸ τῶν ΕΗ, ΗΖ τετράγωνα διπλάσιά ἐστι τοῦ ἀπὸ τῆς ΗΖ τετραγώνου. τοῖς δὲ ἀπὸ τῶν ΕΗ, ΗΖ τετραγώνοις ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΖ τετράγωνον: τὸ ἄρα ἀπὸ τῆς ΕΖ τετράγωνον διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΗΖ. ἴση δὲ ἡ ΗΖ τῇ ΓΔ: τὸ ἄρα ἀπὸ τῆς ΕΖ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΓΔ. ἔστι δὲ καὶ τὸ ἀπὸ τῆς ΕΑ διπλάσιον τοῦ ἀπὸ τῆς ΑΓ: τὰ ἄρα ἀπὸ τῶν ΑΕ, ΕΖ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. τοῖς δὲ ἀπὸ τῶν ΑΕ, ΕΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΑΖ τετράγωνον: ὀρθὴ γάρ ἐστιν ἡ ὑπὸ ΑΕΖ γωνία: τὸ ἄρα ἀπὸ τῆς ΑΖ τετράγωνον διπλάσιόν ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ. τῷ δὲ ἀπὸ τῆς ΑΖ ἴσα τὰ ἀπὸ τῶν ΑΔ, ΔΖ: ὀρθὴ γὰρ ἡ πρὸς τῷ Δ γωνία: τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΖ διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἴση δὲ ἡ ΔΖ τῇ ΔΒ: τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ εἰς ἴσα καὶ ἄνισα, τὰ ἀπὸ τῶν ἀνίσων τῆς ὅλης τμημάτων τετράγωνα διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς μεταξὺ τῶν τομῶν τετραγώνου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 534|>","<|""VertexLabel"" -> ""2.10"", ""Text"" -> ""If a straight line be bisected, and a straight line be added to it in a straight line, the square on the whole with the added straight line and the square on the added straight line both together are double of the square on the half and of the square described on the straight line made up of the half and the added straight line as on one straight line."", ""TextWordCount"" -> 70, ""GreekText"" -> ""ἐὰν εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας, τὸ ἀπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ καὶ τὸ ἀπὸ τῆς προσκειμένης τὰ συναμφότερα τετράγωνα διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης ὡς ἀπὸ μιᾶς ἀναγραφέντος τετραγώνου."", ""GreekTextWordCount"" -> 52, ""References"" -> {{""Common Notion"" -> 1}, {""Postulate"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 6}, {""Book"" -> 1, ""Theorem"" -> 11}, {""Book"" -> 1, ""Theorem"" -> 15}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 1, ""Theorem"" -> 47}}, ""Proof"" -> ""For let a straight line AB be bisected at C, and let a straight line BD be added to it in a straight line; I say that the squares on AD, DB are double of the squares on AC, CD. For let CE be drawn from the point C at right angles to AB [I. 11], and let it be made equal to either AC or CB [I. 3]; let EA, EB be joined; through E let EF be drawn parallel to AD, and through D let FD be drawn parallel to CE. [I. 31] Then, since a straight line EF falls on the parallel straight lines EC, FD, the angles CEF, EFD are equal to two right angles; [I. 29]therefore the angles FEB, EFD are less than two right angles. But straight lines produced from angles less than two right angles meet; [I. Post. 5] therefore EB, FD, if produced in the direction B, D, will meet. Let them be produced and meet at G, and let AG be joined. Then, since AC is equal to CE, the angle EAC is also equal to the angle AEC; [I. 5] and the angle at C is right; therefore each of the angles EAC, AEC is half a right angle. [I. 32] For the same reason each of the angles CEB, EBC is also half a right angle; therefore the angle AEB is right. And, since the angle EBC is half a right angle, the angle DBG is also half a right angle. [I. 15] But the angle BDG is also right, for it is equal to the angle DCE, they being alternate; [I. 29] therefore the remaining angle DGB is half a right angle; [I. 32]therefore the angle DGB is equal to the angle DBG, so that the side BD is also equal to the side GD. [I. 6] Again, since the angle EGF is half a right angle, and the angle at F is right, for it is equal to the opposite angle, the angle at C, [I. 34] the remaining angle FEG is half a right angle; [I. 32] therefore the angle EGF is equal to the angle FEG, so that the side GF is also equal to the side EF. [I. 6] Now, since the square on EC is equal to the square on CA, the squares on EC, CA are double of the square on CA. But the square on EA is equal to the squares on EC, CA; [I. 47] therefore the square on EA is double of the square on AC. [C. N. 1] Again, since FG is equal to EF, the square on FG is also equal to the square on FE; therefore the squares on GF, FE are double of the square on EF. But the square on EG is equal to the squares on GF, FE; [I. 47] therefore the square on EG is double of the square on EF. And EF is equal to CD; [I. 34] therefore the square on EG is double of the square on CD. But the square on EA was also proved double of the square on AC; therefore the squares on AE, EG are double of the squares on AC, CD. And the square on AG is equal to the squares on AE, EG; [I. 47] therefore the square on AG is double of the squares on AC, CD. But the squares on AD, DG are equal to the square on AG; [I. 47] therefore the squares on AD, DG are double of the squares on AC, CD. And DG is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD."", ""ProofWordCount"" -> 616, ""GreekProof"" -> ""εὐθεῖα γάρ τις ἡ ΑΒ τετμήσθω δίχα κατὰ τὸ Γ, προσκείσθω δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας ἡ ΒΔ: λέγω, ὅτι τὰ ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἤχθω γὰρ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ πρὸς ὀρθὰς ἡ ΓΕ, καὶ κείσθω ἴση ἑκατέρᾳ, τῶν ΑΓ, ΓΒ, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, ΕΒ: καὶ διὰ μὲν τοῦ Ε τῇ ΑΔ παράλληλος ἤχθω ἡ ΕΖ, διὰ δὲ τοῦ Δ τῇ ΓΕ παράλληλος ἤχθω ἡ ΖΔ. καὶ ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΕΓ, ΖΔ εὐθεῖά τις ἐνέπεσεν ἡ ΕΖ, αἱ ὑπὸ ΓΕΖ, ΕΖΔ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν: αἱ ἄρα ὑπὸ ΖΕΒ, ΕΖΔ δύο ὀρθῶν ἐλάσσονές εἰσιν: αἱ δὲ ἀπ᾽ ἐλασσόνων ἢ δύο ὀρθῶν ἐκβαλλόμεναι συμπίπτουσιν: αἱ ἄρα ΕΒ, ΖΔ ἐκβαλλόμεναι ἐπὶ τὰ Β, Δ μέρη συμπεσοῦνται. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν κατὰ τὸ Η, καὶ ἐπεζεύχθω ἡ ΑΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΕ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΕΑΓ τῇ ὑπὸ ΑΕΓ: καὶ ὀρθὴ ἡ πρὸς τῷ Γ: ἡμίσεια ἄρα ὀρθῆς ἐστιν ἑκατέρα τῶν ὑπὸ ΕΑΓ, ΑΕΓ. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ὑπὸ ΓΕΒ, ΕΒΓ ἡμίσειά ἐστιν ὀρθῆς: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ. καὶ ἐπεὶ ἡμίσεια ὀρθῆς ἐστιν ἡ ὑπὸ ΕΒΓ, ἡμίσεια ἄρα ὀρθῆς καὶ ἡ ὑπὸ ΔΒΗ. ἔστι δὲ καὶ ἡ ὑπὸ ΒΔΗ ὀρθή: ἴση γάρ ἐστι τῇ ὑπὸ ΔΓΕ: ἐναλλὰξ γάρ: λοιπὴ ἄρα ἡ ὑπὸ ΔΗΒ ἡμίσειά ἐστιν ὀρθῆς: ἡ ἄρα ὑπὸ ΔΗΒ τῇ ὑπὸ ΔΒΗ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΒΔ πλευρᾷ τῇ ΗΔ ἐστιν ἴση. πάλιν, ἐπεὶ ἡ ὑπὸ ΕΗΖ ἡμίσειά ἐστιν ὀρθῆς, ὀρθὴ δὲ ἡ πρὸς τῷ Ζ: ἴση γάρ ἐστι τῇ ἀπεναντίον τῇ πρὸς τῷ Γ: λοιπὴ ἄρα ἡ ὑπὸ ΖΕΗ ἡμίσειά ἐστιν ὀρθῆς: ἴση ἄρα ἡ ὑπὸ ΕΗΖ γωνία τῇ ὑπὸ ΖΕΗ: ὥστε καὶ πλευρὰ ἡ ΗΖ πλευρᾷ τῇ ΕΖ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΕΓ τῇ ΓΑ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΕΓ τετράγωνον τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ: τὰ ἄρα ἀπὸ τῶν ΕΓ, ΓΑ τετράγωνα διπλάσιά ἐστι τοῦ ἀπὸ τῆς ΓΑ τετραγώνου. τοῖς δὲ ἀπὸ τῶν ΕΓ, ΓΑ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΑ: τὸ ἄρα ἀπὸ τῆς ΕΑ τετράγωνον διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΑΓ τετραγώνου. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΖΗ τῇ ΕΖ, ἴσον ἐστὶ καὶ τὸ ἀπὸ τῆς ΖΗ τῷ ἀπὸ τῆς ΖΕ: τὰ ἄρα ἀπὸ τῶν ΗΖ, ΖΕ διπλάσιά ἐστι τοῦ ἀπὸ τῆς ΕΖ. τοῖς δὲ ἀπὸ τῶν ΗΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΗ: τὸ ἄρα ἀπὸ τῆς ΕΗ διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΕΖ. ἴση δὲ ἡ ΕΖ τῇ ΓΔ: τὸ ἄρα ἀπὸ τῆς ΕΗ τετράγωνον διπλάσιόν ἐστι τοῦ ἀπὸ τῆς ΓΔ. ἐδείχθη δὲ καὶ τὸ ἀπὸ τῆς ΕΑ διπλάσιον τοῦ ἀπὸ τῆς ΑΓ: τὰ ἄρα ἀπὸ τῶν ΑΕ, ΕΗ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. τοῖς δὲ ἀπὸ τῶν ΑΕ, ΕΗ τετραγώνοις ἴσον ἐστὶ τὸ ἀπὸ τῆς ΑΗ τετράγωνον: τὸ ἄρα ἀπὸ τῆς ΑΗ διπλάσιόν ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ. τῷ δὲ ἀπὸ τῆς ΑΗ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΑΔ, ΔΗ: τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΗ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἴση δὲ ἡ ΔΗ τῇ ΔΒ: τὰ ἄρα ἀπὸ τῶν ΑΔ, ΔΒ τετράγωνα διπλάσιά ἐστι τῶν ἀπὸ τῶν ΑΓ, ΓΔ τετραγώνων. ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ δίχα, προστεθῇ δέ τις αὐτῇ εὐθεῖα ἐπ᾽ εὐθείας, τὸ ἀπὸ τῆς ὅλης σὺν τῇ προσκειμένῃ καὶ τὸ ἀπὸ τῆς προσκειμένης τὰ συναμφότερα τετράγωνα διπλάσιά ἐστι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ τοῦ ἀπὸ τῆς συγκειμένης ἔκ τε τῆς ἡμισείας καὶ τῆς προσκειμένης ὡς ἀπὸ μιᾶς ἀναγραφέντος τετραγώνου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 591|>","<|""VertexLabel"" -> ""2.11"", ""Text"" -> ""To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment."", ""TextWordCount"" -> 28, ""GreekText"" -> ""τὴν δοθεῖσαν εὐθεῖαν τεμεῖν ὥστε τὸ ὑπὸ τῆς ὅλης καὶ τοῦ ἑτέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ."", ""GreekTextWordCount"" -> 24, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 46}, {""Book"" -> 1, ""Theorem"" -> 47}, {""Book"" -> 2, ""Theorem"" -> 6}}, ""Proof"" -> ""Let AB be the given straight line; thus it is required to cut AB so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment. For let the square ABDC be described on AB; [I. 46] let AC be bisected at the point E, and let BE be joined; let CA be drawn through to F, and let EF be made equal to BE; let the square FH be described on AF, and let GH be drawn through to K. I say that AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on AH. For, since the straight line AC has been bisected at E, and FA is added to it, the rectangle contained by CF, FA together with the square on AE is equal to the square on EF. [II. 6] But EF is equal to EB; therefore the rectangle CF, FA together with the square on AE is equal to the square on EB. But the squares on BA, AE are equal to the square on EB, for the angle at A is right; [I. 47] therefore the rectangle CF, FA together with the square on AE is equal to the squares on BA, AE. Let the square on AE be subtracted from each; therefore the rectangle CF, FA which remains is equal to the square on AB. Now the rectangle CF, FA is FK, for AF is equal to FG; and the square on AB is AD; therefore FK is equal to AD. Let AK be subtracted from each; therefore FH which remains is equal to HD. And HD is the rectangle AB, BH, for AB is equal to BD; and FH is the square on AH; therefore the rectangle contained by AB, BH is equal to the square on HA. therefore the given straight line AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on HA. Q. E. F."", ""ProofWordCount"" -> 346, ""GreekProof"" -> ""ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ: δεῖ δὴ τὴν ΑΒ τεμεῖν ὥστε τὸ ὑπὸ τῆς ὅλης καὶ τοῦ ἑτέρου τῶν τμημάτων περιεχόμενον ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τοῦ λοιποῦ τμήματος τετραγώνῳ. Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΒΔΓ, καὶ τετμήσθω ἡ ΑΓ δίχα κατὰ τὸ Ε σημεῖον, καὶ ἐπεζεύχθω ἡ ΒΕ, καὶ διήχθω ἡ ΓΑ ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΒΕ ἴση ἡ ΕΖ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΑΖ τετράγωνον τὸ ΖΘ, καὶ διήχθω ἡ ΗΘ ἐπὶ τὸ Κ: λέγω, ὅτι ἡ ΑΒ τέτμηται κατὰ τὸ Θ, ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΘ περιεχόμενον ὀρθογώνιον ἴσον ποιεῖν τῷ ἀπὸ τῆς ΑΘ τετραγώνῳ. ἐπεὶ γὰρ εὐθεῖα ἡ ΑΓ τέτμηται δίχα κατὰ τὸ Ε, πρόσκειται δὲ αὐτῇ ἡ ΖΑ, τὸ ἄρα ὑπὸ τῶν ΓΖ, ΖΑ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΑΕ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΖ τετραγώνῳ. ἴση δὲ ἡ ΕΖ τῇ ΕΒ: τὸ ἄρα ὑπὸ τῶν ΓΖ, ΖΑ μετὰ τοῦ ἀπὸ τῆς ΑΕ ἴσον ἐστὶ τῷ ἀπὸ ΕΒ. ἀλλὰ τῷ ἀπὸ ΕΒ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΒΑ, ΑΕ: ὀρθὴ γὰρ ἡ πρὸς τῷ Α γωνία: τὸ ἄρα ὑπὸ τῶν ΓΖ, ΖΑ μετὰ τοῦ ἀπὸ τῆς ΑΕ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΕ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΑΕ: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΓΖ, ΖΑ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ τετραγώνῳ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΓΖ, ΖΑ τὸ ΖΚ: ἴση γὰρ ἡ ΑΖ τῇ ΖΗ: τὸ δὲ ἀπὸ τῆς ΑΒ τὸ ΑΔ: τὸ ἄρα ΖΚ ἴσον ἐστὶ τῷ ΑΔ. κοινὸν ἀφῃρήσθω τὸ ΑΚ: λοιπὸν ἄρα τὸ ΖΘ τῷ ΘΔ ἴσον ἐστίν. καί ἐστι τὸ μὲν ΘΔ τὸ ὑπὸ τῶν ΑΒ, ΒΘ: ἴση γὰρ ἡ ΑΒ τῇ ΒΔ: τὸ δὲ ΖΘ τὸ ἀπὸ τῆς ΑΘ: τὸ ἄρα ὑπὸ τῶν ΑΒ, ΒΘ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ ΘΑ τετραγώνῳ. ἡ ἄρα δοθεῖσα εὐθεῖα ἡ ΑΒ τέτμηται κατὰ τὸ Θ ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΘ περιεχόμενον ὀρθογώνιον ἴσον ποιεῖν τῷ ἀπὸ τῆς ΘΑ τετραγώνῳ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 327|>","<|""VertexLabel"" -> ""2.12"", ""Text"" -> ""In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the perpendicular falls, and the straight line cut off outside by the perpendicular towards the obtuse angle."", ""TextWordCount"" -> 59, ""GreekText"" -> ""ἐν τοῖς ἀμβλυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ἀμβλεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον μεῖζόν ἐστι τῶν ἀπὸ τῶν τὴν ἀμβλεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ἀμβλεῖαν γωνίαν, ἐφ᾽ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐκτὸς ὑπὸ τῆς καθέτου πρὸς τῇ ἀμβλείᾳ γωνίᾳ."", ""GreekTextWordCount"" -> 52, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 47}, {""Book"" -> 2, ""Theorem"" -> 4}}, ""Proof"" -> ""Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and let BD be drawn from the point B perpendicular to CA produced; I say that the square on BC is greater than the squares on BA, AC by twice the rectangle contained by CA, AD. For, since the straight line CD has been cut at random at the point A, the square on DC is equal to the squares on CA, AD and twice the rectangle contained by CA, AD. [II. 4] Let the square on DB be added to each; therefore the squares on CD, DB are equal to the squares on CA, AD, DB and twice the rectangle CA, AD. But the square on CB is equal to the squares on CD, DB, for the angle at D is right; [I. 47] and the square on AB is equal to the squares on AD, DB; [I. 47]therefore the square on CB is equal to the squares on CA, AB and twice the rectangle contained by CA, AD; so that the square on CB is greater than the squares on CA, AB by twice the rectangle contained by CA, AD."", ""ProofWordCount"" -> 194, ""GreekProof"" -> ""ἔστω ἀμβλυγώνιον τρίγωνον τὸ ΑΒΓ ἀμβλεῖαν ἔχον τὴν ὑπὸ ΒΑΓ, καὶ ἤχθω ἀπὸ τοῦ Β σημείου ἐπὶ τὴν ΓΑ ἐκβληθεῖσαν κάθετος ἡ ΒΔ. λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ τετράγωνον μεῖζόν ἐστι τῶν ἀπὸ τῶν ΒΑ, ΑΓ τετραγώνων τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. ἐπεὶ γὰρ εὐθεῖα ἡ ΓΑ τέτμηται, ὡς ἔτυχεν, κατὰ τὸ Α σημεῖον, τὸ ἄρα ἀπὸ τῆς ΔΓ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΓΑ, ΑΔ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΔΒ: τὰ ἄρα ἀπὸ τῶν ΓΔ, ΔΒ ἴσα ἐστὶ τοῖς τε ἀπὸ τῶν ΓΑ, ΑΔ, ΔΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΓΔ, ΔΒ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΓΒ: ὀρθὴ γὰρ ἡ πρὸς τῷ Δ γωνία: τοῖς δὲ ἀπὸ τῶν ΑΔ, ΔΒ ἴσον τὸ ἀπὸ τῆς ΑΒ: τὸ ἄρα ἀπὸ τῆς ΓΒ τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΓΑ, ΑΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ: ὥστε τὸ ἀπὸ τῆς ΓΒ τετράγωνον τῶν ἀπὸ τῶν ΓΑ, ΑΒ τετραγώνων μεῖζόν ἐστι τῷ δὶς ὑπὸ τῶν ΓΑ, ΑΔ περιεχομένῳ ὀρθογωνίῳ. ἐν ἄρα τοῖς ἀμβλυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ἀμβλεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον μεῖζόν ἐστι τῶν ἀπὸ τῶν τὴν ἀμβλεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ἀμβλεῖαν γωνίαν, ἐφ᾽ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐκτὸς ὑπὸ τῆς καθέτου πρὸς τῇ ἀμβλείᾳ γωνίᾳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 244|>","<|""VertexLabel"" -> ""2.13"", ""Text"" -> ""In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off within by the perpendicular towards the acute angle."", ""TextWordCount"" -> 59, ""GreekText"" -> ""ἐν τοῖς ὀξυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀξεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι τῶν ἀπὸ τῶν τὴν ὀξεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ὀξεῖαν γωνίαν, ἐφ᾽ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐντὸς ὑπὸ τῆς καθέτου πρὸς τῇ ὀξείᾳ γωνίᾳ."", ""GreekTextWordCount"" -> 52, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 47}, {""Book"" -> 2, ""Theorem"" -> 7}}, ""Proof"" -> ""Let ABC be an acute-angled triangle having the angle at B acute, and let AD be drawn from the point A perpendicular to BC; I say that the square on AC is less than the squares on CB, BA by twice the rectangle contained by CB, BD. For, since the straight line CB has been cut at random at D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on DC. [II. 7] Let the square on DA be added to each; therefore the squares on CB, BD, DA are equal to twice the rectangle contained by CB, BD and the squares on AD, DC. But the square on AB is equal to the squares on BD, DA, for the angle at D is right; [I. 47] and the square on AC is equal to the squares on AD, DC; therefore the squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD, so that the square on AC alone is less than the squares on CB, BA by twice the rectangle contained by CB, BD."", ""ProofWordCount"" -> 191, ""GreekProof"" -> ""ἔστω ὀξυγώνιον τρίγωνον τὸ ΑΒΓ ὀξεῖαν ἔχον τὴν πρὸς τῷ Β γωνίαν, καὶ ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὴν ΒΓ κάθετος ἡ ΑΔ: λέγω, ὅτι τὸ ἀπὸ τῆς ΑΓ τετράγωνον ἔλαττόν ἐστι τῶν ἀπὸ τῶν ΓΒ, ΒΑ τετραγώνων τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ. ἐπεὶ γὰρ εὐθεῖα ἡ ΓΒ τέτμηται, ὡς ἔτυχεν, κατὰ τὸ Δ, τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΔ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΔΓ τετραγώνῳ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΔΑ τετράγωνον: τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΔ, ΔΑ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τοῖς ἀπὸ τῶν ΑΔ, ΔΓ τετραγώνοις. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΒΔ, ΔΑ ἴσον τὸ ἀπὸ τῆς ΑΒ: ὀρθὴ γὰρ ἡ πρὸς τῷ Δ γωνίᾳ: τοῖς δὲ ἀπὸ τῶν ΑΔ, ΔΓ ἴσον τὸ ἀπὸ τῆς ΑΓ: τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΑ ἴσα ἐστὶ τῷ τε ἀπὸ τῆς ΑΓ καὶ τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ: ὥστε μόνον τὸ ἀπὸ τῆς ΑΓ ἔλαττόν ἐστι τῶν ἀπὸ τῶν ΓΒ, ΒΑ τετραγώνων τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ. ἐν ἄρα τοῖς ὀξυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀξεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι τῶν ἀπὸ τῶν τὴν ὀξεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ὀξεῖαν γωνίαν, ἐφ᾽ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐντὸς ὑπὸ τῆς καθέτου πρὸς τῇ ὀξείᾳ γωνίᾳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 243|>","<|""VertexLabel"" -> ""2.14"", ""Text"" -> ""To construct a square equal to a given rectilineal figure."", ""TextWordCount"" -> 10, ""GreekText"" -> ""τῷ δοθέντι εὐθυγράμμῳ ἴσον τετράγωνον συστήσασθαι."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 45}, {""Book"" -> 1, ""Theorem"" -> 47}, {""Book"" -> 2, ""Theorem"" -> 5}}, ""Proof"" -> ""Let A be the given rectilineal figure; thus it is required to construct a square equal to the rectilineal figure A. For let there be constructed the rectangular parallelogram BD equal to the rectilineal figure A. [I. 45] Then, if BE is equal to ED, that which was enjoined will have been done; for a square BD has been constructed equal to the rectilineal figure A. But, if not, one of the straight lines BE, ED is greater. Let BE be greater, and let it be produced to F; let EF be made equal to ED, and let BF be bisected at G. With centre G and distance one of the straight lines GB, GF let the semicircle BHF be described; let DE be produced to H, and let GH be joined. Then, since the straight line BF has been cut into equal segments at G, and into unequal segments at E, the rectangle contained by BE, EF together with the square on EG is equal to the square on GF. [II. 5] But GF is equal to GH; therefore the rectangle BE, EF together with the square on GE is equal to the square on GH. But the squares on HE, EG are equal to the square on GH; [I. 47] therefore the rectangle BE, EF together with the square on GE is equal to the squares on HE, EG. Let the square on GE be subtracted from each; therefore the rectangle contained by BE, EF which remains is equal to the square on EH. But the rectangle BE, EF is BD, for EF is equal to ED; therefore the parallelogram BD is equal to the square on HE. And BD is equal to the rectilineal figure A. Therefore the rectilineal figure A is also equal to the square which can be described on EH. Therefore a square, namely that which can be described on EH, has been constructed equal to the given rectilineal figure A."", ""ProofWordCount"" -> 327, ""GreekProof"" -> ""ἔστω τὸ δοθὲν εὐθύγραμμον τὸ Α: δεῖ δὴ τῷ Α εὐθυγράμμῳ ἴσον τετράγωνον συστήσασθαι. συνεστάτω γὰρ τῷ Α εὐθυγράμμῳ ἴσον παραλληλόγραμμον ὀρθογώνιον τὸ ΒΔ: εἰ μὲν οὖν ἴση ἐστὶν ἡ ΒΕ τῇ ΕΔ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. συνέσταται γὰρ τῷ Α εὐθυγράμμῳ ἴσον τετράγωνον τὸ ΒΔ: εἰ δὲ οὔ, μία τῶν ΒΕ, ΕΔ μείζων ἐστίν. ἔστω μείζων ἡ ΒΕ, καὶ ἐκβεβλήσθω ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΕΔ ἴση ἡ ΕΖ, καὶ τετμήσθω ἡ ΒΖ δίχα κατὰ τὸ Η, καὶ κέντρῳ τῷ Η, διαστήματι δὲ ἑνὶ τῶν ΗΒ, ΗΖ ἡμικύκλιον γεγράφθω τὸ ΒΘΖ, καὶ ἐκβεβλήσθω ἡ ΔΕ ἐπὶ τὸ Θ, καὶ ἐπεζεύχθω ἡ ΗΘ. ἐπεὶ οὖν εὐθεῖα ἡ ΒΖ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Η, εἰς δὲ ἄνισα κατὰ τὸ Ε, τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΕΗ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΖ τετραγώνῳ. ἴση δὲ ἡ ΗΖ τῇ ΗΘ: τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ μετὰ τοῦ ἀπὸ τῆς ΗΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΘ. τῷ δὲ ἀπὸ τῆς ΗΘ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΘΕ, ΕΗ τετράγωνα: τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ μετὰ τοῦ ἀπὸ ΗΕ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΘΕ, ΕΗ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΗΕ τετράγωνον: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΒΕ, ΕΖ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ τετραγώνῳ. ἀλλὰ τὸ ὑπὸ τῶν ΒΕ, ΕΖ τὸ ΒΔ ἐστιν: ἴση γὰρ ἡ ΕΖ τῇ ΕΔ: τὸ ἄρα ΒΔ παραλληλόγραμμον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΘΕ τετραγώνῳ. ἴσον δὲ τὸ ΒΔ τῷ Α εὐθυγράμμῳ. καὶ τὸ Α ἄρα εὐθύγραμμον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ ἀναγραφησομένῳ τετραγώνῳ. τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ Α ἴσον τετράγωνον συνέσταται τὸ ἀπὸ τῆς ΕΘ ἀναγραφησόμενον: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 283|>","<|""VertexLabel"" -> ""3.1"", ""Text"" -> ""To find the centre of a given circle."", ""TextWordCount"" -> 8, ""GreekText"" -> ""τοῦ δοθέντος κύκλου τὸ κέντρον εὑρεῖν."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Definition"" -> 10}, {""Book"" -> 1, ""Theorem"" -> 8}}, ""Proof"" -> ""Let ABC be the given circle; thus it is required to find the centre of the circle ABC. Let a straight line AB be drawnthrough it at random, and let it be bisected at the point D; from D let DC be drawn at right angles to AB and let it be drawn through to E; let CE be bisected at F;I say that F is the centre of the circle ABC. For suppose it is not, but, if possible, let G be the centre, and let GA, GD, GB be joined. Then, since AD is equal to DB, and DG is common, the two sides AD, DG are equal to the two sides BD, DG respectively; and the base GA is equal to the base GB, for they areradii; therefore the angle ADG is equal to the angle GDB. [I. 8] But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10]therefore the angle GDB is right. But the angle FDB is also right; therefore the angle FDB is equal to the angle GDB, the greater to the less: which is impossible. Therefore G is not the centre of the circle ABC. Similarly we can prove that neither is any other point except F. Therefore the point F is the centre of the circle ABC."", ""ProofWordCount"" -> 236, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ: δεῖ δὴ τοῦ ΑΒΓ κύκλου τὸ κέντρον εὑρεῖν. διήχθω τις εἰς αὐτόν, ὡς ἔτυχεν, εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω δίχα κατὰ τὸ Δ σημεῖον, καὶ ἀπὸ τοῦ Δ τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΔΓ καὶ διήχθω ἐπὶ τὸ Ε, καὶ τετμήσθω ἡ ΓΕ δίχα κατὰ τὸ Ζ: λέγω, ὅτι τὸ Ζ κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἔστω τὸ Η, καὶ ἐπεζεύχθωσαν αἱ ΗΑ, ΗΔ, ΗΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ ἡ ΔΗ, δύο δὴ αἱ ΑΔ, ΔΗ δύο ταῖς ΗΔ, ΔΒ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ βάσις ἡ ΗΑ βάσει τῇ ΗΒ ἐστιν ἴση: ἐκ κέντρου γάρ: γωνία ἄρα ἡ ὑπὸ ΑΔΗ γωνίᾳ τῇ ὑπὸ ΗΔΒ ἴση ἐστίν. ὅταν δὲ εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΗΔΒ. ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΔΒ ὀρθή: ἴση ἄρα ἡ ὑπὸ ΖΔΒ τῇ ὑπὸ ΗΔΒ, ἡ μείζων τῇ ἐλάττονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Η κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλο τι πλὴν τοῦ Ζ. τὸ Ζ ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἐν κύκλῳ εὐθεῖά τις εὐθεῖάν τινα δίχα καὶ πρὸς ὀρθὰς τέμνῃ, ἐπὶ τῆς τεμνούσης ἐστὶ τὸ κέντρον τοῦ κύκλου: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 231|>","<|""VertexLabel"" -> ""3.2"", ""Text"" -> ""If on the circumference of a circle two points be taken at random, the straight line joining the points will fall within the circle."", ""TextWordCount"" -> 24, ""GreekText"" -> ""ἐὰν κύκλου ἐπὶ τῆς περιφερείας ληφθῇ δύο τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ κύκλου."", ""GreekTextWordCount"" -> 19, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 16}, {""Book"" -> 1, ""Theorem"" -> 19}, {""Book"" -> 3, ""Theorem"" -> 1}}, ""Proof"" -> ""Let ABC be a circle, and let two points A, B be taken at random on its circumference; I say that the straight line joined from A to B will fall within the circle. For suppose it does not, but, if possible, let it fall outside, as AEB; let the centre of the circle ABC be taken [III. 1], and let it be D; let DA, DB be joined, and let DFE be drawn through. Then, since DA is equal to DB, the angle DAE is also equal to the angle DBE. [I. 5]And, since one side AEB of the triangle DAE is produced, the angle DEB is greater than the angle DAE. [I. 16] But the angle DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE. And the greater angle is subtended by the greater side; [I. 19] therefore DB is greater than DE. But DB is equal to DF; therefore DF is greater than DE, the less than the greater: which is impossible. Therefore the straight line joined from A to B will not fall outside the circle. Similarly we can prove that neither will it fall on the circumference itself; therefore it will fall within."", ""ProofWordCount"" -> 207, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓ, καὶ ἐπὶ τῆς περιφερείας αὐτοῦ εἰλήφθω δύο τυχόντα σημεῖα τὰ Α, Β: λέγω, ὅτι ἡ ἀπὸ τοῦ Α ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ κύκλου. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, πιπτέτω ἐκτὸς ὡς ἡ ΑΕΒ, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ δ, καὶ ἐπεζεύχθωσαν αἱ ΔΑ, ΔΒ, καὶ διήχθω ἡ ΔΖΕ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΔΒ, ἴση ἄρα καὶ γωνία ἡ ὑπὸ ΔΑΕ τῇ ὑπὸ ΔΒΕ: καὶ ἐπεὶ τριγώνου τοῦ ΔΑΕ μία πλευρὰ προσεκβέβληται ἡ ΑΕΒ, μείζων ἄρα ἡ ὑπὸ ΔΕΒ γωνία τῆς ὑπὸ ΔΑΕ. ἴση δὲ ἡ ὑπὸ ΔΑΕ τῇ ὑπὸ ΔΒΕ: μείζων ἄρα ἡ ὑπὸ ΔΕΒ τῆς ὑπὸ ΔΒΕ. ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει: μείζων ἄρα ἡ ΔΒ τῆς ΔΕ. ἴση δὲ ἡ ΔΒ τῇ ΔΖ. μείζων ἄρα ἡ ΔΖ τῆς ΔΕ ἡ ἐλάττων τῆς μείζονος: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Α ἐπὶ τὸ Β ἐπιζευγνυμένη εὐθεῖα ἐκτὸς πεσεῖται τοῦ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ ἐπ᾽ αὐτῆς τῆς περιφερείας: ἐντὸς ἄρα. ἐὰν ἄρα κύκλου ἐπὶ τῆς περιφερείας ληφθῇ δύο τυχόντα σημεῖα, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς πεσεῖται τοῦ κύκλου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 199|>","<|""VertexLabel"" -> ""3.3"", ""Text"" -> ""If in a circle a straight line through the centre bisect a straight line not through the centre, it also cuts it at right angles; and if it cut it at right angles, it also bisects it."", ""TextWordCount"" -> 37, ""GreekText"" -> ""ἐὰν ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν τέμνει."", ""GreekTextWordCount"" -> 31, ""References"" -> {{""Book"" -> 1, ""Definition"" -> 10}, {""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 8}, {""Book"" -> 1, ""Theorem"" -> 26}}, ""Proof"" -> ""Let ABC be a circle, and in it let a straight line CD through the centre bisect a straight line AB not through the centre at the point F; I say that it also cuts it at right angles. For let the centre of the circle ABC be taken, and let it be E; let EA, EB be joined. Then, since AF is equal to FB, and FE is common, two sides are equal to two sides; and the base EA is equal to the base EB; therefore the angle AFE is equal to the angle BFE. [I. 8] But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10]therefore each of the angles AFE, BFE is right. Therefore CD, which is through the centre, and bisects AB which is not through the centre, also cuts it at right angles. Again, let CD cut AB at right angles;I say that it also bisects it. that is, that AF is equal to FB. For, with the same construction, since EA is equal to EB, the angle EAF is also equal to the angle EBF. [I. 5] But the right angle AFE is equal to the right angle BFE,therefore EAF, EBF are two triangles having two angles equal to two angles and one side equal to one side, namely EF, which is common to them, and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26]therefore AF is equal to FB."", ""ProofWordCount"" -> 272, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓ, καὶ ἐν αὐτῷ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΓΔ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΒ δίχα τεμνέτω κατὰ τὸ Ζ σημεῖον: λέγω, ὅτι καὶ πρὸς ὀρθὰς αὐτὴν τέμνει. εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ Ε, καὶ ἐπεζεύχθωσαν αἱ ΕΑ, ΕΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, κοινὴ δὲ ἡ ΖΕ, δύο δυσὶν ἴσαι εἰσίν. καὶ βάσις ἡ ΕΑ βάσει τῇ ΕΒ ἴση: γωνία ἄρα ἡ ὑπὸ ΑΖΕ γωνίᾳ τῇ ὑπὸ ΒΖΕ ἴση ἐστίν. ὅταν δὲ εὐθεῖα ἐπ᾽ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν: ἑκατέρα ἄρα τῶν ὑπὸ ΑΖΕ, ΒΖΕ ὀρθή ἐστιν. ἡ ΓΔ ἄρα διὰ τοῦ κέντρου οὖσα τὴν ΑΒ μὴ διὰ τοῦ κέντρου οὖσαν δίχα τέμνουσα καὶ πρὸς ὀρθὰς τέμνει. ἀλλὰ δὴ ἡ ΓΔ τὴν ΑΒ πρὸς ὀρθὰς τεμνέτω: λέγω, ὅτι καὶ δίχα αὐτὴν τέμνει, τουτέστιν, ὅτι ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ ἴση ἐστὶν ἡ ΕΑ τῇ ΕΒ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΕΑΖ τῇ ὑπὸ ΕΒΖ. ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΑΖΕ ὀρθῇ τῇ ὑπὸ ΒΖΕ ἴση: δύο ἄρα τρίγωνά ἐστι τὰ ΕΑΖ, ΕΖΒ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΕΖ ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει: ἴση ἄρα ἡ ΑΖ τῇ ΖΒ. ἐὰν ἄρα ἐν κύκλῳ εὐθεῖά τις διὰ τοῦ κέντρου εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου δίχα τέμνῃ, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: καὶ ἐὰν πρὸς ὀρθὰς αὐτὴν τέμνῃ, καὶ δίχα αὐτὴν τέμνει: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 270|>","<|""VertexLabel"" -> ""3.4"", ""Text"" -> ""If in a circle two straight lines cut one another which are not through the centre, they do not bisect one another."", ""TextWordCount"" -> 22, ""GreekText"" -> ""ἐὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ διὰ τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα."", ""GreekTextWordCount"" -> 16, ""References"" -> {{""Book"" -> 3, ""Theorem"" -> 1}, {""Book"" -> 3, ""Theorem"" -> 3}}, ""Proof"" -> ""Let ABCD be a circle, and in it let the two straight lines AC, BD, which are not through the centre, cut one another at E; I say that they do not bisect one another. For, if possible, let them bisect one another, so that AE is equal to EC, and BE to ED; let the centre of the circle ABCD be taken [III. 1], and let it be F; let FE be joined. Then, since a straight line FE through the centre bisects a straight line AC not through the centre, it also cuts it at right angles; [III. 3]therefore the angle FEA is right. Again, since a straight line FE bisects a straight line BD, it also cuts it at right angles; [III. 3]therefore the angle FEB is right. But the angle FEA was also proved right; therefore the angle FEA is equal to the angle FEB, the less to the greater: which is impossible. Therefore AC, BD do not bisect one another."", ""ProofWordCount"" -> 167, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε μὴ διὰ τοῦ κέντρου οὖσαι: λέγω, ὅτι οὐ τέμνουσιν ἀλλήλας δίχα. εἰ γὰρ δυνατόν, τεμνέτωσαν ἀλλήλας δίχα ὥστε ἴσην εἶναι τὴν μὲν ΑΕ τῇ ΕΓ, τὴν δὲ ΒΕ τῇ ΕΔ: καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓΔ κύκλου, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΖΕ. ἐπεὶ οὖν εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΖΕ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ δίχα τέμνει, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΕΑ: πάλιν, ἐπεὶ εὐθεῖά τις ἡ ΖΕ εὐθεῖάν τινα τὴν ΒΔ δίχα τέμνει, καὶ πρὸς ὀρθὰς αὐτὴν τέμνει: ὀρθὴ ἄρα ἡ ὑπὸ ΖΕΒ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΖΕΑ ὀρθή: ἴση ἄρα ἡ ὑπὸ ΖΕΑ τῇ ὑπὸ ΖΕΒ ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα αἱ ΑΓ, ΒΔ τέμνουσιν ἀλλήλας δίχα. ἐὰν ἄρα ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας μὴ διὰ τοῦ κέντρου οὖσαι, οὐ τέμνουσιν ἀλλήλας δίχα: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 164|>","<|""VertexLabel"" -> ""3.5"", ""Text"" -> ""If two circles cut one another, they will not have the same centre."", ""TextWordCount"" -> 13, ""GreekText"" -> ""ἐὰν δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Book"" -> 1, ""Definition"" -> 15}}, ""Proof"" -> ""For let the circles ABC, CDG cut one another at the points B, C; I say that they will not have the same centre. For, if possible, let it be E; let EC be joined, and let EFG be drawn through at random. Then, since the point E is the centre of the circle ABC, EC is equal to EF. [I. Def. 15] Again, since the point E is the centre of the circle CDG, EC is equal to EG. But EC was proved equal to EF also; therefore EF is also equal to EG, the less to the greater: which is impossible. Therefore the point E is not the centre of the circles ABC, CDG."", ""ProofWordCount"" -> 116, ""GreekProof"" -> ""δύο γὰρ κύκλοι οἱ ΑΒΓ, ΓΔΗ τεμνέτωσαν ἀλλήλους κατὰ τὰ Β, Γ σημεῖα. λέγω, ὅτι οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον. εἰ γὰρ δυνατόν, ἔστω τὸ Ε, καὶ ἐπεζεύχθω ἡ ΕΓ, καὶ διήχθω ἡ ΕΖΗ, ὡς ἔτυχεν. καὶ ἐπεὶ τὸ Ε σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, ἴση ἐστὶν ἡ ΕΓ τῇ ΕΖ. πάλιν, ἐπεὶ τὸ Ε σημεῖον κέντρον ἐστὶ τοῦ ΓΔΗ κύκλου, ἴση ἐστὶν ἡ ΕΓ τῇ ΕΗ: ἐδείχθη δὲ ἡ ΕΓ καὶ τῇ ΕΖ ἴση: καὶ ἡ ΕΖ ἄρα τῇ ΕΗ ἐστιν ἴση ἡ ἐλάσσων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Ε σημεῖον κέντρον ἐστὶ τῶν ΑΒΓ, ΓΔΗ κύκλων. ἐὰν ἄρα δύο κύκλοι τέμνωσιν ἀλλήλους, οὐκ ἔστιν αὐτῶν τὸ αὐτὸ κέντρον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 118|>","<|""VertexLabel"" -> ""3.6"", ""Text"" -> ""If two circles touch one another, they will not have the same centre."", ""TextWordCount"" -> 13, ""GreekText"" -> ""ἐὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον."", ""GreekTextWordCount"" -> 11, ""References"" -> {}, ""Proof"" -> ""For let the two circles ABC, CDE touch one another at the point C; I say that they will not have the same centre. For, if possible, let it be F; let FC be joined, and let FEB be drawn through at random. Then, since the point F is the centre of the circle ABC, FC is equal to FB. Again, since the point F is the centre of the circle CDE, FC is equal to FE. But FC was proved equal to FB; therefore FE is also equal to FB, the less to the greater: which is impossible. Therefore F is not the centre of the circles ABC, CDE."", ""ProofWordCount"" -> 110, ""GreekProof"" -> ""δύο γὰρ κύκλοι οἱ ΑΒΓ, ΓΔΕ ἐφαπτέσθωσαν ἀλλήλων κατὰ τὸ Γ σημεῖον: λέγω, ὅτι οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον. εἰ γὰρ δυνατόν, ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΖΓ, καὶ διήχθω, ὡς ἔτυχεν, ἡ ΖΕΒ. ἐπεὶ οὖν τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, ἴση ἐστὶν ἡ ΖΓ τῇ ΖΒ. πάλιν, ἐπεὶ τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΓΔΕ κύκλου, ἴση ἐστὶν ἡ ΖΓ τῇ ΖΕ. ἐδείχθη δὲ ἡ ΖΓ τῇ ΖΒ ἴση: καὶ ἡ ΖΕ ἄρα τῇ ΖΒ ἐστιν ἴση, ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Ζ σημεῖον κέντρον ἐστὶ τῶν ΑΒΓ, ΓΔΕ κύκλων. ἐὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων, οὐκ ἔσται αὐτῶν τὸ αὐτὸ κέντρον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 116|>","<|""VertexLabel"" -> ""3.7"", ""Text"" -> ""If on the diameter of a circle a point be taken which is not the centre of the circle, and from the point straight lines fall upon the circle, that will be greatest on which the centre is, the remainder of the same diameter will be least, and of the rest the nearer to the straight line through the centre is always greater than the more remote, and only two equal straight lines will fall from the point on the circle, one on each side of the least straight line."", ""TextWordCount"" -> 90, ""GreekText"" -> ""ἐὰν κύκλου ἐπὶ τῆς διαμέτρου ληφθῇ τι σημεῖον, ὃ μή ἐστι κέντρον τοῦ κύκλου, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσιν εὐθεῖαί τινες, μεγίστη μὲν ἔσται, ἐφ᾽ ἧς τὸ κέντρον, ἐλαχίστη δὲ ἡ λοιπή, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ἐλαχίστης."", ""GreekTextWordCount"" -> 66, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 20}, {""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 24}}, ""Proof"" -> ""Let ABCD be a circle, and let AD be a diameter of it;on AD let a point F be taken which is not the centre of the circle, let E be the centre of the circle, and from F let straight lines FB, FC, FG fall upon the circle ABCD; I say that FA is greatest, FD is least, and of the rest FB isgreater than FC, and FC than FG. For let BE, CE, GE be joined. Then, since in any triangle two sides are greater than the remaining one, [I. 20]EB, EF are greater than BF. But AE is equal to BE; therefore AF is greater than BF. Again, since BE is equal to CE, and FE is common,the two sides BE, EF are equal to the two sides CE, EF. But the angle BEF is also greater than the angle CEF; therefore the base BF is greater than the base CF. [I. 24] For the same reason CF is also greater than FG. Again, since GF, FE are greater than EG, and EG is equal to ED, GF, FE are greater than ED. Let EF be subtracted from each; therefore the remainder GF is greater than the remainder FD. Therefore FA is greatest, FD is least, and FB is greater than FC, and FC than FG. I say also that from the point F only two equal straight lines will fall on the circle ABCD, one on each side of theleast FD. For on the straight line EF, and at the point E on it, let the angle FEH be constructed equal to the angle GEF [I. 23], and let FH be joined. Then, since GE is equal to EH,and EF is common, the two sides GE, EF are equal to the two sides HE, EF; and the angle GEF is equal to the angle HEF; therefore the base FG is equal to the base FH. [I. 4] I say again that another straight line equal to FG will notfall on the circle from the point F. For, if possible, let FK so fall. Then, since FK is equal to FG, and FH to FG, FK is also equal to FH, the nearer to the straight line through the centre being thus equal to the more remote: which is impossible. Therefore another straight line equal to GF will not fall from the point F upon the circle; therefore only one straight line will so fall."", ""ProofWordCount"" -> 412, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἔστω ἡ ΑΔ, καὶ ἐπὶ τῆς ΑΔ εἰλήφθω τι σημεῖον τὸ Ζ, ὃ μή ἐστι κέντρον τοῦ κύκλου, κέντρον δὲ τοῦ κύκλου ἔστω τὸ Ε, καὶ ἀπὸ τοῦ Ζ πρὸς τὸν ΑΒΓΔ κύκλον προσπιπτέτωσαν εὐθεῖαί τινες αἱ ΖΒ, ΖΓ, ΖΗ: λέγω, ὅτι μεγίστη μέν ἐστιν ἡ ΖΑ, ἐλαχίστη δὲ ἡ ΖΔ, τῶν δὲ ἄλλων ἡ μὲν ΖΒ τῆς ΖΓ μείζων, ἡ δὲ ΖΓ τῆς ΖΗ. ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΕ, ΗΕ. καὶ ἐπεὶ παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσιν, αἱ ἄρα ΕΒ, ΕΖ τῆς ΒΖ μείζονές εἰσιν. ἴση δὲ ἡ ΑΕ τῇ ΒΕ αἱ ἄρα ΒΕ, ΕΖ ἴσαι εἰσὶ τῇ ΑΖ: μείζων ἄρα ἡ ΑΖ τῆς ΒΖ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΓΕ, κοινὴ δὲ ἡ ΖΕ, δύο δὴ αἱ ΒΕ, ΕΖ δυσὶ ταῖς ΓΕ, ΕΖ ἴσαι εἰσίν. ἀλλὰ καὶ γωνία ἡ ὑπὸ ΒΕΖ γωνίας τῆς ὑπὸ ΓΕΖ μείζων. βάσις ἄρα ἡ ΒΖ βάσεως τῆς ΓΖ μείζων ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΖ τῆς ΖΗ μείζων ἐστίν. πάλιν, ἐπεὶ αἱ ΗΖ, ΖΕ τῆς ΕΗ μείζονές εἰσιν, ἴση δὲ ἡ ΕΗ τῇ ΕΔ, αἱ ἄρα ΗΖ, ΖΕ τῆς ΕΔ μείζονές εἰσιν. κοινὴ ἀφῃρήσθω ἡ ΕΖ: λοιπὴ ἄρα ἡ ΗΖ λοιπῆς τῆς ΖΔ μείζων ἐστίν. μεγίστη μὲν ἄρα ἡ ΖΑ, ἐλαχίστη δὲ ἡ ΖΔ, μείζων δὲ ἡ μὲν ΖΒ τῆς ΖΓ, ἡ δὲ ΖΓ τῆς ΖΗ. λέγω, ὅτι καὶ ἀπὸ τοῦ Ζ σημείου δύο μόνον ἴσαι προσπεσοῦνται πρὸς τὸν ΑΒΓΔ κύκλον ἐφ᾽ ἑκάτερα τῆς ΖΔ ἐλαχίστης. συνεστάτω γὰρ πρὸς τῇ ΕΖ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε τῇ ὑπὸ ΗΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΖΕΘ, καὶ ἐπεζεύχθω ἡ ΖΘ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΗΕ τῇ ΕΘ, κοινὴ δὲ ἡ ΕΖ, δύο δὴ αἱ ΗΕ, ΕΖ δυσὶ ταῖς ΘΕ, ΕΖ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΗΕΖ γωνίᾳ τῇ ὑπὸ ΘΕΖ ἴση: βάσις ἄρα ἡ ΖΗ βάσει τῇ ΖΘ ἴση ἐστίν. λέγω δή, ὅτι τῇ ΖΗ ἄλλη ἴση οὐ προσπεσεῖται πρὸς τὸν κύκλον ἀπὸ τοῦ Ζ σημείου. εἰ γὰρ δυνατόν, προσπιπτέτω ἡ ΖΚ. καὶ ἐπεὶ ἡ ΖΚ τῇ ΖΗ ἴση ἐστίν, ἀλλὰ ἡ ΖΘ τῇ ΖΗ ἴση ἐστίν, καὶ ἡ ΖΚ ἄρα τῇ ΖΘ ἐστιν ἴση, ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῇ ἀπώτερον ἴση: ὅπερ ἀδύνατον. οὐκ ἄρα ἀπὸ τοῦ Ζ σημείου ἑτέρα τις προσπεσεῖται πρὸς τὸν κύκλον ἴση τῇ ΗΖ: μία ἄρα μόνη. ἐὰν ἄρα κύκλου ἐπὶ τῆς διαμέτρου ληφθῇ τι σημεῖον, ὃ μή ἐστι κέντρον τοῦ κύκλου, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσιν εὐθεῖαί τινες, μεγίστη μὲν ἔσται, ἐφ᾽ ἧς τὸ κέντρον, ἐλαχίστη δὲ ἡ λοιπή, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ αὐτοῦ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ἐλαχίστης: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 461|>","<|""VertexLabel"" -> ""3.8"", ""Text"" -> ""If a point be taken outside a circle and from the point straight lines be drawn through to the circle, one of which is through the centre and the others are drawn at random, then, of the straight lines which fall on the concave circumference, that through the centre is greatest, while of the rest the nearer to that through the centre is always greater than the more remote, but, of the straight lines falling on the convex circumference, that between the point and the diameter is least, while of the rest the nearer to the least is always less than the more remote, and only two equal straight lines will fall on the circle from the point, one on each side of the least."", ""TextWordCount"" -> 125, ""GreekText"" -> ""ἐὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον διαχθῶσιν εὐθεῖαί τινες, ὧν μία μὲν διὰ τοῦ κέντρου, αἱ δὲ λοιπαί, ὡς ἔτυχεν, τῶν μὲν πρὸς τὴν κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη μέν ἐστιν ἡ διὰ τοῦ κέντρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, τῶν δὲ πρὸς τὴν κυρτὴν περιφέρειαν προσπιπτουσῶν εὐθειῶν ἐλαχίστη μέν ἐστιν ἡ μεταξὺ τοῦ τε σημείου καὶ τῆς διαμέτρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς ἐλαχίστης τῆς ἀπώτερόν ἐστιν ἐλάττων, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ἐλαχίστης."", ""GreekTextWordCount"" -> 103, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 20}, {""Book"" -> 1, ""Theorem"" -> 21}, {""Book"" -> 1, ""Theorem"" -> 24}, {""Book"" -> 3, ""Theorem"" -> 1}}, ""Proof"" -> ""Let ABC be a circle, and let a point D be taken outside ABC; let there be drawn through from it straight lines DA, DE, DF, DC, and let DA be through the centre; I say that, of the straight lines falling on the concave circumference AEFC, the straight line DA through the centre is greatest, while DE is greater than DF and DF than DC.; but, of the straight lines falling on the convex circumference HLKG, the straight line DG between the point and the diameter AG is least; and the nearer to the least DG is always less than the more remote, namely DK than DL, and DL than DH. For let the centre of the circle ABC be taken [III. 1], and let it be M; let ME, MF, MC, MK, ML, MH be joined. Then, since AM is equal to EM, let MD be added to each; therefore AD is equal to EM, MD. But EM, MD are greater than ED; [I. 20] therefore AD is also greater than ED. Again, since ME is equal to MF, and MD is common, therefore EM, MD are equal to FM, MD; and the angle EMD is greater than the angle FMD; therefore the base ED is greater than the base FD. [I. 24] Similarly we can prove that FD is greater than CD; therefore DA is greatest, while DE is greater than DF, and DF than DC. Next, since MK, KD are greater than MD, [I. 20] and MG is equal to MK, therefore the remainder KD is greater than the remainder GD, so that GD is less than KD. And, since on MD, one of the sides of the triangle MLD, two straight lines MK, KD were constructed meeting within the triangle, therefore MK, KD are less than ML, LD; [I. 21] and MK is equal to ML; therefore the remainder DK is less than the remainder DL. Similarly we can prove that DL is also less than DH; therefore DG is least, while DK is less than DL, and DL than DH. I say also that only two equal straight lines will fall from the point D on the circle, one on each side of the least DG. On the straight line MD, and at the point M on it, let the angle DMB be constructed equal to the angle KMD, and let DB be joined. Then, since MK is equal to MB, and MD is common, the two sides KM, MD are equal to the two sides BM, MD respectively; and the angle KMD is equal to the angle BMD; therefore the base DK is equal to the base DB. [I. 4] I say that no other straight line equal to the straight line DK will fall on the circle from the point D. For, if possible, let a straight line so fall, and let it be DN. Then, since DK is equal to DN, while DK is equal to DB, DB is also equal to DN, that is, the nearer to the least DG equal to the more remote: which was proved impossible. Therefore no more than two equal straight lines will fall on the circle ABC from the point D, one on each side of DG the least."", ""ProofWordCount"" -> 545, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓ, καὶ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, καὶ ἀπ᾽ αὐτοῦ διήχθωσαν εὐθεῖαί τινες αἱ ΔΑ, ΔΕ, ΔΖ, ΔΓ, ἔστω δὲ ἡ ΔΑ διὰ τοῦ κέντρου. λέγω, ὅτι τῶν μὲν πρὸς τὴν ΑΕΖΓ κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη μέν ἐστιν ἡ διὰ τοῦ κέντρου ἡ ΔΑ, μείζων δὲ ἡ μὲν ΔΕ τῆς ΔΖ ἡ δὲ ΔΖ τῆς ΔΓ, τῶν δὲ πρὸς τὴν ΘΛΚΗ κυρτὴν περιφέρειαν προσπιπτουσῶν εὐθειῶν ἐλαχίστη μέν ἐστιν ἡ ΔΗ ἡ μεταξὺ τοῦ σημείου καὶ τῆς διαμέτρου τῆς ΑΗ, ἀεὶ δὲ ἡ ἔγγιον τῆς ΔΗ ἐλαχίστης ἐλάττων ἐστὶ τῆς ἀπώτερον, ἡ μὲν ΔΚ τῆς ΔΛ, ἡ δὲ ΔΛ τῆς ΔΘ. εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓ κύκλου καὶ ἔστω τὸ Μ: καὶ ἐπεζεύχθωσαν αἱ ΜΕ, ΜΖ, ΜΓ, ΜΚ, ΜΛ, ΜΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΜ τῇ ΕΜ, κοινὴ προσκείσθω ἡ ΜΔ: ἡ ἄρα ΑΔ ἴση ἐστὶ ταῖς ΕΜ, ΜΔ. ἀλλ᾽ αἱ ΕΜ, ΜΔ τῆς ΕΔ μείζονές εἰσιν: καὶ ἡ ΑΔ ἄρα τῆς ΕΔ μείζων ἐστίν. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΜΕ τῇ ΜΖ, κοινὴ δὲ ἡ ΜΔ, αἱ ΕΜ, ΜΔ ἄρα ταῖς ΖΜ, ΜΔ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΕΜΔ γωνίας τῆς ὑπὸ ΖΜΔ μείζων ἐστίν. βάσις ἄρα ἡ ΕΔ βάσεως τῆς ΖΔ μείζων ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΖΔ τῆς ΓΔ μείζων ἐστίν: μεγίστη μὲν ἄρα ἡ ΔΑ, μείζων δὲ ἡ μὲν ΔΕ τῆς ΔΖ, ἡ δὲ ΔΖ τῆς ΔΓ. καὶ ἐπεὶ αἱ ΜΚ, ΚΔ τῆς ΜΔ μείζονές εἰσιν, ἴση δὲ ἡ ΜΗ τῇ ΜΚ, λοιπὴ ἄρα ἡ ΚΔ λοιπῆς τῆς ΗΔ μείζων ἐστίν: ὥστε ἡ ΗΔ τῆς ΚΔ ἐλάττων ἐστίν: καὶ ἐπεὶ τριγώνου τοῦ ΜΛΔ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΜΔ δύο εὐθεῖαι ἐντὸς συνεστάθησαν αἱ ΜΚ, ΚΔ, αἱ ἄρα ΜΚ, ΚΔ τῶν ΜΛ, ΛΔ ἐλάττονές εἰσιν: ἴση δὲ ἡ ΜΚ τῇ ΜΛ: λοιπὴ ἄρα ἡ ΔΚ λοιπῆς τῆς ΔΛ ἐλάττων ἐστίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΔΛ τῆς ΔΘ ἐλάττων ἐστίν: ἐλαχίστη μὲν ἄρα ἡ ΔΗ, ἐλάττων δὲ ἡ μὲν ΔΚ τῆς ΔΛ ἡ δὲ ΔΛ τῆς ΔΘ. λέγω, ὅτι καὶ δύο μόνον ἴσαι ἀπὸ τοῦ Δ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ΔΗ ἐλαχίστης: συνεστάτω πρὸς τῇ ΜΔ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Μ τῇ ὑπὸ ΚΜΔ γωνίᾳ ἴση γωνία ἡ ὑπὸ ΔΜΒ καὶ ἐπεζεύχθω ἡ ΔΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΜΚ τῇ ΜΒ, κοινὴ δὲ ἡ ΜΔ, δύο δὴ αἱ ΚΜ, ΜΔ δύο ταῖς ΒΜ, ΜΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΚΜΔ γωνίᾳ τῇ ὑπὸ ΒΜΔ ἴση: βάσις ἄρα ἡ ΔΚ βάσει τῇ ΔΒ ἴση ἐστίν. λέγω δή, ὅτι τῇ ΔΚ εὐθείᾳ ἄλλη ἴση οὐ προσπεσεῖται πρὸς τὸν κύκλον ἀπὸ τοῦ Δ σημείου. εἰ γὰρ δυνατόν, προσπιπτέτω καὶ ἔστω ἡ ΔΝ. ἐπεὶ οὖν ἡ ΔΚ τῇ ΔΝ ἐστιν ἴση, ἀλλ᾽ ἡ ΔΚ τῇ ΔΒ ἐστιν ἴση, καὶ ἡ ΔΒ ἄρα τῇ ΔΝ ἐστιν ἴση, ἡ ἔγγιον τῆς ΔΗ ἐλαχίστης τῇ ἀπώτερον ἐστιν ἴση: ὅπερ ἀδύνατον ἐδείχθη. οὐκ ἄρα πλείους ἢ δύο ἴσαι πρὸς τὸν ΑΒΓ κύκλον ἀπὸ τοῦ Δ σημείου ἐφ᾽ ἑκάτερα τῆς ΔΗ ἐλαχίστης προσπεσοῦνται. ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον διαχθῶσιν εὐθεῖαί τινες, ὧν μία μὲν διὰ τοῦ κέντρου αἱ δὲ λοιπαί, ὡς ἔτυχεν, τῶν μὲν πρὸς τὴν κοίλην περιφέρειαν προσπιπτουσῶν εὐθειῶν μεγίστη μέν ἐστιν ἡ διὰ τοῦ κέντρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς διὰ τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν, τῶν δὲ πρὸς τὴν κυρτὴν περιφέρειαν προσπιπτουσῶν εὐθειῶν ἐλαχίστη μέν ἐστιν ἡ μεταξὺ τοῦ τε σημείου καὶ τῆς διαμέτρου, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τῆς ἐλαχίστης τῆς ἀπώτερόν ἐστιν ἐλάττων, δύο δὲ μόνον ἴσαι ἀπὸ τοῦ σημείου προσπεσοῦνται πρὸς τὸν κύκλον ἐφ᾽ ἑκάτερα τῆς ἐλαχίστης: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 613|>","<|""VertexLabel"" -> ""3.9"", ""Text"" -> ""If a point be taken within a circle, and more than two equal straight lines fall from the point on the circle, the point taken is the centre of the circle."", ""TextWordCount"" -> 31, ""GreekText"" -> ""ἐὰν κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους ἢ δύο ἴσαι εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου."", ""GreekTextWordCount"" -> 26, ""References"" -> {{""Book"" -> 1, ""Definition"" -> 10}, {""Book"" -> 1, ""Theorem"" -> 8}, {""Book"" -> 3, ""Theorem"" -> 1}}, ""Proof"" -> ""Let ABC be a circle and D a point within it, and from D let more than two equal straight lines, namely DA, DB, DC, fall on the circle ABC; I say that the point D is the centre of the circle ABC. For let AB, BC be joined and bisected at the points E, F, and let ED, FD be joined and drawn through to the points G, K, H, L. Then, since AE is equal to EB, and ED is common, the two sides AE, ED are equal to the two sides BE, ED; and the base DA is equal to the base DB; therefore the angle AED is equal to the angle BED. [I. 8] Therefore each of the angles AED, BED is right; [I. Def. 10] therefore GK cuts AB into two equal parts and at right angles. And since, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line, [III. 1] the centre of the circle is on GK. For the same reason the centre of the circle ABC is also on HL. And the straight lines GK, HL have no other point common but the point D; therefore the point D is the centre of the circle ABC."", ""ProofWordCount"" -> 224, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓ, ἐντὸς δὲ αὐτοῦ σημεῖον τὸ Δ, καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν πλείους ἢ δύο ἴσαι εὐθεῖαι αἱ ΔΑ, ΔΒ, ΔΓ: λέγω, ὅτι τὸ Δ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ἐπεζεύχθωσαν γὰρ αἱ ΑΒ, ΒΓ καὶ τετμήσθωσαν δίχα κατὰ τὰ Ε, Ζ σημεῖα, καὶ ἐπιζευχθεῖσαι αἱ ΕΔ, ΖΔ διήχθωσαν ἐπὶ τὰ Η, Κ, Θ, Λ σημεῖα. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, κοινὴ δὲ ἡ ΕΔ, δύο δὴ αἱ ΑΕ, ΕΔ δύο ταῖς ΒΕ, ΕΔ ἴσαι εἰσίν: καὶ βάσις ἡ ΔΑ βάσει τῇ ΔΒ ἴση: γωνία ἄρα ἡ ὑπὸ ΑΕΔ γωνίᾳ τῇ ὑπὸ ΒΕΔ ἴση ἐστίν: ὀρθὴ ἄρα ἑκατέρα τῶν ὑπὸ ΑΕΔ, ΒΕΔ γωνιῶν: ἡ ΗΚ ἄρα τὴν ΑΒ τέμνει δίχα καὶ πρὸς ὀρθάς. καὶ ἐπεί, ἐὰν ἐν κύκλῳ εὐθεῖά τις εὐθεῖάν τινα δίχα τε καὶ πρὸς ὀρθὰς τέμνῃ, ἐπὶ τῆς τεμνούσης ἐστὶ τὸ κέντρον τοῦ κύκλου, ἐπὶ τῆς ΗΚ ἄρα ἐστὶ τὸ κέντρον τοῦ κύκλου. διὰ τὰ αὐτὰ δὴ καὶ ἐπὶ τῆς ΘΛ ἐστι τὸ κέντρον τοῦ ΑΒΓ κύκλου. καὶ οὐδὲν ἕτερον κοινὸν ἔχουσιν αἱ ΗΚ, ΘΛ εὐθεῖαι ἢ τὸ Δ σημεῖον: τὸ Δ ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐντός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι πλείους ἢ δύο ἴσαι εὐθεῖαι, τὸ ληφθὲν σημεῖον κέντρον ἐστὶ τοῦ κύκλου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 222|>","<|""VertexLabel"" -> ""3.10"", ""Text"" -> ""A circle does not cut a circle at more points than two."", ""TextWordCount"" -> 12, ""GreekText"" -> ""κύκλος κύκλον οὐ τέμνει κατὰ πλείονα σημεῖα ἢ δύο."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 3, ""Theorem"" -> 1}, {""Book"" -> 3, ""Theorem"" -> 5}}, ""Proof"" -> ""For, if possible, let the circle ABC cut the circle DEF at more points than two, namely B, C, F, H; let BH, BG be joined and bisected at the points K, L, and from K, L let KC, LM be drawn at right angles to BH, BG and carried through to the points A, E. Then, since in the circle ABC a straight line AC cuts a straight line BH into two equal parts and at right angles, the centre of the circle ABC is on AC. [III. 1] Again, since in the same circle ABC a straight line NO cuts a straight line BG into two equal parts and at right angles, the centre of the circle ABC is on NO. But it was also proved to be on AC, and the straight lines AC, NO meet at no point except at P; therefore the point P is the centre of the circle ABC. Similarly we can prove that P is also the centre of the circle DEF; therefore the two circles ABC, DEF which cut one another have the same centre P: which is impossible. [III. 5]"", ""ProofWordCount"" -> 190, ""GreekProof"" -> ""εἰ γὰρ δυνατόν, κύκλος ὁ ΑΒΓ κύκλον τὸν ΔΕΖ τεμνέτω κατὰ πλείονα σημεῖα ἢ δύο τὰ Β, Η, Ζ, Θ, καὶ ἐπιζευχθεῖσαι αἱ ΒΘ, ΒΗ δίχα τεμνέσθωσαν κατὰ τὰ κ, Λ σημεῖα: καὶ ἀπὸ τῶν Κ, Λ ταῖς ΒΘ, ΒΗ πρὸς ὀρθὰς ἀχθεῖσαι αἱ ΚΓ, ΛΜ διήχθωσαν ἐπὶ τὰ Α, Ε σημεῖα. ἐπεὶ οὖν ἐν κύκλῳ τῷ ΑΒΓ εὐθεῖά τις ἡ ΑΓ εὐθεῖάν τινα τὴν ΒΘ δίχα καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΑΓ ἄρα ἐστὶ τὸ κέντρον τοῦ ΑΒΓ κύκλου. πάλιν, ἐπεὶ ἐν κύκλῳ τῷ αὐτῷ τῷ ΑΒΓ εὐθεῖά τις ἡ ΝΞ εὐθεῖάν τινα τὴν ΒΗ δίχα καὶ πρὸς ὀρθὰς τέμνει, ἐπὶ τῆς ΝΞ ἄρα ἐστὶ τὸ κέντρον τοῦ ΑΒΓ κύκλου. ἐδείχθη δὲ καὶ ἐπὶ τῆς ΑΓ, καὶ κατ᾽ οὐδὲν συμβάλλουσιν αἱ ΑΓ, ΝΞ εὐθεῖαι ἢ κατὰ τὸ Ο: τὸ Ο ἄρα σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι καὶ τοῦ ΔΕΖ κύκλου κέντρον ἐστὶ τὸ Ο: δύο ἄρα κύκλων τεμνόντων ἀλλήλους τῶν ΑΒΓ, ΔΕΖ τὸ αὐτό ἐστι κέντρον τὸ Ο: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα κύκλος κύκλον τέμνει κατὰ πλείονα σημεῖα ἢ δύο: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 182|>","<|""VertexLabel"" -> ""3.11"", ""Text"" -> ""If two circles touch one another internally, and their centres be taken, the straight line joining their centres, if it be also produced, will fall on the point of contact of the circles."", ""TextWordCount"" -> 33, ""GreekText"" -> ""ἐὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐντός, καὶ ληφθῇ αὐτῶν τὰ κέντρα, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη εὐθεῖα καὶ ἐκβαλλομένη ἐπὶ τὴν συναφὴν πεσεῖται τῶν κύκλων."", ""GreekTextWordCount"" -> 26, ""References"" -> {}, ""Proof"" -> ""For let the two circles ABC, ADE touch one another internally at the point A, and let the centre F of the circle ABC, and the centre G of ADE, be taken; I say that the straight line joined from G to F and produced will fall on A. For suppose it does not, but, if possible, let it fall as FGH, and let AF, AG be joined. Then, since AG, GF are greater than FA, that is, than FH, let FG be subtracted from each; therefore the remainder AG is greater than the remainder GH. But AG is equal to GD; therefore GD is also greater than GH, the less than the greater: which is impossible. Therefore the straight line joined from F to G will not fall outside; therefore it will fall at A on the point of contact."", ""ProofWordCount"" -> 141, ""GreekProof"" -> ""δύο γὰρ κύκλοι οἱ ΑΒΓ, ΑΔΕ ἐφαπτέσθωσαν ἀλλήλων ἐντὸς κατὰ τὸ Α σημεῖον, καὶ εἰλήφθω τοῦ μὲν ΑΒΓ κύκλου κέντρον τὸ Ζ, τοῦ δὲ ΑΔΕ τὸ Η: λέγω, ὅτι ἡ ἀπὸ τοῦ Η ἐπὶ τὸ Ζ ἐπιζευγνυμένη εὐθεῖα ἐκβαλλομένη ἐπὶ τὸ Α πεσεῖται. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, πιπτέτω ὡς ἡ ΖΗΘ, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΑΗ. ἐπεὶ οὖν αἱ ΑΗ, ΗΖ τῆς ΖΑ, τουτέστι τῆς ΖΘ, μείζονές εἰσιν, κοινὴ ἀφῃρήσθω ἡ ΖΗ: λοιπὴ ἄρα ἡ ΑΗ λοιπῆς τῆς ΗΘ μείζων ἐστίν. ἴση δὲ ἡ ΑΗ τῇ ΗΔ: καὶ ἡ ΗΔ ἄρα τῆς ΗΘ μείζων ἐστὶν ἡ ἐλάττων τῆς μείζονος: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα ἡ ἀπὸ τοῦ Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη εὐθεῖα ἐκτὸς πεσεῖται: κατὰ τὸ Α ἄρα ἐπὶ τῆς συναφῆς πεσεῖται. ἐὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐντός, καὶ ληφθῇ αὐτῶν τὰ κέντρα, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη εὐθεῖα καὶ ἐκβαλλομένη ἐπὶ τὴν συναφὴν πεσεῖται τῶν κύκλων: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 155|>","<|""VertexLabel"" -> ""3.12"", ""Text"" -> ""If two circles touch one another externally, the straight line joining their centres will pass through the point of contact."", ""TextWordCount"" -> 20, ""GreekText"" -> ""ἐὰν δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη διὰ τῆς ἐπαφῆς ἐλεύσεται."", ""GreekTextWordCount"" -> 16, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 20}}, ""Proof"" -> ""For let the two circles ABC, ADE touch one anotherexternally at the point A, and let the centre F of ABC, and the centre G of ADE, be taken; I say that the straight line joined from F to G will pass through the point of contact at A. For suppose it does not,but, if possible, let it pass as FCDG, and let AF, AG be joined. Then, since the point F is the centre of the circle ABC,FA is equal to FC. Again, since the point G is the centre of the circle ADE, GA is equal to GD. But FA was also proved equal to FC;therefore FA, AG are equal to FC, GD, so that the whole FG is greater than FA, AG; but it is also less [I. 20]: which is impossible. Therefore the straight line joined from F to G will not fail to pass through the point of contact at A; therefore it will pass through it."", ""ProofWordCount"" -> 165, ""GreekProof"" -> ""δύο γὰρ κύκλοι οἱ ΑΒΓ, ΑΔΕ ἐφαπτέσθωσαν ἀλλήλων ἐκτὸς κατὰ τὸ Α σημεῖον, καὶ εἰλήφθω τοῦ μὲν ΑΒΓ κέντρον τὸ Ζ, τοῦ δὲ ΑΔΕ τὸ Η: λέγω, ὅτι ἡ ἀπὸ τοῦ Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη εὐθεῖα διὰ τῆς κατὰ τὸ Α ἐπαφῆς ἐλεύσεται. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἐρχέσθω ὡς ἡ ΖΓΔΗ, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΑΗ. ἐπεὶ οὖν τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου, ἴση ἐστὶν ἡ ΖΑ τῇ ΖΓ. πάλιν, ἐπεὶ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΔΕ κύκλου, ἴση ἐστὶν ἡ ΗΑ τῇ ΗΔ. ἐδείχθη δὲ καὶ ἡ ΖΑ τῇ ΖΓ ἴση: αἱ ἄρα ΖΑ, ΑΗ ταῖς ΖΓ, ΗΔ ἴσαι εἰσίν: ὥστε ὅλη ἡ ΖΗ τῶν ΖΑ, ΑΗ μείζων ἐστίν: ἀλλὰ καὶ ἐλάττων: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Ζ ἐπὶ τὸ Η ἐπιζευγνυμένη εὐθεῖα διὰ τῆς κατὰ τὸ Α ἐπαφῆς οὐκ ἐλεύσεται: δι᾽ αὐτῆς ἄρα. ἐὰν ἄρα δύο κύκλοι ἐφάπτωνται ἀλλήλων ἐκτός, ἡ ἐπὶ τὰ κέντρα αὐτῶν ἐπιζευγνυμένη εὐθεῖα διὰ τῆς ἐπαφῆς ἐλεύσεται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 167|>","<|""VertexLabel"" -> ""3.13"", ""Text"" -> ""A circle does not touch a circle at more points than one, whether it touch it internally or externally."", ""TextWordCount"" -> 19, ""GreekText"" -> ""κύκλος κύκλου οὐκ ἐφάπτεται κατὰ πλείονα σημεῖα ἢ καθ᾽ ἕν, ἐάν τε ἐντὸς ἐάν τε ἐκτὸς ἐφάπτηται."", ""GreekTextWordCount"" -> 18, ""References"" -> {{""Book"" -> 3, ""Definition"" -> 3}, {""Book"" -> 3, ""Theorem"" -> 2}, {""Book"" -> 3, ""Theorem"" -> 11}}, ""Proof"" -> ""For, if possible, let the circle ABDC touch the circle EBFD, first internally, at morepoints than one, namely D, B. Let the centre G of the circle ABDC, and the centre H of EBFD, be taken. Therefore the straight linejoined from G to H will fall on B, D. [III. 11] Let it so fall, as BGHD. Then, since the point G is the centre of the circle ABCD,BG is equal to GD; therefore BG is greater than HD; therefore BH is much greater than HD. Again, since the point H is the centre of the circle EBFD,BH is equal to HD; but it was also proved much greater than it: which is impossible. Therefore a circle does not touch a circle internally at more points than one. I say further that neither does it so touch it externally. For, if possible, let the circle ACK touch the circle ABDC at more points than one, namely A, C, and let AC be joined. Then, since on the circumference of each of the circlesABDC, ACK two points A, C have been taken at random, the straight line joining the points will fall within each circle; [III. 2] but it fell within the circle ABCD and outside ACK [III. Def. 3]: which is absurd. Therefore a circle does not touch a circle externally at more points than one. And it was proved that neither does it so touch it internally."", ""ProofWordCount"" -> 240, ""GreekProof"" -> ""εἰ γὰρ δυνατόν, κύκλος ὁ ΑΒΓΔ κύκλου τοῦ ΕΒΖΔ ἐφαπτέσθω πρότερον ἐντὸς κατὰ πλείονα σημεῖα ἢ ἓν τὰ δ, Β. καὶ εἰλήφθω τοῦ μὲν ΑΒΓΔ κύκλου κέντρον τὸ Η, τοῦ δὲ ΕΒΖΔ τὸ Θ. ἡ ἄρα ἀπὸ τοῦ Η ἐπὶ τὸ Θ ἐπιζευγνυμένη ἐπὶ τὰ Β, Δ πεσεῖται. πιπτέτω ὡς ἡ ΒΗΘΔ. καὶ ἐπεὶ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓΔ κύκλου, ἴση ἐστὶν ἡ ΒΗ τῇ ΗΔ: μείζων ἄρα ἡ ΒΗ τῆς ΘΔ: πολλῷ ἄρα μείζων ἡ ΒΘ τῆς ΘΔ. πάλιν, ἐπεὶ τὸ Θ σημεῖον κέντρον ἐστὶ τοῦ ΕΒΖΔ κύκλου, ἴση ἐστὶν ἡ ΒΘ τῇ ΘΔ: ἐδείχθη δὲ αὐτῆς καὶ πολλῷ μείζων: ὅπερ ἀδύνατον: οὐκ ἄρα κύκλος κύκλου ἐφάπτεται ἐντὸς κατὰ πλείονα σημεῖα ἢ ἕν. λέγω δή, ὅτι οὐδὲ ἐκτός. εἰ γὰρ δυνατόν, κύκλος ὁ ΑΓΚ κύκλου τοῦ ΑΒΓΔ ἐφαπτέσθω ἐκτὸς κατὰ πλείονα σημεῖα ἢ ἓν τὰ Α, Γ, καὶ ἐπεζεύχθω ἡ ΑΓ. ἐπεὶ οὖν κύκλων τῶν ΑΒΓΔ, ΑΓΚ εἴληπται ἐπὶ τῆς περιφερείας ἑκατέρου δύο τυχόντα σημεῖα τὰ Α, Γ, ἡ ἐπὶ τὰ σημεῖα ἐπιζευγνυμένη εὐθεῖα ἐντὸς ἑκατέρου πεσεῖται: ἀλλὰ τοῦ μὲν ΑΒΓΔ ἐντὸς ἔπεσεν, τοῦ δὲ ΑΓΚ ἐκτός: ὅπερ ἄτοπον: οὐκ ἄρα κύκλος κύκλου ἐφάπτεται ἐκτὸς κατὰ πλείονα σημεῖα ἢ ἕν. ἐδείχθη δέ, ὅτι οὐδὲ ἐντός. κύκλος ἄρα κύκλου οὐκ ἐφάπτεται κατὰ πλείονα σημεῖα ἢ καθ᾽ ἕν, ἐάν τε ἐντὸς ἐάν τε ἐκτὸς ἐφάπτηται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 220|>","<|""VertexLabel"" -> ""3.14"", ""Text"" -> ""In a circle equal straight lines are equally distant from the centre, and those which are equally distant from the centre are equal to one another."", ""TextWordCount"" -> 26, ""GreekText"" -> ""ἐν κύκλῳ αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ τοῦ κέντρου, καὶ αἱ ἴσον ἀπέχουσαι ἀπὸ τοῦ κέντρου ἴσαι ἀλλήλαις εἰσίν."", ""GreekTextWordCount"" -> 20, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 47}, {""Book"" -> 3, ""Definition"" -> 4}, {""Book"" -> 3, ""Theorem"" -> 1}, {""Book"" -> 3, ""Theorem"" -> 3}}, ""Proof"" -> ""Let ABDC be a circle, and let AB, CD be equal straight lines in it; I say that AB, CD are equally distant from the centre. For let the centre of the circle ABDC be taken [III. 1], and let it be E; from E let EF, EG be drawn perpendicular to AB, CD, and let AE, EC be joined. Then, since a straight line EF through the centre cuts a straight line AB not through the centre at right angles, it also bisects it. [III. 3] Therefore AF is equal to FB; therefore AB is double of AF. For the same reason CD is also double of CG; and AB is equal to CD; therefore AF is also equal to CG. And, since AE is equal to EC, the square on AE is also equal to the square on EC. But the squares on AF, EF are equal to the square on AE, for the angle at F is right; and the squares on EG, GC are equal to the square on EC, for the angle at G is right; [I. 47] therefore the squares on AF, FE are equal to the squares on CG, GE, of which the square on AF is equal to the square on CG, for AF is equal to CG; therefore the square on FE which remains is equal to the square on EG, therefore EF is equal to EG. But in a circle straight lines are said to be equally distant from the centre when the perpendiculars drawn to them from the centre are equal; [III. Def. 4] therefore AB, CD are equally distant from the centre. Next, let the straight lines AB, CD be equally distant from the centre; that is, let EF be equal to EG. I say that AB is also equal to CD. For, with the same construction, we can prove, similarly, that AB is double of AF, and CD of CG. And, since AE is equal to CE, the square on AE is equal to the square on CE. But the squares on EF, FA are equal to the square on AE, and the squares on EG, GC equal to the square on CE. [I. 47] Therefore the squares on EF, FA are equal to the squares on EG, GC, of which the square on EF is equal to the square on EG, for EF is equal to EG; therefore the square on AF which remains is equal to the square on CG; therefore AF is equal to CG. And AB is double of AF, and CD double of CG; therefore AB is equal to CD."", ""ProofWordCount"" -> 438, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ ἴσαι εὐθεῖαι ἔστωσαν αἱ ΑΒ, ΓΔ: λέγω, ὅτι αἱ ΑΒ, ΓΔ ἴσον ἀπέχουσιν ἀπὸ τοῦ κέντρου. εἰλήφθω γὰρ τὸ κέντρον τοῦ ΑΒΓΔ κύκλου καὶ ἔστω τὸ Ε, καὶ ἀπὸ τοῦ Ε ἐπὶ τὰς ΑΒ, ΓΔ κάθετοι ἤχθωσαν αἱ ΕΖ, ΕΗ, καὶ ἐπεζεύχθωσαν αἱ ΑΕ, ΕΓ. ἐπεὶ οὖν εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΕΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΒ πρὸς ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει. ἴση ἄρα ἡ ΑΖ τῇ ΖΒ: διπλῆ ἄρα ἡ ΑΒ τῆς ΑΖ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΓΔ τῆς ΓΗ ἐστι διπλῆ: καί ἐστιν ἴση ἡ ΑΒ τῇ ΓΔ: ἴση ἄρα καὶ ἡ ΑΖ τῇ ΓΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΕΓ, ἴσον καὶ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ τῆς ΕΓ. ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΕ ἴσα τὰ ἀπὸ τῶν ΑΖ, ΕΖ: ὀρθὴ γὰρ ἡ πρὸς τῷ Ζ γωνία: τῷ δὲ ἀπὸ τῆς ΕΓ ἴσα τὰ ἀπὸ τῶν ΕΗ, ΗΓ: ὀρθὴ γὰρ ἡ πρὸς τῷ Η γωνία: τὰ ἄρα ἀπὸ τῶν ΑΖ, ΖΕ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΓΗ, ΗΕ, ὧν τὸ ἀπὸ τῆς ΑΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΗ: ἴση γάρ ἐστιν ἡ ΑΖ τῇ ΓΗ: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΖΕ τῷ ἀπὸ τῆς ΕΗ ἴσον ἐστίν: ἴση ἄρα ἡ ΕΖ τῇ ΕΗ. ἐν δὲ κύκλῳ ἴσον ἀπέχειν ἀπὸ τοῦ κέντρου εὐθεῖαι λέγονται, ὅταν αἱ ἀπὸ τοῦ κέντρου ἐπ᾽ αὐτὰς κάθετοι ἀγόμεναι ἴσαι ὦσιν: αἱ ἄρα ΑΒ, ΓΔ ἴσον ἀπέχουσιν ἀπὸ τοῦ κέντρου. ἀλλὰ δὴ αἱ ΑΒ, ΓΔ εὐθεῖαι ἴσον ἀπεχέτωσαν ἀπὸ τοῦ κέντρου, τουτέστιν ἴση ἔστω ἡ ΕΖ τῇ ΕΗ. λέγω, ὅτι ἴση ἐστὶ καὶ ἡ ΑΒ τῇ ΓΔ. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι διπλῆ ἐστιν ἡ μὲν ΑΒ τῆς ΑΖ, ἡ δὲ ΓΔ τῆς ΓΗ: καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΓΕ, ἴσον ἐστὶ τὸ ἀπὸ τῆς ΑΕ τῷ ἀπὸ τῆς ΓΕ: ἀλλὰ τῷ μὲν ἀπὸ τῆς ΑΕ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΕΖ, ΖΑ, τῷ δὲ ἀπὸ τῆς ΓΕ ἴσα τὰ ἀπὸ τῶν ΕΗ, ΗΓ. τὰ ἄρα ἀπὸ τῶν ΕΖ, ΖΑ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΕΗ, ΗΓ: ὧν τὸ ἀπὸ τῆς ΕΖ τῷ ἀπὸ τῆς ΕΗ ἐστιν ἴσον: ἴση γὰρ ἡ ΕΖ τῇ ΕΗ: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΑΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΓΗ: ἴση ἄρα ἡ ΑΖ τῇ ΓΗ: καί ἐστι τῆς μὲν ΑΖ διπλῆ ἡ ΑΒ, τῆς δὲ ΓΗ διπλῆ ἡ ΓΔ: ἴση ἄρα ἡ ΑΒ τῇ ΓΔ. ἐν κύκλῳ ἄρα αἱ ἴσαι εὐθεῖαι ἴσον ἀπέχουσιν ἀπὸ τοῦ κέντρου, καὶ αἱ ἴσον ἀπέχουσαι ἀπὸ τοῦ κέντρου ἴσαι ἀλλήλαις εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 428|>","<|""VertexLabel"" -> ""3.15"", ""Text"" -> ""Of straight lines in a circle the diameter is greatest, and of the rest the nearer to the centre is always greater than the more remote."", ""TextWordCount"" -> 26, ""GreekText"" -> ""ἐν κύκλῳ μεγίστη μὲν ἡ διάμετρος τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν."", ""GreekTextWordCount"" -> 18, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 20}, {""Book"" -> 1, ""Theorem"" -> 24}, {""Book"" -> 3, ""Definition"" -> 5}, {""Book"" -> 3, ""Theorem"" -> 14}}, ""Proof"" -> ""Let ABCD be a circle, let AD be its diameter and E the centre; and let BC be nearer to the diameter AD, and FG more remote; I say that AD is greatest and BC greater than FG. For from the centre E let EH, EK be drawn perpendicular to BC, FG. Then, since BC is nearer to the centre and FG more remote, EK is greater than EH. [III. Def. 5] Let EL be made equal to EH, through L let LM be drawn at right angles to EK and carried through to N, and let ME, EN, FE, EG be joined. Then, since EH is equal to EL, BC is also equal to MN. [III. 14] Again, since AE is equal to EM, and ED to EN, AD is equal to ME, EN. But ME, EN are greater than MN, [I. 20] and MN is equal to BC; therefore AD is greater than BC. And, since the two sides ME, EN are equal to the two sides FE, EG, and the angle MEN greater than the angle FEG, therefore the base MN is greater than the base FG. [I. 24] But MN was proved equal to BC. Therefore the diameter AD is greatest and BC greater than FG."", ""ProofWordCount"" -> 210, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἔστω ἡ ΑΔ, κέντρον δὲ τὸ Ε, καὶ ἔγγιον μὲν τῆς ΑΔ διαμέτρου ἔστω ἡ ΒΓ, ἀπώτερον δὲ ἡ ΖΗ: λέγω, ὅτι μεγίστη μέν ἐστιν ἡ ΑΔ, μείζων δὲ ἡ ΒΓ τῆς ΖΗ. ἤχθωσαν γὰρ ἀπὸ τοῦ Ε κέντρου ἐπὶ τὰς ΒΓ, ΖΗ κάθετοι αἱ ΕΘ, ΕΚ. καὶ ἐπεὶ ἔγγιον μὲν τοῦ κέντρου ἐστὶν ἡ ΒΓ, ἀπώτερον δὲ ἡ ΖΗ, μείζων ἄρα ἡ ΕΚ τῆς ΕΘ. κείσθω τῇ ΕΘ ἴση ἡ ΕΛ, καὶ διὰ τοῦ Λ τῇ ΕΚ πρὸς ὀρθὰς ἀχθεῖσα ἡ ΛΜ διήχθω ἐπὶ τὸ Ν, καὶ ἐπεζεύχθωσαν αἱ ΜΕ, ΕΝ, ΖΕ, ΕΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΕΘ τῇ ΕΛ, ἴση ἐστὶ καὶ ἡ ΒΓ τῇ ΜΝ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΜ, ἡ δὲ ΕΔ τῇ ΕΝ, ἡ ἄρα ΑΔ ταῖς ΜΕ, ΕΝ ἴση ἐστίν. ἀλλ᾽ αἱ μὲν ΜΕ, ΕΝ τῆς ΜΝ μείζονές εἰσιν [καὶ ἡ ΑΔ τῆς ΜΝ μείζων ἐστίν, ἴση δὲ ἡ ΜΝ τῇ ΒΓ: ἡ ΑΔ ἄρα τῆς ΒΓ μείζων ἐστίν. καὶ ἐπεὶ δύο αἱ ΜΕ, ΕΝ δύο ταῖς ΖΕ, ΕΗ ἴσαι εἰσίν, καὶ γωνία ἡ ὑπὸ ΜΕΝ γωνίας τῆς ὑπὸ ΖΕΗ μείζων ἐστίν, βάσις ἄρα ἡ ΜΝ βάσεως τῆς ΖΗ μείζων ἐστίν. ἀλλὰ ἡ ΜΝ τῇ ΒΓ ἐδείχθη ἴση καὶ ἡ ΒΓ τῆς ΖΗ μείζων ἐστίν. μεγίστη μὲν ἄρα ἡ ΑΔ διάμετρος, μείζων δὲ ἡ ΒΓ τῆς ΖΗ. ἐν κύκλῳ ἄρα μεγίστη μέν ἐστιν ἡ διάμετρος, τῶν δὲ ἄλλων ἀεὶ ἡ ἔγγιον τοῦ κέντρου τῆς ἀπώτερον μείζων ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 249|>","<|""VertexLabel"" -> ""3.16"", ""Text"" -> ""The straight line drawn at right angles to the diameter of a circle from its extremity will fall outside the circle, and into the space between the straight line and the circumference another straight line cannot be interposed; further the angle of the semicircle is greater, and the remaining angle less, than any acute rectilineal angle."", ""TextWordCount"" -> 56, ""GreekText"" -> ""ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐκτὸς πεσεῖται τοῦ κύκλου, καὶ εἰς τὸν μεταξὺ τόπον τῆς τε εὐθείας καὶ τῆς περιφερείας ἑτέρα εὐθεῖα οὐ παρεμπεσεῖται, καὶ ἡ μὲν τοῦ ἡμικυκλίου γωνία ἁπάσης γωνίας ὀξείας εὐθυγράμμου μείζων ἐστίν, ἡ δὲ λοιπὴ ἐλάττων."", ""GreekTextWordCount"" -> 46, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 17}, {""Book"" -> 1, ""Theorem"" -> 19}}, ""Proof"" -> ""Let ABC be a circle about D as centre and AB as diameter; I say that the straight line drawn from A at right angles to AB from its extremity will fall outside the circle. For suppose it does not, but, if possible, let it fall within as CA, and let DC be joined. Since DA is equal to DC, the angle DAC is also equal to the angle ACD. [I. 5] But the angle DAC is right; therefore the angle ACD is also right: thus, in the triangle ACD, the two angles DAC, ACD are equal to two right angles: which is impossible. [I. 17] Therefore the straight line drawn from the point A at right angles to BA will not fall within the circle. Similarly we can prove that neither will it fall on the circumference; therefore it will fall outside. Let it fall as AE; I say next that into the space between the straight line AE and the circumference CHA another straight line cannot be interposed. For, if possible, let another straight line be so interposed, as FA, and let DG be drawn from the point D perpendicular to FA. Then, since the angle AGD is right, and the angle DAG is less than a right angle, AD is greater than DG. [I. 19] But DA is equal to DH; therefore DH is greater than DG, the less than the greater: which is impossible. Therefore another straight line cannot be interposed into the space between the straight line and the circumference. I say further that the angle of the semicircle contained by the straight line BA and the circumference CHA is greater than any acute rectilineal angle, and the remaining angle contained by the circumference CHA and the straight line AE is less than any acute rectilineal angle. For, if there is any rectilineal angle greater than the angle contained by the straight line BA and the circumference CHA, and any rectilineal angle less than the angle contained by the circumference CHA and the straight line AE, then into the space between the circumference and the straight line AE a straight line will be interposed such as will make an angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, and another angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE. But such a straight line cannot be interposed; therefore there will not be any acute angle contained by straight lines which is greater than the angle contained by the straight line BA and the circumference CHA, nor yet any acute angle contained by straight lines which is less than the angle contained by the circumference CHA and the straight line AE."", ""ProofWordCount"" -> 469, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓ περὶ κέντρον τὸ Δ καὶ διάμετρον τὴν ΑΒ: λέγω, ὅτι ἡ ἀπὸ τοῦ Α τῇ ΑΒ πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐκτὸς πεσεῖται τοῦ κύκλου. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, πιπτέτω ἐντὸς ὡς ἡ ΓΑ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΔΓ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ ΑΓΔ. ὀρθὴ δὲ ἡ ὑπὸ ΔΑΓ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΓΔ: τριγώνου δὴ τοῦ ΑΓΔ αἱ δύο γωνίαι αἱ ὑπὸ ΔΑΓ, ΑΓΔ δύο ὀρθαῖς ἴσαι εἰσίν: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ἀπὸ τοῦ Α σημείου τῇ ΒΑ πρὸς ὀρθὰς ἀγομένη ἐντὸς πεσεῖται τοῦ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἐπὶ τῆς περιφερείας: ἐκτὸς ἄρα. πιπτέτω ὡς ἡ ΑΕ: λέγω δή, ὅτι εἰς τὸν μεταξὺ τόπον τῆς τε ΑΕ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ἑτέρα εὐθεῖα οὐ παρεμπεσεῖται. εἰ γὰρ δυνατόν, παρεμπιπτέτω ὡς ἡ ΖΑ, καὶ ἤχθω ἀπὸ τοῦ Δ σημείου ἐπὶ τὴν ΖΑ κάθετος ἡ ΔΗ. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΑΗΔ, ἐλάττων δὲ ὀρθῆς ἡ ὑπὸ ΔΑΗ, μείζων ἄρα ἡ ΑΔ τῆς ΔΗ. ἴση δὲ ἡ ΔΑ τῇ ΔΘ: μείζων ἄρα ἡ ΔΘ τῆς ΔΗ, ἡ ἐλάττων τῆς μείζονος: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα εἰς τὸν μεταξὺ τόπον τῆς τε εὐθείας καὶ τῆς περιφερείας ἑτέρα εὐθεῖα παρεμπεσεῖται. λέγω, ὅτι καὶ ἡ μὲν τοῦ ἡμικυκλίου γωνία ἡ περιεχομένη ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ἁπάσης γωνίας ὀξείας εὐθυγράμμου μείζων ἐστίν, ἡ δὲ λοιπὴ ἡ περιεχομένη ὑπό τε τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας ἁπάσης γωνίας ὀξείας εὐθυγράμμου ἐλάττων ἐστίν. εἰ γὰρ ἐστί τις γωνία εὐθύγραμμος μείζων μὲν τῆς περιεχομένης ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας, ἐλάττων δὲ τῆς περιεχομένης ὑπό τε τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας, εἰς τὸν μεταξὺ τόπον τῆς τε ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας εὐθεῖα περεμπεσεῖται, ἥτις ποιήσει μείζονα μὲν τῆς περιεχομένης ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ὑπὸ εὐθειῶν περιεχομένην, ἐλάττονα δὲ τῆς περιεχομένης ὑπό τε τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας. οὐ παρεμπίπτει δέ: οὐκ ἄρα τῆς περιεχομένης γωνίας ὑπό τε τῆς ΒΑ εὐθείας καὶ τῆς ΓΘΑ περιφερείας ἔσται μείζων ὀξεῖα ὑπὸ εὐθειῶν περιεχομένη, οὐδὲ μὴν ἐλάττων τῆς περιεχομένης ὑπό τε τῆς ΓΘΑ περιφερείας καὶ τῆς ΑΕ εὐθείας. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου καὶ ὅτι εὐθεῖα κύκλου καθ᾽ ἓν μόνον ἐφάπτεται σημεῖον, ἐπειδήπερ καὶ ἡ κατὰ δύο αὐτῷ συμβάλλουσα ἐντὸς αὐτοῦ πίπτουσα ἐδείχθη. ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 417|>","<|""VertexLabel"" -> ""3.17"", ""Text"" -> ""From a given point to draw a straight line touching a given circle."", ""TextWordCount"" -> 13, ""GreekText"" -> ""ἀπὸ τοῦ δοθέντος σημείου τοῦ δοθέντος κύκλου ἐφαπτομένην εὐθεῖαν γραμμὴν ἀγαγεῖν."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 3, ""Theorem"" -> 1}, {""Book"" -> 3, ""Theorem"" -> 16}}, ""Proof"" -> ""Let A be the given point, and BCD the given circle; thus it is required to draw from the point A a straight line touching the circle BCD. For let the centre E of the circle be taken; [III. 1] let AE be joined, and with centre E and distance EA let the circle AFG be described; from D let DF be drawn at right angles to EA, and let EF, AB be joined; I say that AB has been drawn from the point A touching the circle BCD. For, since E is the centre of the circles BCD, AFG, EA is equal to EF, and ED to EB; therefore the two sides AE, EB are equal to the two sides FE, ED: and they contain a common angle, the angle at E; therefore the base DF is equal to the base AB, and the triangle DEF is equal to the triangle BEA, and the remaining angles to the remaining angles; [I. 4]therefore the angle EDF is equal to the angle EBA. But the angle EDF is right; therefore the angle EBA is also right. Now EB is a radius; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16] therefore AB touches the circle BCD. Therefore from the given point A the straight line AB has been drawn touching the circle BCD."", ""ProofWordCount"" -> 236, ""GreekProof"" -> ""ἔστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ὁ δὲ δοθεὶς κύκλος ὁ ΒΓΔ: δεῖ δὴ ἀπὸ τοῦ Α σημείου τοῦ ΒΓΔ κύκλου ἐφαπτομένην εὐθεῖαν γραμμὴν ἀγαγεῖν. εἰλήφθω γὰρ τὸ κέντρον τοῦ κύκλου τὸ Ε, καὶ ἐπεζεύχθω ἡ ΑΕ, καὶ κέντρῳ μὲν τῷ Ε διαστήματι δὲ τῷ ΕΑ κύκλος γεγράφθω ὁ ΑΖΗ, καὶ ἀπὸ τοῦ Δ τῇ ΕΑ πρὸς ὀρθὰς ἤχθω ἡ ΔΖ, καὶ ἐπεζεύχθωσαν αἱ ΕΖ, ΑΒ: λέγω, ὅτι ἀπὸ τοῦ Α σημείου τοῦ ΒΓΔ κύκλου ἐφαπτομένη ἦκται ἡ ΑΒ. ἐπεὶ γὰρ τὸ Ε κέντρον ἐστὶ τῶν ΒΓΔ, ΑΖΗ κύκλων, ἴση ἄρα ἐστὶν ἡ μὲν ΕΑ τῇ ΕΖ, ἡ δὲ ΕΔ τῇ ΕΒ: δύο δὴ αἱ ΑΕ, ΕΒ δύο ταῖς ΖΕ, ΕΔ ἴσαι εἰσίν: καὶ γωνίαν κοινὴν περιέχουσι τὴν πρὸς τῷ Ε: βάσις ἄρα ἡ ΔΖ βάσει τῇ ΑΒ ἴση ἐστίν, καὶ τὸ ΔΕΖ τρίγωνον τῷ ΕΒΑ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις: ἴση ἄρα ἡ ὑπὸ ΕΔΖ τῇ ὑπὸ ΕΒΑ. ὀρθὴ δὲ ἡ ὑπὸ ΕΔΖ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΕΒΑ. καί ἐστιν ἡ ΕΒ ἐκ τοῦ κέντρου: ἡ δὲ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου: ἡ ΑΒ ἄρα ἐφάπτεται τοῦ ΒΓΔ κύκλου. ἀπὸ τοῦ ἄρα δοθέντος σημείου τοῦ Α τοῦ δοθέντος κύκλου τοῦ ΒΓΔ ἐφαπτομένη εὐθεῖα γραμμὴ ἦκται ἡ ΑΒ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 216|>","<|""VertexLabel"" -> ""3.18"", ""Text"" -> ""If a straight line touch a circle, and a straight line be joined from the centre to the point of contact, the straight line so joined will be perpendicular to the tangent."", ""TextWordCount"" -> 32, ""GreekText"" -> ""ἐὰν κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τοῦ κέντρου ἐπὶ τὴν ἁφὴν ἐπιζευχθῇ τις εὐθεῖα, ἡ ἐπιζευχθεῖσα κάθετος ἔσται ἐπὶ τὴν ἐφαπτομένην."", ""GreekTextWordCount"" -> 22, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 17}, {""Book"" -> 1, ""Theorem"" -> 19}}, ""Proof"" -> ""For let a straight line DE touch the circle ABC at the point C, let the centre F of the circle ABC be taken, and let FC be joined from F to C; I say that FC is perpendicular to DE. For, if not, let FG be drawn from F perpendicular to DE. Then, since the angle FGC is right, the angle FCG is acute; [I. 17]and the greater angle is subtended by the greater side; [I. 19] therefore FC is greater than FG. But FC is equal to FB; therefore FB is also greater than FG, the less than the greater: which is impossible. Therefore FG is not perpendicular to DE. Similarly we can prove that neither is any other straight line except FC; therefore FC is perpendicular to DE."", ""ProofWordCount"" -> 132, ""GreekProof"" -> ""κύκλου γὰρ τοῦ ΑΒΓ ἐφαπτέσθω τις εὐθεῖα ἡ ΔΕ κατὰ τὸ Γ σημεῖον, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓ κύκλου τὸ Ζ, καὶ ἀπὸ τοῦ Ζ ἐπὶ τὸ Γ ἐπεζεύχθω ἡ ΖΓ: λέγω, ὅτι ἡ ΖΓ κάθετός ἐστιν ἐπὶ τὴν ΔΕ. εἰ γὰρ μή, ἤχθω ἀπὸ τοῦ Ζ ἐπὶ τὴν ΔΕ κάθετος ἡ ΖΗ. ἐπεὶ οὖν ἡ ὑπὸ ΖΗΓ γωνία ὀρθή ἐστιν, ὀξεῖα ἄρα ἐστὶν ἡ ὑπὸ ΖΓΗ: ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει: μείζων ἄρα ἡ ΖΓ τῆς ΖΗ: ἴση δὲ ἡ ΖΓ τῇ ΖΒ: μείζων ἄρα καὶ ἡ ΖΒ τῆς ΖΗ ἡ ἐλάττων τῆς μείζονος: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἡ ΖΗ κάθετός ἐστιν ἐπὶ τὴν ΔΕ. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλη τις πλὴν τῆς ΖΓ: ἡ ΖΓ ἄρα κάθετός ἐστιν ἐπὶ τὴν ΔΕ. ἐὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τοῦ κέντρου ἐπὶ τὴν ἁφὴν ἐπιζευχθῇ τις εὐθεῖα, ἡ ἐπιζευχθεῖσα κάθετος ἔσται ἐπὶ τὴν ἐφαπτομένην: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 157|>","<|""VertexLabel"" -> ""3.19"", ""Text"" -> ""If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the tangent, the centre of the circle will be on the straight line so drawn."", ""TextWordCount"" -> 37, ""GreekText"" -> ""ἐὰν κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς τῇ ἐφαπτομένῃ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἀχθῇ, ἐπὶ τῆς ἀχθείσης ἔσται τὸ κέντρον τοῦ κύκλου."", ""GreekTextWordCount"" -> 25, ""References"" -> {{""Book"" -> 3, ""Theorem"" -> 18}}, ""Proof"" -> ""For let a straight line DE touch the circle ABC at the point C, and from C let CA be drawn at right angles to DE; I say that the centre of the circle is on AC. For suppose it is not, but, if possible, let F be the centre, and let CF be joined. Since a straight line DE touches the circle ABC, and FC has been joined from the centre to the point of contact, FC is perpendicular to DE; [III. 18]therefore the angle FCE is right. But the angle ACE is also right; therefore the angle FCE is equal to the angle ACE, the less to the greater: which is impossible. Therefore F is not the centre of the circle ABC. Similarly we can prove that neither is any other point except a point on AC."", ""ProofWordCount"" -> 140, ""GreekProof"" -> ""κύκλου γὰρ τοῦ ΑΒΓ ἐφαπτέσθω τις εὐθεῖα ἡ ΔΕ κατὰ τὸ Γ σημεῖον, καὶ ἀπὸ τοῦ Γ τῇ ΔΕ πρὸς ὀρθὰς ἤχθω ἡ ΓΑ: λέγω, ὅτι ἐπὶ τῆς ΑΓ ἐστι τὸ κέντρον τοῦ κύκλου. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἔστω τὸ Ζ, καὶ ἐπεζεύχθω ἡ ΓΖ. ἐπεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΔΕ, ἀπὸ δὲ τοῦ κέντρου ἐπὶ τὴν ἁφὴν ἐπέζευκται ἡ ΖΓ, ἡ ΖΓ ἄρα κάθετός ἐστιν ἐπὶ τὴν ΔΕ: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΓΕ. ἐστὶ δὲ καὶ ἡ ὑπὸ ΑΓΕ ὀρθή: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΖΓΕ τῇ ὑπὸ ΑΓΕ ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τὸ Ζ κέντρον ἐστὶ τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δείξομεν, ὅτι οὐδ᾽ ἄλλο τι πλὴν ἐπὶ τῆς ΑΓ. ἐὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς τῇ ἐφαπτομένῃ πρὸς ὀρθὰς εὐθεῖα γραμμὴ ἀχθῇ, ἐπὶ τῆς ἀχθείσης ἔσται τὸ κέντρον τοῦ κύκλου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 153|>","<|""VertexLabel"" -> ""3.20"", ""Text"" -> ""In a circle the angle at the centre is double of the angle at the circumference, when the angles have the same circumference as base."", ""TextWordCount"" -> 25, ""GreekText"" -> ""ἐν κύκλῳ ἡ πρὸς τῷ κέντρῳ γωνία διπλασίων ἐστὶ τῆς πρὸς τῇ περιφερείᾳ, ὅταν τὴν αὐτὴν περιφέρειαν βάσιν ἔχωσιν αἱ γωνίαι."", ""GreekTextWordCount"" -> 21, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 32}}, ""Proof"" -> ""Let ABC be a circle, let the angle BEC be an angleat its centre, and the angle BAC an angle at the circumference, and let them have the same circumference BC as base; I say that the angle BEC is double ofthe angle BAC. For let AE be joined and drawn through to F. Then, since EA is equal to EB, the angle EAB is also equal to the angle EBA; [I. 5]therefore the angles EAB, EBA are double of the angle EAB. But the angle BEF is equal to the angles EAB, EBA; [I. 32] therefore the angle BEF is also double of the angleEAB. For the same reason the angle FEC is also double of the angle EAC. Therefore the whole angle BEC is double of the whole angle BAC. Again let another straight line be inflected, and let there be another angle BDC; let DE be joined and produced to G. Similarly then we can prove that the angle GEC is double of the angle EDC,of which the angle GEB is double of the angle EDB; therefore the angle BEC which remains is double of the angle BDC."", ""ProofWordCount"" -> 193, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓ, καὶ πρὸς μὲν τῷ κέντρῳ αὐτοῦ γωνία ἔστω ἡ ὑπὸ ΒΕΓ, πρὸς δὲ τῇ περιφερείᾳ ἡ ὑπὸ ΒΑΓ, ἐχέτωσαν δὲ τὴν αὐτὴν περιφέρειαν βάσιν τὴν ΒΓ: λέγω, ὅτι διπλασίων ἐστὶν ἡ ὑπὸ ΒΕΓ γωνία τῆς ὑπὸ ΒΑΓ. ἐπιζευχθεῖσα γὰρ ἡ ΑΕ διήχθω ἐπὶ τὸ Ζ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΕΑ τῇ ΕΒ, ἴση καὶ γωνία ἡ ὑπὸ ΕΑΒ τῇ ὑπὸ ΕΒΑ: αἱ ἄρα ὑπὸ ΕΑΒ, ΕΒΑ γωνίαι τῆς ὑπὸ ΕΑΒ διπλασίους εἰσίν. ἴση δὲ ἡ ὑπὸ ΒΕΖ ταῖς ὑπὸ ΕΑΒ, ΕΒΑ: καὶ ἡ ὑπὸ ΒΕΖ ἄρα τῆς ὑπὸ ΕΑΒ ἐστι διπλῆ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΖΕΓ τῆς ὑπὸ ΕΑΓ ἐστι διπλῆ. ὅλη ἄρα ἡ ὑπὸ ΒΕΓ ὅλης τῆς ὑπὸ ΒΑΓ ἐστι διπλῆ. Κεκλάσθω δὴ πάλιν, καὶ ἔστω ἑτέρα γωνία ἡ ὑπὸ ΒΔΓ, καὶ ἐπιζευχθεῖσα ἡ ΔΕ ἐκβεβλήσθω ἐπὶ τὸ Η. ὁμοίως δὴ δείξομεν, ὅτι διπλῆ ἐστιν ἡ ὑπὸ ΗΕΓ γωνία τῆς ὑπὸ ΕΔΓ, ὧν ἡ ὑπὸ ΗΕΒ διπλῆ ἐστι τῆς ὑπὸ ΕΔΒ: λοιπὴ ἄρα ἡ ὑπὸ ΒΕΓ διπλῆ ἐστι τῆς ὑπὸ ΒΔΓ. ἐν κύκλῳ ἄρα ἡ πρὸς τῷ κέντρῳ γωνία διπλασίων ἐστὶ τῆς πρὸς τῇ περιφερείᾳ, ὅταν τὴν αὐτὴν περιφέρειαν βάσιν ἔχωσιν αἱ γωνίαι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 195|>","<|""VertexLabel"" -> ""3.21"", ""Text"" -> ""In a circle the angles in the same segment are equal to one another."", ""TextWordCount"" -> 14, ""GreekText"" -> ""ἐν κύκλῳ αἱ ἐν τῷ αὐτῷ τμήματι γωνίαι ἴσαι ἀλλήλαις εἰσίν."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Book"" -> 3, ""Theorem"" -> 20}}, ""Proof"" -> ""Let ABCD be a circle, and let the angles BAD, BED be angles in the same segment BAED; I say that the angles BAD, BED are equal to one another. For let the centre of the circle ABCD be taken, and let it be F; let BF, FD be joined. Now, since the angle BFD is at the centre, and the angle BAD at the circumference, and they have the same circumference BCD as base, therefore the angle BFD is double of the angle BAD. [III. 20] For the same reason the angle BFD is also double of the angle BED; therefore the angle BAD is equal to the angle BED."", ""ProofWordCount"" -> 111, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν τῷ αὐτῷ τμήματι τῷ ΒΑΕΔ γωνίαι ἔστωσαν αἱ ὑπὸ ΒΑΔ, ΒΕΔ: λέγω, ὅτι αἱ ὑπὸ ΒΑΔ, ΒΕΔ γωνίαι ἴσαι ἀλλήλαις εἰσίν. εἰλήφθω γὰρ τοῦ ΑΒΓΔ κύκλου τὸ κέντρον, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΒΖ, ΖΔ. καὶ ἐπεὶ ἡ μὲν ὑπὸ ΒΖΔ γωνία πρὸς τῷ κέντρῳ ἐστίν, ἡ δὲ ὑπὸ ΒΑΔ πρὸς τῇ περιφερείᾳ, καὶ ἔχουσι τὴν αὐτὴν περιφέρειαν βάσιν τὴν ΒΓΔ, ἡ ἄρα ὑπὸ ΒΖΔ γωνία διπλασίων ἐστὶ τῆς ὑπὸ ΒΑΔ. διὰ τὰ αὐτὰ δὴ ἡ ὑπὸ ΒΖΔ καὶ τῆς ὑπὸ ΒΕΔ ἐστι διπλασίων: ἴση ἄρα ἡ ὑπὸ ΒΑΔ τῇ ὑπὸ ΒΕΔ. ἐν κύκλῳ ἄρα αἱ ἐν τῷ αὐτῷ τμήματι γωνίαι ἴσαι ἀλλήλαις εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 115|>","<|""VertexLabel"" -> ""3.22"", ""Text"" -> ""The opposite angles of quadrilaterals in circles are equal to two right angles."", ""TextWordCount"" -> 13, ""GreekText"" -> ""τῶν ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 3, ""Theorem"" -> 21}}, ""Proof"" -> ""Let ABCD be a circle, and let ABCD be a quadrilateral in it; I say that the opposite angles are equal to two right angles. Let AC, BD be joined. Then, since in any triangle the three angles are equal to two right angles, [I. 32] the three angles CAB, ABC, BCA of the triangle ABC are equal to two right angles. But the angle CAB is equal to the angle BDC, for they are in the same segment BADC; [III. 21] and the angle ACB is equal to the angle ADB, for they are in the same segment ADCB; therefore the whole angle ADC is equal to the angles BAC, ACB. Let the angle ABC be added to each; therefore the angles ABC, BAC, ACB are equal to the angles ABC, ADC. But the angles ABC, BAC, ACB are equal to two right angles; therefore the angles ABC, ADC are also equal to two right angles. Similarly we can prove that the angles BAD, DCB are also equal to two right angles."", ""ProofWordCount"" -> 173, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓΔ, καὶ ἐν αὐτῷ τετράπλευρον ἔστω τὸ ΑΒΓΔ: λέγω, ὅτι αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἐπεζεύχθωσαν αἱ ΑΓ, ΒΔ. ἐπεὶ οὖν παντὸς τριγώνου αἱ τρεῖς γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν, τοῦ ΑΒΓ ἄρα τριγώνου αἱ τρεῖς γωνίαι αἱ ὑπὸ ΓΑΒ, ΑΒΓ, ΒΓΑ δυσὶν ὀρθαῖς ἴσαι εἰσίν. ἴση δὲ ἡ μὲν ὑπὸ ΓΑΒ τῇ ὑπὸ ΒΔΓ: ἐν γὰρ τῷ αὐτῷ τμήματί εἰσι τῷ ΒΑΔΓ: ἡ δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΑΔΒ: ἐν γὰρ τῷ αὐτῷ τμήματί εἰσι τῷ ΑΔΓΒ: ὅλη ἄρα ἡ ὑπὸ ΑΔΓ ταῖς ὑπὸ ΒΑΓ, ΑΓΒ ἴση ἐστίν. κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ: αἱ ἄρα ὑπὸ ΑΒΓ, ΒΑΓ, ΑΓΒ ταῖς ὑπὸ ΑΒΓ, ΑΔΓ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΑΒΓ, ΒΑΓ, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. καὶ αἱ ὑπὸ ΑΒΓ, ΑΔΓ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ὑπὸ ΒΑΔ, ΔΓΒ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. τῶν ἄρα ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 162|>","<|""VertexLabel"" -> ""3.23"", ""Text"" -> ""On the same straight line there cannot be constructed two similar and unequal segments of circles on the same side."", ""TextWordCount"" -> 20, ""GreekText"" -> ""ἐπὶ τῆς αὐτῆς εὐθείας δύο τμήματα κύκλων ὅμοια καὶ ἄνισα οὐ συσταθήσεται ἐπὶ τὰ αὐτὰ μέρη."", ""GreekTextWordCount"" -> 16, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 16}, {""Book"" -> 3, ""Definition"" -> 11}}, ""Proof"" -> ""For, if possible, on the same straight line AB let two similar and unequal segments of circles ACB, ADB be constructed on the same side; let ACD be drawn through, and let CB, DB be joined. Then, since the segment ACB is similar to the segment ADB, and similar segments of circles are those which admit equal angles, [III. Def. 11] the angle ACB is equal to the angle ADB, the exterior to the interior: which is impossible. [I. 16]"", ""ProofWordCount"" -> 80, ""GreekProof"" -> ""εἰ γὰρ δυνατόν, ἐπὶ τῆς αὐτῆς εὐθείας τῆς ΑΒ δύο τμήματα κύκλων ὅμοια καὶ ἄνισα συνεστάτω ἐπὶ τὰ αὐτὰ μέρη τὰ ΑΓΒ, ΑΔΒ, καὶ διήχθω ἡ ΑΓΔ, καὶ ἐπεζεύχθωσαν αἱ ΓΒ, ΔΒ. ἐπεὶ οὖν ὅμοιόν ἐστι τὸ ΑΓΒ τμῆμα τῷ ΑΔΒ τμήματι, ὅμοια δὲ τμήματα κύκλων ἐστὶ τὰ δεχόμενα γωνίας ἴσας, ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΓΒ γωνία τῇ ὑπὸ ΑΔΒ ἡ ἐκτὸς τῇ ἐντός: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐπὶ τῆς αὐτῆς εὐθείας δύο τμήματα κύκλων ὅμοια καὶ ἄνισα συσταθήσεται ἐπὶ τὰ αὐτὰ μέρη: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 88|>","<|""VertexLabel"" -> ""3.24"", ""Text"" -> ""Similar segments of circles on equal straight lines are equal to one another."", ""TextWordCount"" -> 13, ""GreekText"" -> ""τὰ ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις ἐστίν."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Book"" -> 3, ""Theorem"" -> 10}}, ""Proof"" -> ""For let AEB, CFD be similar segments of circles on equal straight lines AB, CD; I say that the segment AEB is equal to the segment CFD. For, if the segment AEB be applied to CFD, and if the point A be placed on C and the straight line AB on CD, the point B will also coincide with the point D, because AB is equal to CD; and, AB coinciding with CD, the segment AEB will also coincide with CFD. For, if the straight line AB coincide with CD but the segment AEB do not coincide with CFD, it will either fall with it, or outside it;or it will fall awry, as CGD, and a circle cuts a circle at more points than two: which is impossible. [III. 10] Therefore, if the straight line AB be applied to CD, the segment AEB will not fail to coincide with CFD also; therefore it will coincide with it and will be equal to it."", ""ProofWordCount"" -> 164, ""GreekProof"" -> ""ἔστωσαν γὰρ ἐπὶ ἴσων εὐθειῶν τῶν ΑΒ, ΓΔ ὅμοια τμήματα κύκλων τὰ ΑΕΒ, ΓΖΔ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΕΒ τμῆμα τῷ ΓΖΔ τμήματι. Ἐφαρμοζομένου γὰρ τοῦ ΑΕΒ τμήματος ἐπὶ τὸ ΓΖΔ καὶ τιθεμένου τοῦ μὲν Α σημείου ἐπὶ τὸ Γ τῆς δὲ ΑΒ εὐθείας ἐπὶ τὴν ΓΔ, ἐφαρμόσει καὶ τὸ Β σημεῖον ἐπὶ τὸ Δ σημεῖον διὰ τὸ ἴσην εἶναι τὴν ΑΒ τῇ ΓΔ: τῆς δὲ ΑΒ ἐπὶ τὴν ΓΔ ἐφαρμοσάσης ἐφαρμόσει καὶ τὸ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ. εἰ γὰρ ἡ ΑΒ εὐθεῖα ἐπὶ τὴν ΓΔ ἐφαρμόσει, τὸ δὲ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ μὴ ἐφαρμόσει, ἤτοι ἐντὸς αὐτοῦ πεσεῖται ἢ ἐκτὸς ἢ παραλλάξει ὡς τὸ ΓΗΔ, καὶ κύκλος κύκλον τέμνει κατὰ πλείονα σημεῖα ἢ δύο: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἐφαρμοζομένης τῆς ΑΒ εὐθείας ἐπὶ τὴν ΓΔ οὐκ ἐφαρμόσει καὶ τὸ ΑΕΒ τμῆμα ἐπὶ τὸ ΓΖΔ: ἐφαρμόσει ἄρα, καὶ ἴσον αὐτῷ ἔσται. τὰ ἄρα ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 159|>","<|""VertexLabel"" -> ""3.25"", ""Text"" -> ""Given a segment of a circle, to describe the complete circle of which it is a segment."", ""TextWordCount"" -> 17, ""GreekText"" -> ""κύκλου τμήματος δοθέντος προσαναγράψαι τὸν κύκλον, οὗπέρ ἐστι τμῆμα."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 6}, {""Book"" -> 3, ""Theorem"" -> 9}}, ""Proof"" -> ""Let ABC be the given segment of a circle; thus it is required to describe the complete circle belonging to the segment ABC, that is, of which it is a segment. For let AC be bisected at D, let DB be drawn from the point D at right angles to AC, and let AB. be joined; the angle ABD is then greater than, equal to, or less than the angle BAD. First let it be greater; and on the straight line BA, and at the point A on it, let the angle BAE be constructed equal to the angle ABD; let DB be drawn through to E, and let EC be joined. Then, since the angle ABE is equal to the angle BAE, the straight line EB is also equal to EA. [I. 6] And, since AD is equal to DC, and DE is common, the two sides AD, DE are equal to the two sides CD, DE respectively; and the angle ADE is equal to the angle CDE, for each is right; therefore the base AE is equal to the base CE. But AE was proved equal to BE; therefore BE is also equal to CE; therefore the three straight lines AE, EB, EC are equal to one another. Therefore the circle drawn with centre E and distance one of the straight lines AE, EB, EC will also pass through the remaining points and will have been completed. [III. 9] Therefore, given a segment of a circle, the complete circle has been described. And it is manifest that the segment ABC is less than a semicircle, because the centre E happens to be outside it. Similarly, even if the angle ABD be equal to the angle BAD, AD being equal to each of the two BD, DC, the three straight lines DA, DB, DC will be equal to one another, D will be the centre of the completed circle, and ABC will clearly be a semicircle. But, if the angle ABD be less than the angle BAD, and if we construct, on the straight line BA and at the point A on it, an angle equal to the angle ABD, the centre will fall on DB within the segment ABC, and the segment ABC will clearly be greater than a semicircle. Therefore, given a segment of a circle, the complete circle has been described."", ""ProofWordCount"" -> 395, ""GreekProof"" -> ""ἔστω τὸ δοθὲν τμῆμα κύκλου τὸ ΑΒΓ: δεῖ δὴ τοῦ ΑΒΓ τμήματος προσαναγράψαι τὸν κύκλον, οὗπέρ ἐστι τμῆμα. τετμήσθω γὰρ ἡ ΑΓ δίχα κατὰ τὸ Δ, καὶ ἤχθω ἀπὸ τοῦ Δ σημείου τῇ ΑΓ πρὸς ὀρθὰς ἡ ΔΒ, καὶ ἐπεζεύχθω ἡ ΑΒ: ἡ ὑπὸ ΑΒΔ γωνία ἄρα τῆς ὑπὸ ΒΑΔ ἤτοι μείζων ἐστὶν ἢ ἴση ἢ ἐλάττων. ἔστω πρότερον μείζων, καὶ συνεστάτω πρὸς τῇ ΒΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΒΔ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΕ, καὶ διήχθω ἡ ΔΒ ἐπὶ τὸ Ε, καὶ ἐπεζεύχθω ἡ ΕΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΑΒΕ γωνία τῇ ὑπὸ ΒΑΕ, ἴση ἄρα ἐστὶ καὶ ἡ ΕΒ εὐθεῖα τῇ ΕΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΓ, κοινὴ δὲ ἡ ΔΕ, δύο δὴ αἱ ΑΔ, ΔΕ δύο ταῖς ΓΔ, ΔΕ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ: καὶ γωνία ἡ ὑπὸ ΑΔΕ γωνίᾳ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση: ὀρθὴ γὰρ ἑκατέρα: βάσις ἄρα ἡ ΑΕ βάσει τῇ ΓΕ ἐστιν ἴση. ἀλλὰ ἡ ΑΕ τῇ ΒΕ ἐδείχθη ἴση: καὶ ἡ ΒΕ ἄρα τῇ ΓΕ ἐστιν ἴση: αἱ τρεῖς ἄρα αἱ ΑΕ, ΕΒ, ΕΓ ἴσαι ἀλλήλαις εἰσίν: ὁ ἄρα κέντρῳ τῷ Ε διαστήματι δὲ ἑνὶ τῶν ΑΕ, ΕΒ, ΕΓ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται προσαναγεγραμμένος. κύκλου ἄρα τμήματος δοθέντος προσαναγέγραπται ὁ κύκλος. καὶ δῆλον, ὡς τὸ ΑΒΓ τμῆμα ἔλαττόν ἐστιν ἡμικυκλίου διὰ τὸ τὸ Ε κέντρον ἐκτὸς αὐτοῦ τυγχάνειν. ὁμοίως δὲ κἂν ᾖ ἡ ὑπὸ ΑΒΔ γωνία ἴση τῇ ὑπὸ ΒΑΔ, τῆς ΑΔ ἴσης γενομένης ἑκατέρᾳ τῶν ΒΔ, ΔΓ αἱ τρεῖς αἱ ΔΑ, ΔΒ, ΔΓ ἴσαι ἀλλήλαις ἔσονται, καὶ ἔσται τὸ Δ κέντρον τοῦ προσαναπεπληρωμένου κύκλου, καὶ δηλαδὴ ἔσται τὸ ΑΒΓ ἡμικύκλιον. ἐὰν δὲ ἡ ὑπὸ ΑΒΔ ἐλάττων ᾖ τῆς ὑπὸ ΒΑΔ, καὶ συστησώμεθα πρὸς τῇ ΒΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΒΔ γωνίᾳ ἴσην, ἐντὸς τοῦ ΑΒΓ τμήματος πεσεῖται τὸ κέντρον ἐπὶ τῆς ΔΒ, καὶ ἔσται δηλαδὴ τὸ ΑΒΓ τμῆμα μεῖζον ἡμικυκλίου. κύκλου ἄρα τμήματος δοθέντος προσαναγέγραπται ὁ κύκλος: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 331|>","<|""VertexLabel"" -> ""3.26"", ""Text"" -> ""In equal circles equal angles stand on equal circumferences, whether they stand at the centres or at the circumferences."", ""TextWordCount"" -> 19, ""GreekText"" -> ""ἐν τοῖς ἴσοις κύκλοις αἱ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι."", ""GreekTextWordCount"" -> 23, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 3, ""Definition"" -> 11}, {""Book"" -> 3, ""Theorem"" -> 24}}, ""Proof"" -> ""Let ABC, DEF be equal circles, and in them let there be equal angles, namely at the centres the angles BGC, EHF, and at the circumferences the angles BAC, EDF; I say that the circumference BKC is equal to the circumference ELF. For let BC, EF be joined. Now, since the circles ABC, DEF are equal, the radii are equal. Thus the two straight lines BG, GC are equal to the two straight lines EH, HF; and the angle at G is equal to the angle at H; therefore the base BC is equal to the base EF. [I. 4] And, since the angle at A is equal to the angle at D, the segment BAC is similar to the segment EDF; [III. Def. 11]and they are upon equal straight lines. But similar segments of circles on equal straight lines are equal to one another; [III. 24] therefore the segment BAC is equal to EDF. But the whole circle ABC is also equal to the whole circle DEF; therefore the circumference BKC which remains is equal to the circumference ELF."", ""ProofWordCount"" -> 181, ""GreekProof"" -> ""ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ καὶ ἐν αὐτοῖς ἴσαι γωνίαι ἔστωσαν πρὸς μὲν τοῖς κέντροις αἱ ὑπὸ ΒΗΓ, ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ: λέγω, ὅτι ἴση ἐστὶν ἡ ΒΚΓ περιφέρεια τῇ ΕΛΖ περιφερείᾳ. ἐπεζεύχθωσαν γὰρ αἱ ΒΓ, ΕΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΑΒΓ, ΔΕΖ κύκλοι, ἴσαι εἰσὶν αἱ ἐκ τῶν κέντρων: δύο δὴ αἱ ΒΗ, ΗΓ δύο ταῖς ΕΘ, ΘΖ ἴσαι: καὶ γωνία ἡ πρὸς τῷ η γωνίᾳ τῇ πρὸς τῷ Θ ἴση: βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἐστιν ἴση. καὶ ἐπεὶ ἴση ἐστὶν ἡ πρὸς τῷ Α γωνία τῇ πρὸς τῷ Δ, ὅμοιον ἄρα ἐστὶ τὸ ΒΑΓ τμῆμα τῷ ΕΔΖ τμήματι: καί εἰσιν ἐπὶ ἴσων εὐθειῶν τῶν ΒΓ, ΕΖ: τὰ δὲ ἐπὶ ἴσων εὐθειῶν ὅμοια τμήματα κύκλων ἴσα ἀλλήλοις ἐστίν: ἴσον ἄρα τὸ ΒΑΓ τμῆμα τῷ ΕΔΖ. ἔστι δὲ καὶ ὅλος ὁ ΑΒΓ κύκλος ὅλῳ τῷ ΔΕΖ κύκλῳ ἴσος: λοιπὴ ἄρα ἡ ΒΚΓ περιφέρεια τῇ ΕΛΖ περιφερείᾳ ἐστὶν ἴση. ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 185|>","<|""VertexLabel"" -> ""3.27"", ""Text"" -> ""In equal circles angles standing on equal circumferences are equal to one another, whether they stand at the centres or at the circumferences."", ""TextWordCount"" -> 23, ""GreekText"" -> ""ἐν τοῖς ἴσοις κύκλοις αἱ ἐπὶ ἴσων περιφερειῶν βεβηκυῖαι γωνίαι ἴσαι ἀλλήλαις εἰσίν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι."", ""GreekTextWordCount"" -> 25, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 3, ""Theorem"" -> 20}, {""Book"" -> 3, ""Theorem"" -> 26}}, ""Proof"" -> ""For in equal circles ABC, DEF, on equal circumferences BC, EF, let the angles BGC, EHF stand at the centres G, H, and the angles BAC, EDF at the circumferences; I say that the angle BGC is equal to the angle EHF, and the angle BAC is equal to the angle EDF. For, if the angle BGC is unequal to the angle EHF, one of them is greater. Let the angle BGC be greater: and on the straight line BG, and at the point G on it, let the angle BGK be constructed equal to the angle EHF. [I. 23] Now equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference BK is equal to the circumference EF. But EF is equal to BC; therefore BK is also equal to BC, the less to the greater: which is impossible. Therefore the angle BGC is not unequal to the angle EHF; therefore it is equal to it. And the angle at A is half of the angle BGC, and the angle at D half of the angle EHF; [III. 20] therefore the angle at A is also equal to the angle at D."", ""ProofWordCount"" -> 199, ""GreekProof"" -> ""ἐν γὰρ ἴσοις κύκλοις τοῖς ΑΒΓ, ΔΕΖ ἐπὶ ἴσων περιφερειῶν τῶν ΒΓ, ΕΖ πρὸς μὲν τοῖς Η, Θ κέντροις γωνίαι βεβηκέτωσαν αἱ ὑπὸ ΒΗΓ, ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ: λέγω, ὅτι ἡ μὲν ὑπὸ ΒΗΓ γωνία τῇ ὑπὸ ΕΘΖ ἐστιν ἴση, ἡ δὲ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΒΗΓ τῇ ὑπὸ ΕΘΖ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΒΗΓ, καὶ συνεστάτω πρὸς τῇ ΒΗ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Η τῇ ὑπὸ ΕΘΖ γωνίᾳ ἴση ἡ ὑπὸ ΒΗΚ: αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ὅταν πρὸς τοῖς κέντροις ὦσιν: ἴση ἄρα ἡ ΒΚ περιφέρεια τῇ ΕΖ περιφερείᾳ. ἀλλὰ ἡ ΕΖ τῇ ΒΓ ἐστιν ἴση: καὶ ἡ ΒΚ ἄρα τῇ ΒΓ ἐστιν ἴση ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΒΗΓ γωνία τῇ ὑπὸ ΕΘΖ: ἴση ἄρα. καί ἐστι τῆς μὲν ὑπὸ ΒΗΓ ἡμίσεια ἡ πρὸς τῷ Α, τῆς δὲ ὑπὸ ΕΘΖ ἡμίσεια ἡ πρὸς τῷ Δ: ἴση ἄρα καὶ ἡ πρὸς τῷ Α γωνία τῇ πρὸς τῷ Δ. ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ ἐπὶ ἴσων περιφερειῶν βεβηκυῖαι γωνίαι ἴσαι ἀλλήλαις εἰσίν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 211|>","<|""VertexLabel"" -> ""3.28"", ""Text"" -> ""In equal circles equal straight lines cut off equal circumferences, the greater equal to the greater and the less to the less."", ""TextWordCount"" -> 22, ""GreekText"" -> ""ἐν τοῖς ἴσοις κύκλοις αἱ ἴσαι εὐθεῖαι ἴσας περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ ἐλάττονι."", ""GreekTextWordCount"" -> 20, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 8}, {""Book"" -> 3, ""Theorem"" -> 26}}, ""Proof"" -> ""Let ABC, DEF be equal circles, and in the circles let AB, DE be equal straight lines cutting off ACB, DFE as greater circumferences and AGB, DHE as lesser; I say that the greater circumference ACB is equal to the greater circumference DFE, and the less circumference AGB to DHE. For let the centres K, L of the circles be taken, and let AK, KB, DL, LE be joined. Now, since the circles are equal, the radii are also equal; therefore the two sides AK, KB are equal to the two sides DL, LE; and the base AB is equal to the base DE; therefore the angle AKB is equal to the angle DLE. [I. 8] But equal angles stand on equal circumferences, when they are at the centres; [III. 26] therefore the circumference AGB is equal to DHE. And the whole circle ABC is also equal to the whole circle DEF; therefore the circumference ACB which remains is also equal to the circumference DFE which remains."", ""ProofWordCount"" -> 167, ""GreekProof"" -> ""ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ ἐν τοῖς κύκλοις ἴσαι εὐθεῖαι ἔστωσαν αἱ ΑΒ, ΔΕ τὰς μὲν ΑΓΒ, ΔΖΕ περιφερείας μείζονας ἀφαιροῦσαι τὰς δὲ ΑΗΒ, ΔΘΕ ἐλάττονας: λέγω, ὅτι ἡ μὲν ΑΓΒ μείζων περιφέρεια ἴση ἐστὶ τῇ ΔΖΕ μείζονι περιφερείᾳ, ἡ δὲ ΑΗΒ ἐλάττων περιφέρεια τῇ ΔΘΕ. εἰλήφθω γὰρ τὰ κέντρα τῶν κύκλων τὰ Κ, Λ, καὶ ἐπεζεύχθωσαν αἱ ΑΚ, ΚΒ, ΔΛ, ΛΕ. καὶ ἐπεὶ ἴσοι κύκλοι εἰσίν, ἴσαι εἰσὶ καὶ αἱ ἐκ τῶν κέντρων: δύο δὴ αἱ ΑΚ, ΚΒ δυσὶ ταῖς ΔΛ, ΛΕ ἴσαι εἰσίν: καὶ βάσις ἡ ΑΒ βάσει τῇ ΔΕ ἴση: γωνία ἄρα ἡ ὑπὸ ΑΚΒ γωνίᾳ τῇ ὑπὸ ΔΛΕ ἴση ἐστίν. αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν, ὅταν πρὸς τοῖς κέντροις ὦσιν: ἴση ἄρα ἡ ΑΗΒ περιφέρεια τῇ ΔΘΕ. ἐστὶ δὲ καὶ ὅλος ὁ ΑΒΓ κύκλος ὅλῳ τῷ ΔΕΖ κύκλῳ ἴσος: καὶ λοιπὴ ἄρα ἡ ΑΓΒ περιφέρεια λοιπῇ τῇ ΔΖΕ περιφερείᾳ ἴση ἐστίν. ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ ἴσαι εὐθεῖαι ἴσας περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ ἐλάττονι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 174|>","<|""VertexLabel"" -> ""3.29"", ""Text"" -> ""In equal circles equal circumferences are subtended by equal straight lines."", ""TextWordCount"" -> 11, ""GreekText"" -> ""ἐν τοῖς ἴσοις κύκλοις τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 3, ""Theorem"" -> 27}}, ""Proof"" -> ""Let ABC, DEF be equal circles, and in them let equal circumferences BGC, EHF be cut off; and let the straight lines BC, EF be joined; I say that BC is equal to EF. For let the centres of the circles be taken, and let them be K, L; let BK, KC, EL, LF be joined. Now, since the circumference BGC is equal to the circumference EHF, the angle BKC is also equal to the angle ELF. [III. 27] And, since the circles ABC, DEF are equal, the radii are also equal; therefore the two sides BK, KC are equal to the two sides EL, LF; and they contain equal angles; therefore the base BC is equal to the base EF. [I. 4]"", ""ProofWordCount"" -> 123, ""GreekProof"" -> ""ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ ἐν αὐτοῖς ἴσαι περιφέρειαι ἀπειλήφθωσαν αἱ ΒΗΓ, ΕΘΖ, καὶ ἐπεζεύχθωσαν αἱ ΒΓ, ΕΖ εὐθεῖαι: λέγω, ὅτι ἴση ἐστὶν ἡ ΒΓ τῇ ΕΖ. εἰλήφθω γὰρ τὰ κέντρα τῶν κύκλων, καὶ ἔστω τὰ Κ, Λ, καὶ ἐπεζεύχθωσαν αἱ ΒΚ, ΚΓ, ΕΛ, ΛΖ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΗΓ περιφέρεια τῇ ΕΘΖ περιφερείᾳ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΚΓ τῇ ὑπὸ ΕΛΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΑΒΓ, ΔΕΖ κύκλοι, ἴσαι εἰσὶ καὶ αἱ ἐκ τῶν κέντρων: δύο δὴ αἱ ΒΚ, ΚΓ δυσὶ ταῖς ΕΛ, ΛΖ ἴσαι εἰσίν: καὶ γωνίας ἴσας περιέχουσιν: βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΖ ἴση ἐστίν. ἐν ἄρα τοῖς ἴσοις κύκλοις τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 120|>","<|""VertexLabel"" -> ""3.30"", ""Text"" -> ""To bisect a given circumference."", ""TextWordCount"" -> 5, ""GreekText"" -> ""τὴν δοθεῖσαν περιφέρειαν δίχα τεμεῖν."", ""GreekTextWordCount"" -> 5, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 3, ""Theorem"" -> 28}}, ""Proof"" -> ""Let ADB be the given circumference; thus it is required to bisect the circumference ADB. Let AB be joined and bisected at C; from the point C let CD be drawn at right angles to the straight line AB, and let AD, DB be joined. Then, since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD; and the angle ACD is equal to the angle BCD, for each is right; therefore the base AD is equal to the base DB. [I. 4] But equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the less; [III. 28] and each of the circumferences AD, DB is less than a semicircle; therefore the circumference AD is equal to the circumference DB. Therefore the given circumference has been bisected at the point D."", ""ProofWordCount"" -> 150, ""GreekProof"" -> ""ἔστω ἡ δοθεῖσα περιφέρεια ἡ ΑΔΒ: δεῖ δὴ τὴν ΑΔΒ περιφέρειαν δίχα τεμεῖν. ἐπεζεύχθω ἡ ΑΒ, καὶ τετμήσθω δίχα κατὰ τὸ Γ, καὶ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ πρὸς ὀρθὰς ἤχθω ἡ ΓΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, κοινὴ δὲ ἡ ΓΔ, δύο δὴ αἱ ΑΓ, ΓΔ δυσὶ ταῖς ΒΓ, ΓΔ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΑΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση: ὀρθὴ γὰρ ἑκατέρα: βάσις ἄρα ἡ ΑΔ βάσει τῇ ΔΒ ἴση ἐστίν. αἱ δὲ ἴσαι εὐθεῖαι ἴσας περιφερείας ἀφαιροῦσι τὴν μὲν μείζονα τῇ μείζονι τὴν δὲ ἐλάττονα τῇ ἐλάττονι: καί ἐστιν ἑκατέρα τῶν ΑΔ, ΔΒ περιφερειῶν ἐλάττων ἡμικυκλίου: ἴση ἄρα ἡ ΑΔ περιφέρεια τῇ ΔΒ περιφερείᾳ. ἡ ἄρα δοθεῖσα περιφέρεια δίχα τέτμηται κατὰ τὸ Δ σημεῖον: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 132|>","<|""VertexLabel"" -> ""3.31"", ""Text"" -> ""In a circle the angle in the semicircle is right, that in a greater segment less than a right angle, and that in a less segment greater than a right angle; and further the angle of the greater segment is greater than a right angle, and the angle of the less segment less than a right angle."", ""TextWordCount"" -> 57, ""GreekText"" -> ""ἐν κύκλῳ ἡ μὲν ἐν τῷ ἡμικυκλίῳ γωνία ὀρθή ἐστιν, ἡ δὲ ἐν τῷ μείζονι τμήματι ἐλάττων ὀρθῆς, ἡ δὲ ἐν τῷ ἐλάττονι τμήματι μείζων ὀρθῆς: καὶ ἔτι ἡ μὲν τοῦ μείζονος τμήματος γωνία μείζων ἐστὶν ὀρθῆς, ἡ δὲ τοῦ ἐλάττονος τμήματος γωνία ἐλάττων ὀρθῆς."", ""GreekTextWordCount"" -> 45, ""References"" -> {{""Book"" -> 1, ""Definition"" -> 10}, {""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 17}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 3, ""Theorem"" -> 22}}, ""Proof"" -> ""Let ABCD be a circle, let BC be its diameter, and E its centre, and let BA, AC, AD, DC be joined; I say that the angle BAC in the semicircle BAC is right, the angle ABC in the segment ABC greater than the semicircle is less than a right angle, and the angle ADC in the segment ADC less than the semicircle is greater than a right angle. Let AE be joined, and let BA be carried through to F. Then, since BE is equal to EA, the angle ABE is also equal to the angle BAE. [I. 5] Again, since CE is equal to EA, the angle ACE is also equal to the angle CAE. [I. 5] Therefore the whole angle BAC is equal to the two angles ABC, ACB. But the angle FAC exterior to the triangle ABC is also equal to the two angles ABC, ACB; [I. 32] therefore the angle BAC is also equal to the angle FAC; therefore each is right; [I. Def. 10]therefore the angle BAC in the semicircle BAC is right. Next, since in the triangle ABC the two angles ABC, BAC are less than two right angles, [I. 17] and the angle BAC is a right angle, the angle ABC is less than a right angle; and it is the angle in the segment ABC greater than the semicircle. Next, since ABCD is a quadrilateral in a circle, and the opposite angles of quadrilaterals in circles are equal to two right angles, [III. 22] while the angle ABC is less than a right angle, therefore the angle ADC which remains is greater than a right angle; and it is the angle in the segment ADC less than the semicircle. I say further that the angle of the greater segment, namely that contained by the circumference ABC and the straight line AC, is greater than a right angle; and the angle of the less segment, namely that contained by the circumference ADC and the straight line AC, is less than a right angle. This is at once manifest. For, since the angle contained by the straight lines BA, AC is right, the angle contained by the circumference ABC and the straight line AC is greater than a right angle. Again, since the angle contained by the straight lines AC, AF is right, the angle contained by the straight line CA and the circumference ADC is less than a right angle."", ""ProofWordCount"" -> 409, ""GreekProof"" -> ""ἔστω κύκλος ὁ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἔστω ἡ ΒΓ, κέντρον δὲ τὸ Ε, καὶ ἐπεζεύχθωσαν αἱ ΒΑ, ΑΓ, ΑΔ, ΔΓ: λέγω, ὅτι ἡ μὲν ἐν τῷ ΒΑΓ ἡμικυκλίῳ γωνία ἡ ὑπὸ ΒΑΓ ὀρθή ἐστιν, ἡ δὲ ἐν τῷ ΑΒΓ μείζονι τοῦ ἡμικυκλίου τμήματι γωνία ἡ ὑπὸ ΑΒΓ ἐλάττων ἐστὶν ὀρθῆς, ἡ δὲ ἐν τῷ ΑΔΓ ἐλάττονι τοῦ ἡμικυκλίου τμήματι γωνία ἡ ὑπὸ ΑΔΓ μείζων ἐστὶν ὀρθῆς. ἐπεζεύχθω ἡ ΑΕ, καὶ διήχθω ἡ ΒΑ ἐπὶ τὸ Ζ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΑ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΒΕ τῇ ὑπὸ ΒΑΕ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ΓΕ τῇ ΕΑ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΓΑΕ: ὅλη ἄρα ἡ ὑπὸ ΒΑΓ δυσὶ ταῖς ὑπὸ ΑΒΓ, ΑΓΒ ἴση ἐστίν. ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΑΓ ἐκτὸς τοῦ ΑΒΓ τριγώνου δυσὶ ταῖς ὑπὸ ΑΒΓ, ΑΓΒ γωνίαις ἴση: ἴση ἄρα καὶ ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΖΑΓ: ὀρθὴ ἄρα ἑκατέρα: ἡ ἄρα ἐν τῷ ΒΑΓ ἡμικυκλίῳ γωνία ἡ ὑπὸ ΒΑΓ ὀρθή ἐστιν. καὶ ἐπεὶ τοῦ ΑΒΓ τριγώνου δύο γωνίαι αἱ ὑπὸ ΑΒΓ, ΒΑΓ δύο ὀρθῶν ἐλάττονές εἰσιν, ὀρθὴ δὲ ἡ ὑπὸ ΒΑΓ, ἐλάττων ἄρα ὀρθῆς ἐστιν ἡ ὑπὸ ΑΒΓ γωνία: καί ἐστιν ἐν τῷ ΑΒΓ μείζονι τοῦ ἡμικυκλίου τμήματι. καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΑΒΓΔ, τῶν δὲ ἐν τοῖς κύκλοις τετραπλεύρων αἱ ἀπεναντίον γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν αἱ ἄρα ὑπὸ ΑΒΓ, ΑΔΓ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν, καί ἐστιν ἡ ὑπὸ ΑΒΓ ἐλάττων ὀρθῆς: λοιπὴ ἄρα ἡ ὑπὸ ΑΔΓ γωνία μείζων ὀρθῆς ἐστιν: καί ἐστιν ἐν τῷ ΑΔΓ ἐλάττονι τοῦ ἡμικυκλίου τμήματι. λέγω, ὅτι καὶ ἡ μὲν τοῦ μείζονος τμήματος γωνία ἡ περιεχομένη ὑπό τε τῆς ΑΒΓ περιφερείας καὶ τῆς ΑΓ εὐθείας μείζων ἐστὶν ὀρθῆς, ἡ δὲ τοῦ ἐλάττονος τμήματος γωνία ἡ περιεχομένη ὑπό τε τῆς ΑΔΓ περιφερείας καὶ τῆς ΑΓ εὐθείας ἐλάττων ἐστὶν ὀρθῆς. καί ἐστιν αὐτόθεν φανερόν. ἐπεὶ γὰρ ἡ ὑπὸ τῶν ΒΑ, ΑΓ εὐθειῶν ὀρθή ἐστιν, ἡ ἄρα ὑπὸ τῆς ΑΒΓ περιφερείας καὶ τῆς ΑΓ εὐθείας περιεχομένη μείζων ἐστὶν ὀρθῆς. πάλιν, ἐπεὶ ἡ ὑπὸ τῶν ΑΓ, ΑΖ εὐθειῶν ὀρθή ἐστιν, ἡ ἄρα ὑπὸ τῆς ΓΑ εὐθείας καὶ τῆς ΑΔΓ περιφερείας περιεχομένη ἐλάττων ἐστὶν ὀρθῆς. ἐν κύκλῳ ἄρα ἡ μὲν ἐν τῷ ἡμικυκλίῳ γωνία ὀρθή ἐστιν, ἡ δὲ ἐν τῷ μείζονι τμήματι ἐλάττων ὀρθῆς, ἡ δὲ ἐν τῷ ἐλάττονι τμήματι μείζων ὀρθῆς, καὶ ἔτι ἡ μὲν τοῦ μείζονος τμήματος γωνία μείζων ἐστὶν ὀρθῆς, ἡ δὲ τοῦ ἐλάττονος τμήματος γωνία ἐλάττων ὀρθῆς: ὅπερ ἔδει δεῖξαι. πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἡ μία γωνία τριγώνου ταῖς δυσὶν ἴση ᾖ, ὀρθή ἐστιν ἡ γωνία διὰ τὸ καὶ τὴν ἐκείνης ἐκτὸς ταῖς αὐταῖς ἴσην εἶναι: ἐὰν δὲ αἱ ἐφεξῆς ἴσαι ὦσιν, ὀρθαί εἰσιν."", ""GreekProofWordCount"" -> 440|>","<|""VertexLabel"" -> ""3.32"", ""Text"" -> ""If a straight line touch a circle, and from the point of contact there be drawn across, in the circle, a straight line cutting the circle, the angles which it makes with the tangent will be equal to the angles in the alternate segments of the circle."", ""TextWordCount"" -> 47, ""GreekText"" -> ""ἐὰν κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς εἰς τὸν κύκλον διαχθῇ τις εὐθεῖα τέμνουσα τὸν κύκλον, ἃς ποιεῖ γωνίας πρὸς τῇ ἐφαπτομένῃ, ἴσαι ἔσονται ταῖς ἐν τοῖς ἐναλλὰξ τοῦ κύκλου τμήμασι γωνίαις."", ""GreekTextWordCount"" -> 34, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 3, ""Theorem"" -> 19}, {""Book"" -> 3, ""Theorem"" -> 22}, {""Book"" -> 3, ""Theorem"" -> 31}}, ""Proof"" -> ""For let a straight line EF touch the circle ABCD at the point B, and from the point B let there be drawn across, in the circle ABCD, a straight line BD cutting it; I say that the angles which BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle, that is, that the angle FBD is equal to the angle constructed in the segment BAD, and the angle EBD is equal to the angle constructed in the segment DCB. For let BA be drawn from B at right angles to EF, let a point C be taken at random on the circumference BD, and let AD, DC, CB be joined. Then, since a straight line EF touches the circle ABCD at B, and BA has been drawn from the point of contact at right angles to the tangent, the centre of the circle ABCD is on BA. [III. 19] Therefore BA is a diameter of the circle ABCD; therefore the angle ADB, being an angle in a semicircle, is right. [III. 31] Therefore the remaining angles BAD, ABD are equal to one right angle. [I. 32] But the angle ABF is also right; therefore the angle ABF is equal to the angles BAD, ABD. Let the angle ABD be subtracted from each; therefore the angle DBF which remains is equal to the angle BAD in the alternate segment of the circle. Next, since ABCD is a quadrilateral in a circle, its opposite angles are equal to two right angles. [III. 22] But the angles DBF, DBE are also equal to two right angles; therefore the angles DBF, DBE are equal to the angles BAD, BCD, of which the angle BAD was proved equal to the angle DBF; therefore the angle DBE which remains is equal to the angle DCB in the alternate segment DCB of the circle."", ""ProofWordCount"" -> 317, ""GreekProof"" -> ""κύκλου γὰρ τοῦ ΑΒΓΔ ἐφαπτέσθω τις εὐθεῖα ἡ ΕΖ κατὰ τὸ Β σημεῖον, καὶ ἀπὸ τοῦ Β σημείου διήχθω τις εὐθεῖα εἰς τὸν ΑΒΓΔ κύκλον τέμνουσα αὐτὸν ἡ ΒΔ. λέγω, ὅτι ἃς ποιεῖ γωνίας ἡ ΒΔ μετὰ τῆς ΕΖ ἐφαπτομένης, ἴσαι ἔσονται ταῖς ἐν τοῖς ἐναλλὰξ τμήμασι τοῦ κύκλου γωνίαις, τουτέστιν, ὅτι ἡ μὲν ὑπὸ ΖΒΔ γωνία ἴση ἐστὶ τῇ ἐν τῷ ΒΑΔ τμήματι συνισταμένῃ γωνίᾳ, ἡ δὲ ὑπὸ ΕΒΔ γωνία ἴση ἐστὶ τῇ ἐν τῷ ΔΓΒ τμήματι συνισταμένῃ γωνίᾳ. ἤχθω γὰρ ἀπὸ τοῦ Β τῇ ΕΖ πρὸς ὀρθὰς ἡ ΒΑ, καὶ εἰλήφθω ἐπὶ τῆς ΒΔ περιφερείας τυχὸν σημεῖον τὸ Γ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ, ΓΒ. καὶ ἐπεὶ κύκλου τοῦ ΑΒΓΔ ἐφάπτεταί τις εὐθεῖα ἡ ΕΖ κατὰ τὸ Β, καὶ ἀπὸ τῆς ἁφῆς ἦκται τῇ ἐφαπτομένῃ πρὸς ὀρθὰς ἡ ΒΑ, ἐπὶ τῆς ΒΑ ἄρα τὸ κέντρον ἐστὶ τοῦ ΑΒΓΔ κύκλου. ἡ ΒΑ ἄρα διάμετρός ἐστι τοῦ ΑΒΓΔ κύκλου: ἡ ἄρα ὑπὸ ΑΔΒ γωνία ἐν ἡμικυκλίῳ οὖσα ὀρθή ἐστιν. λοιπαὶ ἄρα αἱ ὑπὸ ΒΑΔ, ΑΒΔ μιᾷ ὀρθῇ ἴσαι εἰσίν. ἐστὶ δὲ καὶ ἡ ὑπὸ ΑΒΖ ὀρθή: ἡ ἄρα ὑπὸ ΑΒΖ ἴση ἐστὶ ταῖς ὑπὸ ΒΑΔ, ΑΒΔ. κοινὴ ἀφῃρήσθω ἡ ὑπὸ ΑΒΔ: λοιπὴ ἄρα ἡ ὑπὸ ΔΒΖ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τμήματι τοῦ κύκλου γωνίᾳ τῇ ὑπὸ ΒΑΔ. καὶ ἐπεὶ ἐν κύκλῳ τετράπλευρόν ἐστι τὸ ΑΒΓΔ, αἱ ἀπεναντίον αὐτοῦ γωνίαι δυσὶν ὀρθαῖς ἴσαι εἰσίν. εἰσὶ δὲ καὶ αἱ ὑπὸ ΔΒΖ, ΔΒΕ δυσὶν ὀρθαῖς ἴσαι: αἱ ἄρα ὑπὸ ΔΒΖ, ΔΒΕ ταῖς ὑπὸ ΒΑΔ, ΒΓΔ ἴσαι εἰσίν, ὧν ἡ ὑπὸ ΒΑΔ τῇ ὑπὸ ΔΒΖ ἐδείχθη ἴση: λοιπὴ ἄρα ἡ ὑπὸ ΔΒΕ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι τῷ ΔΓΒ τῇ ὑπὸ ΔΓΒ γωνίᾳ ἐστὶν ἴση. ἐὰν ἄρα κύκλου ἐφάπτηταί τις εὐθεῖα, ἀπὸ δὲ τῆς ἁφῆς εἰς τὸν κύκλον διαχθῇ τις εὐθεῖα τέμνουσα τὸν κύκλον, ἃς ποιεῖ γωνίας πρὸς τῇ ἐφαπτομένῃ, ἴσαι ἔσονται ταῖς ἐν τοῖς ἐναλλὰξ τοῦ κύκλου τμήμασι γωνίαις: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 314|>","<|""VertexLabel"" -> ""3.33"", ""Text"" -> ""On a given straight line to describe a segment of a circle admitting an angle equal to a given rectilineal angle."", ""TextWordCount"" -> 21, ""GreekText"" -> ""ἐπὶ τῆς δοθείσης εὐθείας γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ."", ""GreekTextWordCount"" -> 14, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 3, ""Theorem"" -> 16}, {""Book"" -> 3, ""Theorem"" -> 31}, {""Book"" -> 3, ""Theorem"" -> 32}}, ""Proof"" -> ""Let AB be the given straight line, and the angle at C the given rectilineal angle; thus it is required to describe on the given straight line AB a segment of a circle admitting an angle equal to the angle at C. The angle at C is then acute, or right, or obtuse. First let it be acute, and, as in the first figure, on the straight line AB, and at the point A, let the angle BAD be constructed equal to the angle at C; therefore the angle BAD is also acute. Let AE be drawn at right angles to DA, let AB be bisected at F, let FG be drawn from the point F at right angles to AB, and let GB be joined. Then, since AF is equal to FB, and FG is common, the two sides AF, FG are equal to the two sides BF, FG; and the angle AFG is equal to the angle BFG; therefore the base AG is equal to the base BG. [I. 4] Therefore the circle described with centre G and distance GA will pass through B also. Let it be drawn, and let it be ABE; let EB be joined. Now, since AD is drawn from A, the extremity of the diameter AE, at right angles to AE, therefore AD touches the circle ABE. [III. 16] Since then a straight line AD touches the circle ABE, and from the point of contact at A a straight line AB is drawn across in the circle ABE, the angle DAB is equal to the angle AEB in the alternate segment of the circle. [III. 32] But the angle DAB is equal to the angle at C; therefore the angle at C is also equal to the angle AEB. Therefore on the given straight line AB the segment AEB of a circle has been described admitting the angle AEB equal to the given angle, the angle at C. Next let the angle at C be right; and let it be again required to describe on AB a segment of a circle admitting an angle equal to the right angle at C. Let the angle BAD be constructed equal to the right angle at C, as is the case in the second figure; let AB be bisected at F, and with centre F and distance either FA or FB let the circle AEB be described. Therefore the straight line AD touches the circle ABE, because the angle at A is right. [III. 16] And the angle BAD is equal to the angle in the segment AEB, for the latter too is itself a right angle, being an angle in a semicircle. [III. 31] But the angle BAD is also equal to the angle at C. Therefore the angle AEB is also equal to the angle at C. Therefore again the segment AEB of a circle has been described on AB admitting an angle equal to the angle at C. Next, let the angle at C be obtuse; and on the straight line AB, and at the point A, let the angle BAD be constructed equal to it, as is the case in the third figure; let AE be drawn at right angles to AD, let AB be again bisected at F, let FG be drawn at right angles to AB, and let GB be joined. Then, since AF is again equal to FB, and FG is common, the two sides AF, FG are equal to the two sides BF, FG; and the angle AFG is equal to the angle BFG; therefore the base AG is equal to the base BG. [I. 4] Therefore the circle described with centre G and distance GA will pass through B also; let it so pass, as AEB. Now, since AD is drawn at right angles to the diameter AE from its extremity, AD touches the circle AEB. [III. 16] And AB has been drawn across from the point of contact at A; therefore the angle BAD is equal to the angle constructed in the alternate segment AHB of the circle. [III. 32] But the angle BAD is equal to the angle at C. Therefore the angle in the segment AHB is also equal to the angle at C. Therefore on the given straight line AB the segment AHB of a circle has been described admitting an angle equal to the angle at C."", ""ProofWordCount"" -> 735, ""GreekProof"" -> ""ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ πρὸς τῷ Γ: δεῖ δὴ ἐπὶ τῆς δοθείσης εὐθείας τῆς ΑΒ γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ. ἡ δὴ πρὸς τῷ Γ γωνία ἤτοι ὀξεῖά ἐστιν ἢ ὀρθὴ ἢ ἀμβλεῖα: ἔστω πρότερον ὀξεῖα, καὶ ὡς ἐπὶ τῆς πρώτης καταγραφῆς συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ α σημείῳ τῇ πρὸς τῷ Γ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΔ: ὀξεῖα ἄρα ἐστὶ καὶ ἡ ὑπὸ ΒΑΔ. ἤχθω τῇ ΔΑ πρὸς ὀρθὰς ἡ ΑΕ, καὶ τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ ἤχθω ἀπὸ τοῦ Ζ σημείου τῇ ΑΒ πρὸς ὀρθὰς ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΗΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, κοινὴ δὲ ἡ ΖΗ, δύο δὴ αἱ ΑΖ, ΖΗ δύο ταῖς ΒΖ, ΖΗ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΑΖΗ γωνίᾳ τῇ ὑπὸ ΒΖΗ ἴση: βάσις ἄρα ἡ ΑΗ βάσει τῇ ΒΗ ἴση ἐστίν. ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ τῷ ΗΑ κύκλος γραφόμενος ἥξει καὶ διὰ τοῦ Β. γεγράφθω καὶ ἔστω ὁ ΑΒΕ, καὶ ἐπεζεύχθω ἡ ΕΒ. ἐπεὶ οὖν ἀπ᾽ ἄκρας τῆς ΑΕ διαμέτρου ἀπὸ τοῦ Α τῇ ΑΕ πρὸς ὀρθάς ἐστιν ἡ ΑΔ, ἡ ΑΔ ἄρα ἐφάπτεται τοῦ ΑΒΕ κύκλου: ἐπεὶ οὖν κύκλου τοῦ ΑΒΕ ἐφάπτεταί τις εὐθεῖα ἡ ΑΔ, καὶ ἀπὸ τῆς κατὰ τὸ Α ἁφῆς εἰς τὸν ΑΒΕ κύκλον διῆκταί τις εὐθεῖα ἡ ΑΒ, ἡ ἄρα ὑπὸ ΔΑΒ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι γωνίᾳ τῇ ὑπὸ ΑΕΒ. ἀλλ᾽ ἡ ὑπὸ ΔΑΒ τῇ πρὸς τῷ Γ ἐστιν ἴση: καὶ ἡ πρὸς τῷ Γ ἄρα γωνία ἴση ἐστὶ τῇ ὑπὸ ΑΕΒ. ἐπὶ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τμῆμα κύκλου γέγραπται τὸ ΑΕΒ δεχόμενον γωνίαν τὴν ὑπὸ ΑΕΒ ἴσην τῇ δοθείσῃ τῇ πρὸς τῷ Γ. ἀλλὰ δὴ ὀρθὴ ἔστω ἡ πρὸς τῷ Γ: καὶ δέον πάλιν ἔστω ἐπὶ τῆς ΑΒ γράψαι τμῆμα κύκλου δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ ὀρθῇ γωνίᾳ. συνεστάτω πάλιν τῇ πρὸς τῷ Γ ὀρθῇ γωνίᾳ ἴση ἡ ὑπὸ ΒΑΔ, ὡς ἔχει ἐπὶ τῆς δευτέρας καταγραφῆς, καὶ τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ κέντρῳ τῷ Ζ, διαστήματι δὲ ὁποτέρῳ τῶν ΖΑ, ΖΒ, κύκλος γεγράφθω ὁ ΑΕΒ. ἐφάπτεται ἄρα ἡ ΑΔ εὐθεῖα τοῦ ΑΒΕ κύκλου διὰ τὸ ὀρθὴν εἶναι τὴν πρὸς τῷ Α γωνίαν. καὶ ἴση ἐστὶν ἡ ὑπὸ ΒΑΔ γωνία τῇ ἐν τῷ ΑΕΒ τμήματι: ὀρθὴ γὰρ καὶ αὐτὴ ἐν ἡμικυκλίῳ οὖσα. ἀλλὰ καὶ ἡ ὑπὸ ΒΑΔ τῇ πρὸς τῷ Γ ἴση ἐστίν. καὶ ἡ ἐν τῷ ΑΕΒ ἄρα ἴση ἐστὶ τῇ πρὸς τῷ Γ. γέγραπται ἄρα πάλιν ἐπὶ τῆς ΑΒ τμῆμα κύκλου τὸ ΑΕΒ δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ. ἀλλὰ δὴ ἡ πρὸς τῷ Γ ἀμβλεῖα ἔστω: καὶ συνεστάτω αὐτῇ ἴση πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ Α σημείῳ ἡ ὑπὸ ΒΑΔ, ὡς ἔχει ἐπὶ τῆς τρίτης καταγραφῆς, καὶ τῇ ΑΔ πρὸς ὀρθὰς ἤχθω ἡ ΑΕ, καὶ τετμήσθω πάλιν ἡ ΑΒ δίχα κατὰ τὸ Ζ, καὶ τῇ ΑΒ πρὸς ὀρθὰς ἤχθω ἡ ΖΗ, καὶ ἐπεζεύχθω ἡ ΗΒ. καὶ ἐπεὶ πάλιν ἴση ἐστὶν ἡ ΑΖ τῇ ΖΒ, καὶ κοινὴ ἡ ΖΗ, δύο δὴ αἱ ΑΖ, ΖΗ δύο ταῖς ΒΖ, ΖΗ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΑΖΗ γωνίᾳ τῇ ὑπὸ ΒΖΗ ἴση: βάσις ἄρα ἡ ΑΗ βάσει τῇ ΒΗ ἴση ἐστίν: ὁ ἄρα κέντρῳ μὲν τῷ η διαστήματι δὲ τῷ ΗΑ κύκλος γραφόμενος ἥξει καὶ διὰ τοῦ Β. ἐρχέσθω ὡς ὁ ΑΕΒ. καὶ ἐπεὶ τῇ ΑΕ διαμέτρῳ ἀπ᾽ ἄκρας πρὸς ὀρθάς ἐστιν ἡ ΑΔ, ἡ ΑΔ ἄρα ἐφάπτεται τοῦ ΑΕΒ κύκλου. καὶ ἀπὸ τῆς κατὰ τὸ Α ἐπαφῆς διῆκται ἡ ΑΒ: ἡ ἄρα ὑπὸ ΒΑΔ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι τῷ ΑΘΒ συνισταμένῃ γωνίᾳ. ἀλλ᾽ ἡ ὑπὸ ΒΑΔ γωνία τῇ πρὸς τῷ Γ ἴση ἐστίν. καὶ ἡ ἐν τῷ ΑΘΒ ἄρα τμήματι γωνία ἴση ἐστὶ τῇ πρὸς τῷ Γ. ἐπὶ τῆς ἄρα δοθείσης εὐθείας τῆς ΑΒ γέγραπται τμῆμα κύκλου τὸ ΑΘΒ δεχόμενον γωνίαν ἴσην τῇ πρὸς τῷ Γ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 656|>","<|""VertexLabel"" -> ""3.34"", ""Text"" -> ""From a given circle to cut off a segment admitting an angle equal to a given rectilineal angle."", ""TextWordCount"" -> 18, ""GreekText"" -> ""ἀπὸ τοῦ δοθέντος κύκλου τμῆμα ἀφελεῖν δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ."", ""GreekTextWordCount"" -> 13, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 3, ""Theorem"" -> 32}}, ""Proof"" -> ""Let ABC be the given circle, and the angle at D the given rectilineal angle; thus it is required to cut off from the circle ABC a segment admitting an angle equal to the given rectilineal angle, the angle at D. Let EF be drawn touching ABC at the point B, and on the straight line FB, and at the point B on it, let the angle FBC be constructed equal to the angle at D. [I. 23] Then, since a straight line EF touches the circle ABC, and BC has been drawn across from the point of contact at B, the angle FBC is equal to the angle constructed in the alternate segment BAC. [III. 32] But the angle FBC is equal to the angle at D; therefore the angle in the segment BAC is equal to the angle at D. Therefore from the given circle ABC the segment BAC. has been cut off admitting an angle equal to the given rectilineal angle, the angle at D."", ""ProofWordCount"" -> 168, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ πρὸς τῷ Δ: δεῖ δὴ ἀπὸ τοῦ ΑΒΓ κύκλου τμῆμα ἀφελεῖν δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ πρὸς τῷ Δ. ἤχθω τοῦ ΑΒΓ ἐφαπτομένη ἡ ΕΖ κατὰ τὸ Β σημεῖον, καὶ συνεστάτω πρὸς τῇ ΖΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β τῇ πρὸς τῷ Δ γωνίᾳ ἴση ἡ ὑπὸ ΖΒΓ. ἐπεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΕΖ, καὶ ἀπὸ τῆς κατὰ τὸ Β ἐπαφῆς διῆκται ἡ ΒΓ, ἡ ὑπὸ ΖΒΓ ἄρα γωνία ἴση ἐστὶ τῇ ἐν τῷ ΒΑΓ ἐναλλὰξ τμήματι συνισταμένῃ γωνίᾳ. ἀλλ᾽ ἡ ὑπὸ ΖΒΓ τῇ πρὸς τῷ Δ ἐστιν ἴση: καὶ ἡ ἐν τῷ ΒΑΓ ἄρα τμήματι ἴση ἐστὶ τῇ πρὸς τῷ Δ γωνίᾳ. ἀπὸ τοῦ δοθέντος ἄρα κύκλου τοῦ ΑΒΓ τμῆμα ἀφῄρηται τὸ ΒΑΓ δεχόμενον γωνίαν ἴσην τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ τῇ πρὸς τῷ Δ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 151|>","<|""VertexLabel"" -> ""3.35"", ""Text"" -> ""If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other."", ""TextWordCount"" -> 31, ""GreekText"" -> ""ἐὰν ἐν κύκλῳ δύο εὐθεῖαι τέμνωσιν ἀλλήλας, τὸ ὑπὸ τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθογωνίῳ."", ""GreekTextWordCount"" -> 25, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 47}, {""Book"" -> 2, ""Theorem"" -> 5}, {""Book"" -> 3, ""Theorem"" -> 3}}, ""Proof"" -> ""For in the circle ABCD let the two straight lines AC, BD cut one another at the point E; I say that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB. If now AC, BD are through the centre, so that E is the centre of the circle ABCD, it is manifest that, AE, EC, DE, EB being equal, the rectangle contained by AE, EC is also equal to the rectangle contained by DE, EB. Next let AC, DB not be through the centre; let the centre of ABCD be taken, and let it be F; from F let FG, FH be drawn perpendicular to the straight lines AC, DB, and let FB, FC, FE be joined. Then, since a straight line GF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3]therefore AG is equal to GC. Since, then, the straight line AC has been cut into equal parts at G and into unequal parts at E, the rectangle contained by AE, EC together with the square on EG is equal to the square on GC; [II. 5] Let the square on GF be added; therefore the rectangle AE, EC together with the squares on GE, GF is equal to the squares on CG, GF. But the square on FE is equal to the squares on EG, GF, and the square on FC is equal to the squares on CG, GF; [I. 47] therefore the rectangle AE, EC together with the square on FE is equal to the square on FC. And FC is equal to FB; therefore the rectangle AE, EC together with the square on EF is equal to the square on FB. For the same reason, also, the rectangle DE, EB together with the square on FE is equal to the square on FB. But the rectangle AE, EC together with the square on FE was also proved equal to the square on FB; therefore the rectangle AE, EC together with the square on FE is equal to the rectangle DE, EB together with the square on FE. Let the square on FE be subtracted from each; therefore the rectangle contained by AE, EC which remains is equal to the rectangle contained by DE, EB."", ""ProofWordCount"" -> 387, ""GreekProof"" -> ""ἐν γὰρ κύκλῳ τῷ ΑΒΓΔ δύο εὐθεῖαι αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε σημεῖον: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. εἰ μὲν οὖν αἱ ΑΓ, ΒΔ διὰ τοῦ κέντρου εἰσὶν ὥστε τὸ Ε κέντρον εἶναι τοῦ ΑΒΓΔ κύκλου, φανερόν, ὅτι ἴσων οὐσῶν τῶν ΑΕ, ΕΓ, ΔΕ, ΕΒ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. μὴ ἔστωσαν δὴ αἱ ΑΓ, ΔΒ διὰ τοῦ κέντρου, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓΔ, καὶ ἔστω τὸ Ζ, καὶ ἀπὸ τοῦ Ζ ἐπὶ τὰς ΑΓ, ΔΒ εὐθείας κάθετοι ἤχθωσαν αἱ ΖΗ, ΖΘ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΓ, ΖΕ. καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΗΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει: ἴση ἄρα ἡ ΑΗ τῇ ΗΓ. ἐπεὶ οὖν εὐθεῖα ἡ ΑΓ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Η, εἰς δὲ ἄνισα κατὰ τὸ Ε, τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΕΗ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΓ: κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΗΖ: τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τῶν ἀπὸ τῶν ΗΕ, ΗΖ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΓΗ, ΗΖ. ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΕΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΕ, τοῖς δὲ ἀπὸ τῶν ΓΗ, ΗΖ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΓ: τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΓ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ: τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΕΖ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΒ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΒ. ἐδείχθη δὲ καὶ τὸ ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον τῷ ἀπὸ τῆς ΖΒ: τὸ ἄρα ὑπὸ τῶν ΑΕ, ΕΓ μετὰ τοῦ ἀπὸ τῆς ΖΕ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ μετὰ τοῦ ἀπὸ τῆς ΖΕ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΖΕ: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΕ, ΕΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΔΕ, ΕΒ περιεχομένῳ ὀρθογωνίῳ. ἐὰν ἄρα ἐν κύκλῳ εὐθεῖαι δύο τέμνωσιν ἀλλήλας, τὸ ὑπὸ τῶν τῆς μιᾶς τμημάτων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν τῆς ἑτέρας τμημάτων περιεχομένῳ ὀρθογωνίῳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 390|>","<|""VertexLabel"" -> ""3.36"", ""Text"" -> ""If a point be taken outside a circle and from it there fall on the circle two straight lines, and if one of them cut the circle and the other touch it, the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference will be equal to the square on the tangent."", ""TextWordCount"" -> 70, ""GreekText"" -> ""ἐὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπ᾽ αὐτοῦ πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφαπτομένης τετραγώνῳ."", ""GreekTextWordCount"" -> 50, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 47}, {""Book"" -> 2, ""Theorem"" -> 6}, {""Book"" -> 3, ""Theorem"" -> 3}, {""Book"" -> 3, ""Theorem"" -> 18}}, ""Proof"" -> ""For let a point D be taken outside the circle ABC, and from D let the two straight lines DCA, DB fall on the circle ABC; let DCA cut the circle ABC and let BD touch it; I say that the rectangle contained by AD, DC is equal to the square on DB. Then DCA is either through the centre or not through the centre. First let it be through the centre, and let F be the centre of the circle ABC; let FB be joined; therefore the angle FBD is right. [III. 18] And, since AC has been bisected at F, and CD is added to it, the rectangle AD, DC together with the square on FC is equal to the square on FD. [II. 6] But FC is equal to FB; therefore the rectangle AD, DC together with the square on FB is equal to the square on FD. And the squares on FB, BD are equal to the square on FD; [I. 47] therefore the rectangle AD, DC together with the square on FB is equal to the squares on FB, BD. Let the square on FB be subtracted from each; therefore the rectangle AD, DC which remains is equal to the square on the tangent DB. Again, let DCA not be through the centre of the circle ABC; let the centre E be taken, and from E let EF be drawn perpendicular to AC; let EB, EC, ED be joined. Then the angle EBD is right. [III. 18] And, since a straight line EF through the centre cuts a straight line AC not through the centre at right angles, it also bisects it; [III. 3]therefore AF is equal to FC. Now, since the straight line AC has been bisected at the point F, and CD is added to it, the rectangle contained by AD, DC together with the square on FC is equal to the square on FD. [II. 6] Let the square on FE be added to each; therefore the rectangle AD, DC together with the squares on CF, FE is equal to the squares on FD, FE. But the square on EC is equal to the squares on CF, FE, for the angle EFC is right; [I. 47] and the square on ED is equal to the squares on DF, FE; therefore the rectangle AD, DC together with the square on EC is equal to the square on ED. And EC is equal to EB; therefore the rectangle AD, DC together with the square on EB is equal to the square on ED. But the squares on EB, BD are equal to the square on ED, for the angle EBD is right; [I. 47] therefore the rectangle AD, DC together with the square on EB is equal to the squares on EB, BD. Let the square on EB be subtracted from each; therefore the rectangle AD, DC which remains is equal to the square on DB."", ""ProofWordCount"" -> 493, ""GreekProof"" -> ""κύκλου γὰρ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν δύο εὐθεῖαι αἱ ΔΓΑ, ΔΒ: καὶ ἡ μὲν ΔΓΑ τεμνέτω τὸν ΑΒΓ κύκλον, ἡ δὲ ΒΔ ἐφαπτέσθω: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΔ, ΔΓ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ τετραγώνῳ. ἡ ἄρα ΔΓΑ ἤτοι διὰ τοῦ κέντρου ἐστὶν ἢ οὔ. ἔστω πρότερον διὰ τοῦ κέντρου, καὶ ἔστω τὸ Ζ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἐπεζεύχθω ἡ ΖΒ: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΖΒΔ. καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ δίχα τέτμηται κατὰ τὸ Ζ, πρόσκειται δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. ἴση δὲ ἡ ΖΓ τῇ ΖΒ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. τῷ δὲ ἀπὸ τῆς ΖΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΖΒ, ΒΔ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΒ, ΒΔ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΖΒ: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ ἐφαπτομένης. ἀλλὰ δὴ ἡ ΔΓΑ μὴ ἔστω διὰ τοῦ κέντρου τοῦ ΑΒΓ κύκλου, καὶ εἰλήφθω τὸ κέντρον τὸ Ε, καὶ ἀπὸ τοῦ Ε ἐπὶ τὴν ΑΓ κάθετος ἤχθω ἡ ΕΖ, καὶ ἐπεζεύχθωσαν αἱ ΕΒ, ΕΓ, ΕΔ: ὀρθὴ ἄρα ἐστὶν ἡ ὑπὸ ΕΒΔ. καὶ ἐπεὶ εὐθεῖά τις διὰ τοῦ κέντρου ἡ ΕΖ εὐθεῖάν τινα μὴ διὰ τοῦ κέντρου τὴν ΑΓ πρὸς ὀρθὰς τέμνει, καὶ δίχα αὐτὴν τέμνει: ἡ ΑΖ ἄρα τῇ ΖΓ ἐστιν ἴση. καὶ ἐπεὶ εὐθεῖα ἡ ΑΓ τέτμηται δίχα κατὰ τὸ Ζ σημεῖον, πρόσκειται δὲ αὐτῇ ἡ ΓΔ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΖΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΖΔ. κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΖΕ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τῶν ἀπὸ τῶν ΓΖ, ΖΕ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΔ, ΖΕ. τοῖς δὲ ἀπὸ τῶν ΓΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΓ: ὀρθὴ γὰρ ἐστιν ἡ ὑπὸ ΕΖΓ γωνία: τοῖς δὲ ἀπὸ τῶν ΔΖ, ΖΕ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΕΔ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΔ. ἴση δὲ ἡ ΕΓ τῇ ΕΒ: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΒ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΔ. τῷ δὲ ἀπὸ τῆς ΕΔ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΕΒ, ΒΔ: ὀρθὴ γὰρ ἡ ὑπὸ ΕΒΔ γωνία: τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ μετὰ τοῦ ἀπὸ τῆς ΕΒ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΕΒ, ΒΔ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΕΒ: λοιπὸν ἄρα τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ. ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, καὶ ἀπ᾽ αὐτοῦ πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ ἐφάπτηται, ἔσται τὸ ὑπὸ ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς ἐφαπτομένης τετραγώνῳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 500|>","<|""VertexLabel"" -> ""3.37"", ""Text"" -> ""If a point be taken outside a circle and from the point there fall on the circle two straight lines, if one of them cut the circle, and the other fall on it, and if further the rectangle contained by the whole of the straight line which cuts the circle and the straight line intercepted on it outside between the point and the convex circumference be equal to the square on the straight line which falls on the circle, the straight line which falls on it will touch the circle."", ""TextWordCount"" -> 90, ""GreekText"" -> ""ἐὰν κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ προσπίπτῃ, ᾖ δὲ τὸ ὑπὸ τῆς ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς προσπιπτούσης, ἡ προσπίπτουσα ἐφάψεται τοῦ κύκλου."", ""GreekTextWordCount"" -> 56, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 8}, {""Book"" -> 3, ""Theorem"" -> 16}, {""Book"" -> 3, ""Theorem"" -> 18}, {""Book"" -> 3, ""Theorem"" -> 36}}, ""Proof"" -> ""For let a point D be taken outside the circle ABC; from D let the two straight lines DCA, DB fall on the circle ACB; let DCA cut the circle and DB fall on it; and let the rectangle AD, DC be equal to the square on DB. I say that DB touches the circle ABC. For let DE be drawn touching ABC; let the centre of the circle ABC be taken, and let it be F; let FE, FB, FD be joined. Thus the angle FED is right. [III. 18] Now, since DE touches the circle ABC, and DCA cuts it, the rectangle AD, DC is equal to the square on DE. [III. 36] But the rectangle AD, DC was also equal to the square on DB; therefore the square on DE is equal to the square on DB; therefore DE is equal to DB. And FE is equal to FB; therefore the two sides DE, EF are equal to the two sides DB, BF; and FD is the common base of the triangles; therefore the angle DEF is equal to the angle DBF. [I. 8] But the angle DEF is right; therefore the angle DBF is also right. And FB produced is a diameter; and the straight line drawn at right angles to the diameter of a circle, from its extremity, touches the circle; [III. 16] therefore DB touches the circle. Similarly this can be proved to be the case even if the centre be on AC."", ""ProofWordCount"" -> 249, ""GreekProof"" -> ""κύκλου γὰρ τοῦ ΑΒΓ εἰλήφθω τι σημεῖον ἐκτὸς τὸ Δ, καὶ ἀπὸ τοῦ Δ πρὸς τὸν ΑΒΓ κύκλον προσπιπτέτωσαν δύο εὐθεῖαι αἱ ΔΓΑ, ΔΒ, καὶ ἡ μὲν ΔΓΑ τεμνέτω τὸν κύκλον, ἡ δὲ ΔΒ προσπιπτέτω, ἔστω δὲ τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον τῷ ἀπὸ τῆς ΔΒ. λέγω, ὅτι ἡ ΔΒ ἐφάπτεται τοῦ ΑΒΓ κύκλου. ἤχθω γὰρ τοῦ ΑΒΓ ἐφαπτομένη ἡ ΔΕ, καὶ εἰλήφθω τὸ κέντρον τοῦ ΑΒΓ κύκλου, καὶ ἔστω τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΖΕ, ΖΒ, ΖΔ. ἡ ἄρα ὑπὸ ΖΕΔ ὀρθή ἐστιν. καὶ ἐπεὶ ἡ ΔΕ ἐφάπτεται τοῦ ΑΒΓ κύκλου, τέμνει δὲ ἡ ΔΓΑ, τὸ ἄρα ὑπὸ τῶν ΑΔ, ΔΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΕ. ἦν δὲ καὶ τὸ ὑπὸ τῶν ΑΔ, ΔΓ ἴσον τῷ ἀπὸ τῆς ΔΒ: τὸ ἄρα ἀπὸ τῆς ΔΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΔΒ: ἴση ἄρα ἡ ΔΕ τῇ ΔΒ. ἐστὶ δὲ καὶ ἡ ΖΕ τῇ ΖΒ ἴση: δύο δὴ αἱ ΔΕ, ΕΖ δύο ταῖς ΔΒ, ΒΖ ἴσαι εἰσίν: καὶ βάσις αὐτῶν κοινὴ ἡ ΖΔ: γωνία ἄρα ἡ ὑπὸ ΔΕΖ γωνίᾳ τῇ ὑπὸ ΔΒΖ ἐστιν ἴση. ὀρθὴ δὲ ἡ ὑπὸ ΔΕΖ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΔΒΖ. καί ἐστιν ἡ ΖΒ ἐκβαλλομένη διάμετρος: ἡ δὲ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐφάπτεται τοῦ κύκλου: ἡ ΔΒ ἄρα ἐφάπτεται τοῦ ΑΒΓ κύκλου. ὁμοίως δὴ δειχθήσεται, κἂν τὸ κέντρον ἐπὶ τῆς ΑΓ τυγχάνῃ. ἐὰν ἄρα κύκλου ληφθῇ τι σημεῖον ἐκτός, ἀπὸ δὲ τοῦ σημείου πρὸς τὸν κύκλον προσπίπτωσι δύο εὐθεῖαι, καὶ ἡ μὲν αὐτῶν τέμνῃ τὸν κύκλον, ἡ δὲ προσπίπτῃ, ᾖ δὲ τὸ ὑπὸ ὅλης τῆς τεμνούσης καὶ τῆς ἐκτὸς ἀπολαμβανομένης μεταξὺ τοῦ τε σημείου καὶ τῆς κυρτῆς περιφερείας ἴσον τῷ ἀπὸ τῆς προσπιπτούσης, ἡ προσπίπτουσα ἐφάψεται τοῦ κύκλου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 283|>","<|""VertexLabel"" -> ""4.1"", ""Text"" -> ""Into a given circle to fit a straight line equal to a given straight line which is not greater than the diameter of the circle."", ""TextWordCount"" -> 25, ""GreekText"" -> ""εἰς τὸν δοθέντα κύκλον τῇ δοθείσῃ εὐθείᾳ μὴ μείζονι οὔσῃ τῆς τοῦ κύκλου διαμέτρου ἴσην εὐθεῖαν ἐναρμόσαι."", ""GreekTextWordCount"" -> 17, ""References"" -> {}, ""Proof"" -> ""Let ABC be the given circle, and D the given straight line not greater than the diameter of the circle; thus it is required to fit into the circle ABC a straight line equal to the straight line D. Let a diameter BC of the circle ABC be drawn. Then, if BC is equal to D, that which was enjoined will have been done; for BC has been fitted into the circle ABC equal to the straight line D. But, if BC is greater than D, let CE be made equal to D, and with centre C and distance CE let the circle EAF be described; let CA be joined. Then, since the point C is the centre of the circle EAF, CA is equal to CE. But CE is equal to D; therefore D is also equal to CA. Therefore into the given circle ABC there has been fitted CA equal to the given straight line D."", ""ProofWordCount"" -> 158, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, ἡ δὲ δοθεῖσα εὐθεῖα μὴ μείζων τῆς τοῦ κύκλου διαμέτρου ἡ Δ. δεῖ δὴ εἰς τὸν ΑΒΓ κύκλον τῇ Δ εὐθείᾳ ἴσην εὐθεῖαν ἐναρμόσαι. ἤχθω τοῦ ΑΒΓ κύκλου διάμετρος ἡ ΒΓ. εἰ μὲν οὖν ἴση ἐστὶν ἡ ΒΓ τῇ Δ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν: ἐνήρμοσται γὰρ εἰς τὸν ΑΒΓ κύκλον τῇ Δ εὐθείᾳ ἴση ἡ ΒΓ. εἰ δὲ μείζων ἐστὶν ἡ ΒΓ τῆς Δ, κείσθω τῇ Δ ἴση ἡ ΓΕ, καὶ κέντρῳ τῷ Γ διαστήματι δὲ τῷ ΓΕ κύκλος γεγράφθω ὁ ΕΑΖ, καὶ ἐπεζεύχθω ἡ ΓΑ. ἐπεὶ οὖν τὸ Γ σημεῖον κέντρον ἐστὶ τοῦ ΕΑΖ κύκλου, ἴση ἐστὶν ἡ ΓΑ τῇ ΓΕ. ἀλλὰ τῇ Δ ἡ ΓΕ ἐστιν ἴση: καὶ ἡ Δ ἄρα τῇ ΓΑ ἐστιν ἴση. εἰς ἄρα τὸν δοθέντα κύκλον τὸν ΑΒΓ τῇ δοθείσῃ εὐθείᾳ τῇ Δ ἴση ἐνήρμοσται ἡ ΓΑ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 143|>","<|""VertexLabel"" -> ""4.2"", ""Text"" -> ""In a given circle to inscribe a triangle equiangular with a given triangle."", ""TextWordCount"" -> 13, ""GreekText"" -> ""εἰς τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον ἐγγράψαι."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 3, ""Theorem"" -> 16}, {""Book"" -> 3, ""Theorem"" -> 32}}, ""Proof"" -> ""Let ABC be the given circle, and DEF the given triangle; thus it is required to inscribe in the circle ABC a triangle equiangular with the triangle DEF. Let GH be drawn touching the circle ABC at A [III. 16]; on the straight line AH, and at the point A on it, let the angle HAC be constructed equal to the angle DEF, and on the straight line AG, and at the point A on it, let the angle GAB be constructed equal to the angle DFE; [I. 23] let BC be joined. Then, since a straight line AH touches the circle ABC, and from the point of contact at A the straight line AC is drawn across in the circle, therefore the angle HAC is equal to the angle ABC in the alternate segment of the circle. [III. 32] But the angle HAC is equal to the angle DEF; therefore the angle ABC is also equal to the angle DEF. For the same reason the angle ACB is also equal to the angle DFE; therefore the remaining angle BAC is also equal to the remaining angle EDF. [I. 32] Therefore in the given circle there has been inscribed a triangle equiangular with the given triangle."", ""ProofWordCount"" -> 206, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, τὸ δὲ δοθὲν τρίγωνον τὸ ΔΕΖ: δεῖ δὴ εἰς τὸν ΑΒΓ κύκλον τῷ ΔΕΖ τριγώνῳ ἰσογώνιον τρίγωνον ἐγγράψαι. ἤχθω τοῦ ΑΒΓ κύκλου ἐφαπτομένη ἡ ΗΘ κατὰ τὸ Α, καὶ συνεστάτω πρὸς τῇ ΑΘ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΔΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΘΑΓ, πρὸς δὲ τῇ ΑΗ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΔΖΕ γωνίᾳ ἴση ἡ ὑπὸ ΗΑΒ, καὶ ἐπεζεύχθω ἡ ΒΓ. ἐπεὶ οὖν κύκλου τοῦ ΑΒΓ ἐφάπτεταί τις εὐθεῖα ἡ ΑΘ, καὶ ἀπὸ τῆς κατὰ τὸ Α ἐπαφῆς εἰς τὸν κύκλον διῆκται εὐθεῖα ἡ ΑΓ, ἡ ἄρα ὑπὸ ΘΑΓ ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι γωνίᾳ τῇ ὑπὸ ΑΒΓ. ἀλλ᾽ ἡ ὑπὸ ΘΑΓ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση: καὶ ἡ ὑπὸ ΑΒΓ ἄρα γωνία τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση: καὶ λοιπὴ ἄρα ἡ ὑπὸ ΒΑΓ λοιπῇ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ, καὶ ἐγγέγραπται εἰς τὸν ΑΒΓ κύκλον. εἰς τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 195|>","<|""VertexLabel"" -> ""4.3"", ""Text"" -> ""About a given circle to circumscribe a triangle equiangular with a given triangle."", ""TextWordCount"" -> 13, ""GreekText"" -> ""περὶ τὸν δοθέντα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον περιγράψαι."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 13}, {""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 3, ""Theorem"" -> 1}, {""Book"" -> 3, ""Theorem"" -> 16}, {""Book"" -> 3, ""Theorem"" -> 18}}, ""Proof"" -> ""Let ABC be the given circle, and DEF the given triangle;thus it is required to circumscribe about the circle ABC a triangle equiangular with the triangle DEF. Let EF be produced in both directions to the points G, H, let the centre K of the circle ABC be taken [III. 1], and letthe straight line KB be drawn across at random; on the straight line KB, and at the point K on it, let the angle BKA be constructed equal to the angle DEG, and the angle BKC equal to the angle DFH; [I. 23] and through the points A, B, C let LAM, MBN, NCL bedrawn touching the circle ABC. [III. 16] Now, since LM, MN, NL touch the circle ABC at the points A, B, C, and KA, KB, KC have been joined from the centre K to the points A, B, C,therefore the angles at the points A, B, C are right. [III. 18] And, since the four angles of the quadrilateral AMBK are equal to four right angles, inasmuch as AMBK is in fact divisible into two triangles, and the angles KAM, KBM are right, therefore the remaining angles AKB, AMB are equal to two right angles. But the angles DEG, DEF are also equal to two right angles; [I. 13] therefore the angles AKB, AMB are equal to the anglesDEG, DEF, of which the angle AKB is equal to the angle DEG; therefore the angle AMB which remains is equal to the angle DEF which remains. Similarly it can be proved that the angle LNB is alsoequal to the angle DFE; therefore the remaining angle MLN is equal to the angle EDF. [I. 32] Therefore the triangle LMN is equiangular with the triangle DEF; and it has been circumscribed about thecircle ABC. Therefore about a given circle there has been circumscribed a triangle equiangular with the given triangle."", ""ProofWordCount"" -> 315, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓ, τὸ δὲ δοθὲν τρίγωνον τὸ ΔΕΖ: δεῖ δὴ περὶ τὸν ΑΒΓ κύκλον τῷ ΔΕΖ τριγώνῳ ἰσογώνιον τρίγωνον περιγράψαι. Ἐκβεβλήσθω ἡ ΕΖ ἐφ᾽ ἑκάτερα τὰ μέρη κατὰ τὰ Η, Θ σημεῖα, καὶ εἰλήφθω τοῦ ΑΒΓ κύκλου κέντρον τὸ Κ, καὶ διήχθω, ὡς ἔτυχεν, εὐθεῖα ἡ ΚΒ, καὶ συνεστάτω πρὸς τῇ ΚΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Κ τῇ μὲν ὑπὸ ΔΕΗ γωνίᾳ ἴση ἡ ὑπὸ ΒΚΑ, τῇ δὲ ὑπὸ ΔΖΘ ἴση ἡ ὑπὸ ΒΚΓ, καὶ διὰ τῶν Α, Β, Γ σημείων ἤχθωσαν ἐφαπτόμεναι τοῦ ΑΒΓ κύκλου αἱ ΛΑΜ, ΜΒΝ, ΝΓΛ. καὶ ἐπεὶ ἐφάπτονται τοῦ ΑΒΓ κύκλου αἱ ΛΜ, ΜΝ, ΝΛ κατὰ τὰ Α, Β, Γ σημεῖα, ἀπὸ δὲ τοῦ Κ κέντρου ἐπὶ τὰ Α, Β, Γ σημεῖα ἐπεζευγμέναι εἰσὶν αἱ ΚΑ, ΚΒ, ΚΓ, ὀρθαὶ ἄρα εἰσὶν αἱ πρὸς τοῖς Α, Β, Γ σημείοις γωνίαι. καὶ ἐπεὶ τοῦ ΑΜΒΚ τετραπλεύρου αἱ τέσσαρες γωνίαι τέτρασιν ὀρθαῖς ἴσαι εἰσίν, ἐπειδήπερ καὶ εἰς δύο τρίγωνα διαιρεῖται τὸ ΑΜΒΚ, καί εἰσιν ὀρθαὶ αἱ ὑπὸ ΚΑΜ, ΚΒΜ γωνίαι, λοιπαὶ ἄρα αἱ ὑπὸ ΑΚΒ, ΑΜΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν. εἰσὶ δὲ καὶ αἱ ὑπὸ ΔΕΗ, ΔΕΖ δυσὶν ὀρθαῖς ἴσαι: αἱ ἄρα ὑπὸ ΑΚΒ, ΑΜΒ ταῖς ὑπὸ ΔΕΗ, ΔΕΖ ἴσαι εἰσίν, ὧν ἡ ὑπὸ ΑΚΒ τῇ ὑπὸ ΔΕΗ ἐστιν ἴση: λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἡ ὑπὸ ΛΝΒ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση: καὶ λοιπὴ ἄρα ἡ ὑπὸ ΜΛΝ λοιπῇ τῇ ὑπὸ ΕΔΖ ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΛΜΝ τρίγωνον τῷ ΔΕΖ τριγώνῳ: καὶ περιγέγραπται περὶ τὸν ΑΒΓ κύκλον. περὶ τὸν δοθέντα ἄρα κύκλον τῷ δοθέντι τριγώνῳ ἰσογώνιον τρίγωνον περιγέγραπται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 275|>","<|""VertexLabel"" -> ""4.4"", ""Text"" -> ""In a given triangle to inscribe a circle."", ""TextWordCount"" -> 8, ""GreekText"" -> ""εἰς τὸ δοθὲν τρίγωνον κύκλον ἐγγράψαι."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 9}, {""Book"" -> 1, ""Theorem"" -> 26}, {""Book"" -> 3, ""Theorem"" -> 16}, {""Book"" -> 4, ""Definition"" -> 5}}, ""Proof"" -> ""Let ABC be the given triangle; thus it is required to inscribe a circle in the triangle ABC. Let the angles ABC, ACB be bisected by the straight lines BD, CD [I. 9], and let these meet one another at the point D; from D let DE, DF, DG be drawn perpendicular to the straightlines AB, BC, CA. Now, since the angle ABD is equal to the angle CBD, and the right angle BED is also equal to the right angle BFD,EBD, FBD are two triangles having two angles equal to two angles and one side equal to one side, namely that subtending one of the equal angles, which is BD common to the triangles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore DE is equal to DF. For the same reason DG is also equal to DF. Therefore the three straight lines DE, DF, DG are equalto one another; therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will pass also through the remaining points, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right. For, if it cuts them, the straight line drawn at right angles to the diameter of the circle from its extremity will be found to fall within the circle: which was proved absurd; [III. 16] therefore the circle described with centre D and distance one of the straight lines DE, DF, DG will not cut the straight lines AB, BC, CA; therefore it will touch them, and will be the circle inscribed in the triangle ABC. [IV. Def. 5] Let it be inscribed, as FGE. Therefore in the given triangle ABC the circle EFG has been inscribed."", ""ProofWordCount"" -> 300, ""GreekProof"" -> ""ἔστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ: δεῖ δὴ εἰς τὸ ΑΒΓ τρίγωνον κύκλον ἐγγράψαι. τετμήσθωσαν αἱ ὑπὸ ΑΒΓ, ΑΓΒ γωνίαι δίχα ταῖς ΒΔ, ΓΔ εὐθείαις, καὶ συμβαλλέτωσαν ἀλλήλαις κατὰ τὸ Δ σημεῖον, καὶ ἤχθωσαν ἀπὸ τοῦ Δ ἐπὶ τὰς ΑΒ, ΒΓ, ΓΑ εὐθείας κάθετοι αἱ ΔΕ, ΔΖ, ΔΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΑΒΔ γωνία τῇ ὑπὸ ΓΒΔ, ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΒΕΔ ὀρθῇ τῇ ὑπὸ ΒΖΔ ἴση, δύο δὴ τρίγωνά ἐστι τὰ ΕΒΔ, ΖΒΔ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην τὴν ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν κοινὴν αὐτῶν τὴν ΒΔ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξουσιν: ἴση ἄρα ἡ ΔΕ τῇ ΔΖ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΔΗ τῇ ΔΖ ἐστιν ἴση. αἱ τρεῖς ἄρα εὐθεῖαι αἱ ΔΕ, ΔΖ, ΔΗ ἴσαι ἀλλήλαις εἰσίν: ὁ ἄρα κέντρῳ τῷ Δ καὶ διαστήματι ἑνὶ τῶν Ε, Ζ, Η κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Ε, Ζ, Η σημείοις γωνίας. εἰ γὰρ τεμεῖ αὐτάς, ἔσται ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐντὸς πίπτουσα τοῦ κύκλου: ὅπερ ἄτοπον ἐδείχθη: οὐκ ἄρα ὁ κέντρῳ τῷ Δ διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Η γραφόμενος κύκλος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΑ εὐθείας: ἐφάψεται ἄρα αὐτῶν, καὶ ἔσται ὁ κύκλος ἐγγεγραμμένος εἰς τὸ ΑΒΓ τρίγωνον. ἐγγεγράφθω ὡς ὁ ΖΗΕ. εἰς ἄρα τὸ δοθὲν τρίγωνον τὸ ΑΒΓ κύκλος ἐγγέγραπται ὁ ΕΖΗ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 254|>","<|""VertexLabel"" -> ""4.5"", ""Text"" -> ""About a given triangle to circumscribe a circle."", ""TextWordCount"" -> 8, ""GreekText"" -> ""περὶ τὸ δοθὲν τρίγωνον κύκλον περιγράψαι."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 10}}, ""Proof"" -> ""Let ABC be the given triangle; thus it is required to circumscribe a circle about the given triangle ABC. Let the straight lines AB, AC be bisected at the points D, E [I. 10], and from the points D, E let DF, EF be drawn at right angles to AB, AC; they will then meet within the triangle ABC, or on the straight line BC, or outside BC. First let them meet within at F, and let FB, FC, FA be joined. Then, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base FB. [I. 4] Similarly we can prove that CF is also equal to AF; so that FB is also equal to FC; therefore the three straight lines FA, FB, FC are equal to one another. Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and the circle will have been circumscribed about the triangle ABC. Let it be circumscribed, as ABC. Next, let DF, EF meet on the straight line BC at F, as is the case in the second figure; and let AF be joined. Then, similarly, we shall prove that the point F is the centre of the circle circumscribed about the triangle ABC. Again, let DF, EF meet outside the triangle ABC at F, as is the case in the third figure, and let AF, BF, CF be joined. Then again, since AD is equal to DB, and DF is common and at right angles, therefore the base AF is equal to the base BF. [I. 4] Similarly we can prove that CF is also equal to AF; so that BF is also equal to FC; therefore the circle described with centre F and distance one of the straight lines FA, FB, FC will pass also through the remaining points, and will have been circumscribed about the triangle ABC. Therefore about the given triangle a circle has been circumscribed."", ""ProofWordCount"" -> 343, ""GreekProof"" -> ""ἔστω τὸ δοθὲν τρίγωνον τὸ ΑΒΓ: δεῖ δὴ περὶ τὸ δοθὲν τρίγωνον τὸ ΑΒΓ κύκλον περιγράψαι. τετμήσθωσαν αἱ ΑΒ, ΑΓ εὐθεῖαι δίχα κατὰ τὰ Δ, Ε σημεῖα, καὶ ἀπὸ τῶν Δ, Ε σημείων ταῖς ΑΒ, ΑΓ πρὸς ὁρθὰς ἤχθωσαν αἱ ΔΖ, ΕΖ: συμπεσοῦνται δὴ ἤτοι ἐντὸς τοῦ ΑΒΓ τριγώνου ἢ ἐπὶ τῆς ΒΓ εὐθείας ἢ ἐκτὸς τῆς ΒΓ. συμπιπτέτωσαν πρότερον ἐντὸς κατὰ τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΓ, ΖΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις ἄρα ἡ ΑΖ βάσει τῇ ΖΒ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση: ὥστε καὶ ἡ ΖΒ τῇ ΖΓ ἐστιν ἴση: αἱ τρεῖς ἄρα αἱ ΖΑ, ΖΒ, ΖΓ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν Α, Β, Γ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων, καὶ ἔσται περιγεγραμμένος ὁ κύκλος περὶ τὸ ΑΒΓ τρίγωνον. περιγεγράφθω ὡς ὁ ΑΒΓ. ἀλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐπὶ τῆς ΒΓ εὐθείας κατὰ τὸ Ζ, ὡς ἔχει ἐπὶ τῆς δευτέρας καταγραφῆς, καὶ ἐπεζεύχθω ἡ ΑΖ. ὁμοίως δὴ δείξομεν, ὅτι τὸ Ζ σημεῖον κέντρον ἐστὶ τοῦ περὶ τὸ ΑΒΓ τρίγωνον περιγραφομένου κύκλου. ἀλλὰ δὴ αἱ ΔΖ, ΕΖ συμπιπτέτωσαν ἐκτὸς τοῦ ΑΒΓ τριγώνου κατὰ τὸ Ζ πάλιν, ὡς ἔχει ἐπὶ τῆς τρίτης καταγραφῆς, καὶ ἐπεζεύχθωσαν αἱ ΑΖ, ΒΖ, ΓΖ. καὶ ἐπεὶ πάλιν ἴση ἐστὶν ἡ ΑΔ τῇ ΔΒ, κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΔΖ, βάσις ἄρα ἡ ΑΖ βάσει τῇ ΒΖ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ΓΖ τῇ ΑΖ ἐστιν ἴση: ὥστε καὶ ἡ ΒΖ τῇ ΖΓ ἐστιν ἴση: ὁ ἄρα πάλιν κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν ΖΑ, ΖΒ, ΖΓ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων, καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓ τρίγωνον. περὶ τὸ δοθὲν ἄρα τρίγωνον κύκλος περιγέγραπται: ὅπερ ἔδει ποιῆσαι. Πόρισμα καὶ φανερόν, ὅτι, ὅτε μὲν ἐντὸς τοῦ τριγώνου πίπτει τὸ κέντρον τοῦ κύκλου, ἡ ὑπὸ ΒΑΓ γωνία ἐν μείζονι τμήματι τοῦ ἡμικυκλίου τυγχάνουσα ἐλάττων ἐστὶν ὀρθῆς: ὅτε δὲ ἐπὶ τῆς ΒΓ εὐθείας τὸ κέντρον πίπτει, ἡ ὑπὸ ΒΑΓ γωνία ἐν ἡμικυκλίῳ τυγχάνουσα ὀρθή ἐστιν: ὅτε δὲ τὸ κέντρον τοῦ κύκλου ἐκτὸς τοῦ τριγώνου πίπτει, ἡ ὑπὸ ΒΑΓ ἐν ἐλάττονι τμήματι τοῦ ἡμικυκλίου τυγχάνουσα μείζων ἐστὶν ὀρθῆς. ὥστε καὶ ὅταν ἐλάττων ὀρθῆς τυγχάνῃ ἡ διδομένη γωνία, ἐντὸς τοῦ τριγώνου πεσοῦνται αἱ ΔΖ, ΕΖ, ὅταν δὲ ὀρθή, ἐπὶ τῆς ΒΓ, ὅταν δὲ μείζων ὀρθῆς, ἐκτὸς τῆς ΒΓ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 402|>","<|""VertexLabel"" -> ""4.6"", ""Text"" -> ""In a given circle to inscribe a square."", ""TextWordCount"" -> 8, ""GreekText"" -> ""εἰς τὸν δοθέντα κύκλον τετράγωνον ἐγγράψαι."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Definition"" -> 22}, {""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 3, ""Theorem"" -> 31}}, ""Proof"" -> ""Let ABCD be the given circle; thus it is required to inscribe a square in the circle ABCD. Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and let AB, BC, CD, DA be joined. Then, since BE is equal to ED, for E is the centre, and EA is common and at right angles, therefore the base AB is equal to the base AD. [I. 4] For the same reason each of the straight lines BC, CD is also equal to each of the straight lines AB, AD; therefore the quadrilateral ABCD is equilateral. I say next that it is also right-angled. For, since the straight line BD is a diameter of the circle ABCD, therefore BAD is a semicircle; therefore the angle BAD is right. [III. 31] For the same reason each of the angles ABC, BCD, CDA is also right; therefore the quadrilateral ABCD is right-angled. But it was also proved equilateral; therefore it is a square; [I. Def. 22] and it has been inscribed in the circle ABCD. Therefore in the given circle the square ABCD has been inscribed."", ""ProofWordCount"" -> 191, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ: δεῖ δὴ εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον ἐγγράψαι. ἤχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΑ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΔ: κέντρον γὰρ τὸ Ε: κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΕΑ, βάσις ἄρα ἡ ΑΒ βάσει τῇ ΑΔ ἴση ἐστίν. διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ΒΓ, ΓΔ ἑκατέρᾳ τῶν ΑΒ, ΑΔ ἴση ἐστίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔ τετράπλευρον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ ἡ ΒΔ εὐθεῖα διάμετρός ἐστι τοῦ ΑΒΓΔ κύκλου, ἡμικύκλιον ἄρα ἐστὶ τὸ ΒΑΔ: ὀρθὴ ἄρα ἡ ὑπὸ ΒΑΔ γωνία. διὰ τὰ αὐτὰ δὴ καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΑ ὀρθή ἐστιν: ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔ τετράπλευρον. ἐδείχθη δὲ καὶ ἰσόπλευρον: τετράγωνον ἄρα ἐστίν. καὶ ἐγγέγραπται εἰς τὸν ΑΒΓΔ κύκλον. εἰς ἄρα τὸν δοθέντα κύκλον τετράγωνον ἐγγέγραπται τὸ ΑΒΓΔ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 152|>","<|""VertexLabel"" -> ""4.7"", ""Text"" -> ""About a given circle to circumscribe a square."", ""TextWordCount"" -> 8, ""GreekText"" -> ""περὶ τὸν δοθέντα κύκλον τετράγωνον περιγράψαι."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 28}, {""Book"" -> 1, ""Theorem"" -> 30}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 3, ""Theorem"" -> 16}, {""Book"" -> 3, ""Theorem"" -> 18}}, ""Proof"" -> ""Let ABCD be the given circle; thus it is required to circumscribe a square about the circle ABCD. Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and through the points A, B, C, D let FG, GH, HK, KF be drawn touching the circle ABCD. [III. 16] Then, since FG touches the circle ABCD, and EA has been joined from the centre E to the point of contact at A, therefore the angles at A are right. [III. 18] For the same reason the angles at the points B, C, D are also right. Now, since the angle AEB is right, and the angle EBG is also right, therefore GH is parailel to AC. [I. 28] For the same reason AC is also parallel to FK, so that GH is also parallel to FK. [I. 30] Similarly we can prove that each of the straight lines GF, HK is parallel to BED. Therefore GK, GC, AK, FB, BK are parallelograms; therefore GF is equal to HK, and GH to FK. [I. 34] And, since AC is equal to BD, and AC is also equal to each of the straight lines GH, FK, while BD is equal to each of the straight lines GF, HK, [I. 34]therefore the quadrilateral FGHK is equilateral. I say next that it is also right-angled. For, since GBEA is a parallelogram, and the angle AEB is right, therefore the angle AGB is also right. [I. 34] Similarly we can prove that the angles at H, K, F are also right. Therefore FGHK is right-angled. But it was also proved equilateral; therefore it is a square; and it has been circumscribed about the circle ABCD. Therefore about the given circle a square has been circumscribed."", ""ProofWordCount"" -> 298, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ: δεῖ δὴ περὶ τὸν ΑΒΓΔ κύκλον τετράγωνον περιγράψαι. ἤχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ διὰ τῶν Α, Β, Γ, Δ σημείων ἤχθωσαν ἐφαπτόμεναι τοῦ ΑΒΓΔ κύκλου αἱ ΖΗ, ΗΘ, ΘΚ, ΚΖ. ἐπεὶ οὖν ἐφάπτεται ἡ ΖΗ τοῦ ΑΒΓΔ κύκλου, ἀπὸ δὲ τοῦ Ε κέντρου ἐπὶ τὴν κατὰ τὸ Α ἐπαφὴν ἐπέζευκται ἡ ΕΑ, αἱ ἄρα πρὸς τῷ Α γωνίαι ὀρθαί εἰσιν. διὰ τὰ αὐτὰ δὴ καὶ αἱ πρὸς τοῖς Β, Γ, Δ σημείοις γωνίαι ὀρθαί εἰσιν. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΑΕΒ γωνία, ἐστὶ δὲ ὀρθὴ καὶ ἡ ὑπὸ ΕΒΗ, παράλληλος ἄρα ἐστὶν ἡ ΗΘ τῇ ΑΓ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΑΓ τῇ ΖΚ ἐστι παράλληλος. ὥστε καὶ ἡ ΗΘ τῇ ΖΚ ἐστι παράλληλος. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΗΖ, ΘΚ τῇ ΒΕΔ ἐστι παράλληλος. παραλληλόγραμμα ἄρα ἐστὶ τὰ ΗΚ, ΗΓ, ΑΚ, ΖΒ, ΒΚ: ἴση ἄρα ἐστὶν ἡ μὲν ΗΖ τῇ ΘΚ, ἡ δὲ ΗΘ τῇ ΖΚ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΒΔ, ἀλλὰ καὶ ἡ μὲν ΑΓ ἑκατέρᾳ τῶν ΗΘ, ΖΚ, ἡ δὲ ΒΔ ἑκατέρᾳ τῶν ΗΖ, ΘΚ ἐστιν ἴση καὶ ἑκατέρα ἄρα τῶν ΗΘ, ΖΚ ἑκατέρᾳ τῶν ΗΖ, ΘΚ ἐστιν ἴση, ἰσόπλευρον ἄρα ἐστὶ τὸ ΖΗΘΚ τετράπλευρον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΗΒΕΑ, καί ἐστιν ὀρθὴ ἡ ὑπὸ ΑΕΒ, ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΗΒ. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ πρὸς τοῖς Θ, Κ, Ζ γωνίαι ὀρθαί εἰσιν. ὀρθογώνιον ἄρα ἐστὶ τὸ ΖΗΘΚ. ἐδείχθη δὲ καὶ ἰσόπλευρον: τετράγωνον ἄρα ἐστίν. καὶ περιγέγραπται περὶ τὸν ΑΒΓΔ κύκλον. περὶ τὸν δοθέντα ἄρα κύκλον τετράγωνον περιγέγραπται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 274|>","<|""VertexLabel"" -> ""4.8"", ""Text"" -> ""In a given square to inscribe a circle."", ""TextWordCount"" -> 8, ""GreekText"" -> ""εἰς τὸ δοθὲν τετράγωνον κύκλον ἐγγράψαι."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 10}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 3, ""Theorem"" -> 16}}, ""Proof"" -> ""Let ABCD be the given square; thus it is required to inscribe a circle in the given square ABCD. Let the straight lines AD, AB be bisected at the points E, F respectively [I. 10], through E let EH be drawn parallel to either AB or CD, and through F let FK be drawn parallel to either AD or BC; [I. 31] therefore each of the figures AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their opposite sides are evidently equal. [I. 34] Now, since AD is equal to AB, and AE is half of AD, and AF half of AB, therefore AE is equal to AF, so that the opposite sides are also equal; therefore FG is equal to GE. Similarly we can prove that each of the straight lines GH, GK is equal to each of the straight lines FG, GE; therefore the four straight lines GE, GF, GH, GK are equal to one another. Therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will pass also through the remaining points. And it will touch the straight lines AB, BC, CD, DA, because the angles at E, F, H, K are right. For, if the circle cuts AB, BC, CD, DA, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle: which was proved absurd; [III. 16] therefore the circle described with centre G and distance one of the straight lines GE, GF, GH, GK will not cut the straight lines AB, BC, CD, DA. Therefore it will touch them, and will have been inscribed in the square ABCD. Therefore in the given square a circle has been inscribed."", ""ProofWordCount"" -> 295, ""GreekProof"" -> ""ἔστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ: δεῖ δὴ εἰς τὸ ΑΒΓΔ τετράγωνον κύκλον ἐγγράψαι. τετμήσθω ἑκατέρα τῶν ΑΔ, ΑΒ δίχα κατὰ τὰ Ε, Ζ σημεῖα, καὶ διὰ μὲν τοῦ Ε ὁποτέρᾳ τῶν ΑΒ, ΓΔ παράλληλος ἤχθω ἡ ΕΘ, διὰ δὲ τοῦ Ζ ὁποτέρᾳ τῶν ΑΔ, ΒΓ παράλληλος ἤχθω ἡ ΖΚ: παραλληλόγραμμον ἄρα ἐστὶν ἕκαστον τῶν ΑΚ, ΚΒ, ΑΘ, ΘΔ, ΑΗ, ΗΓ, ΒΗ, ΗΔ, καὶ αἱ ἀπεναντίον αὐτῶν πλευραὶ δηλονότι ἴσαι εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΒ, καί ἐστι τῆς μὲν ΑΔ ἡμίσεια ἡ ΑΕ, τῆς δὲ ΑΒ ἡμίσεια ἡ ΑΖ, ἴση ἄρα καὶ ἡ ΑΕ τῇ ΑΖ: ὥστε καὶ αἱ ἀπεναντίον: ἴση ἄρα καὶ ἡ ΖΗ τῇ ΗΕ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΗΘ, ΗΚ ἑκατέρᾳ τῶν ΖΗ, ΗΕ ἐστιν ἴση: αἱ τέσσαρες ἄρα αἱ ΗΕ, ΗΖ, ΗΘ, ΗΚ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Θ, Κ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων: καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Ε, Ζ, Θ, Κ γωνίας: εἰ γὰρ τεμεῖ ὁ κύκλος τὰς ΑΒ, ΒΓ, ΓΔ, ΔΑ, ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένη ἐντὸς πεσεῖται τοῦ κύκλου: ὅπερ ἄτοπον ἐδείχθη. οὐκ ἄρα ὁ κέντρῳ τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Θ, Κ κύκλος γραφόμενος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΑ εὐθείας. ἐφάψεται ἄρα αὐτῶν καὶ ἔσται ἐγγεγραμμένος εἰς τὸ ΑΒΓΔ τετράγωνον. εἰς ἄρα τὸ δοθὲν τετράγωνον κύκλος ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 250|>","<|""VertexLabel"" -> ""4.9"", ""Text"" -> ""About a given square to circumscribe a circle."", ""TextWordCount"" -> 8, ""GreekText"" -> ""περὶ τὸ δοθὲν τετράγωνον κύκλον περιγράψαι."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 6}, {""Book"" -> 1, ""Theorem"" -> 8}}, ""Proof"" -> ""Let ABCD be the given square; thus it is required to circumscribe a circle about the square ABCD. For let AC, BD be joined, and let them cut one another at E. Then, since DA is equal to AB, and AC is common, therefore the two sides DA, AC are equal to the two sides BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8] Therefore the angle DAB is bisected by AC. Similarly we can prove that each of the angles ABC, BCD, CDA is bisected by the straight lines AC, DB. Now, since the angle DAB is equal to the angle ABC, and the angle EAB is half the angle DAB, and the angle EBA half the angle ABC, therefore the angle EAB is also equal to the angle EBA; so that the side EA is also equal to EB. [I. 6] Similarly we can prove that each of the straight lines EA, EB is equal to each of the straight lines EC, ED. Therefore the four straight lines EA, EB, EC, ED are equal to one another. Therefore the circle described with centre E and distance one of the straight lines EA, EB, EC, ED will pass also through the remaining points; and it will have been circumscribed about the square ABCD. Let it be circumscribed, as ABCD. Therefore about the given square a circle has been circumscribed."", ""ProofWordCount"" -> 245, ""GreekProof"" -> ""ἔστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ: δεῖ δὴ περὶ τὸ ΑΒΓΔ τετράγωνον κύκλον περιγράψαι. ἐπιζευχθεῖσαι γὰρ αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, κοινὴ δὲ ἡ ΑΓ, δύο δὴ αἱ ΔΑ, ΑΓ δυσὶ ταῖς ΒΑ, ΑΓ ἴσαι εἰσίν: καὶ βάσις ἡ ΔΓ βάσει τῇ ΒΓ ἴση: γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ ἴση ἐστίν: ἡ ἄρα ὑπὸ ΔΑΒ γωνία δίχα τέτμηται ὑπὸ τῆς ΑΓ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΑ δίχα τέτμηται ὑπὸ τῶν ΑΓ, ΔΒ εὐθειῶν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΑΒ γωνία τῇ ὑπὸ ΑΒΓ, καί ἐστι τῆς μὲν ὑπὸ ΔΑΒ ἡμίσεια ἡ ὑπὸ ΕΑΒ, τῆς δὲ ὑπὸ ΑΒΓ ἡμίσεια ἡ ὑπὸ ΕΒΑ, καὶ ἡ ὑπὸ ΕΑΒ ἄρα τῇ ὑπὸ ΕΒΑ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΕΑ τῇ ΕΒ ἐστιν ἴση. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΕΑ, ΕΒ εὐθειῶν ἑκατέρᾳ τῶν ΕΓ, ΕΔ ἴση ἐστίν. αἱ τέσσαρες ἄρα αἱ ΕΑ, ΕΒ, ΕΓ, ΕΔ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ε καὶ διαστήματι ἑνὶ τῶν Α, Β, Γ, Δ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓΔ τετράγωνον. περιγεγράφθω ὡς ὁ ΑΒΓΔ. περὶ τὸ δοθὲν ἄρα τετράγωνον κύκλος περιγέγραπται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 211|>","<|""VertexLabel"" -> ""4.10"", ""Text"" -> ""To construct an isosceles triangle having each of the angles at the base double of the remaining one."", ""TextWordCount"" -> 18, ""GreekText"" -> ""ἰσοσκελὲς τρίγωνον συστήσασθαι ἔχον ἑκατέραν τῶν πρὸς τῇ βάσει γωνιῶν διπλασίονα τῆς λοιπῆς."", ""GreekTextWordCount"" -> 13, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 6}, {""Book"" -> 1, ""Theorem"" -> 32}, {(""Book"" -> 2)*(""Book"" -> 3), (""Theorem"" -> 11)*(""Theorem"" -> 32)}, {""Book"" -> 3, ""Theorem"" -> 37}, {""Book"" -> 4, ""Theorem"" -> 5}}, ""Proof"" -> ""Let any straight line AB be set out, and let it be cut at the point C so that the rectangle contained by AB, BC is equal to the square on CA; [II. 11] with centre A and distance AB let the circle BDE be described, and let there be fitted in the circle BDE the straight line BD equal to the straight line AC which is not greater than the diameter of the circle BDE. [IV. 1] Let AD, DC be joined, and let the circle ACD be circumscribed about the triangle ACD. [IV. 5] Then, since the rectangle AB, BC is equal to the square on AC, and AC is equal to BD, therefore the rectangle AB, BC is equal to the square on BD. And, since a point B has been taken outside the circle ACD, and from B the two straight lines BA, BD have fallen on the circle ACD, and one of them cuts it, while the other falls on it, and the rectangle AB, BC is equal to the square on BD, therefore BD touches the circle ACD. [III. 37] Since, then, BD touches it, and DC is drawn across from the point of contact at D, therefore the angle BDC is equal to the angle DAC in the alternate segment of the circle. [III. 32] Since, then, the angle BDC is equal to the angle DAC, let the angle CDA be added to each; therefore the whole angle BDA is equal to the two angles CDA, DAC. But the exterior angle BCD is equal to the angles CDA, DAC; [I. 32] therefore the angle BDA is also equal to the angle BCD. But the angle BDA is equal to the angle CBD, since the side AD is also equal to AB; [I. 5] so that the angle DBA is also equal to the angle BCD. Therefore the three angles BDA, DBA, BCD are equal to one another. And, since the angle DBC is equal to the angle BCD, the side BD is also equal to the side DC. [I. 6] But BD is by hypothesis equal to CA; therefore CA is also equal to CD, so that the angle CDA is also equal to the angle DAC; [I. 5] therefore the angles CDA, DAC are double of the angle DAC. But the angle BCD is equal to the angles CDA, DAC; therefore the angle BCD is also double of the angle CAD. But the angle BCD is equal to each of the angles BDA, DBA; therefore each of the angles BDA, DBA is also double of the angle DAB. Therefore the isosceles triangle ABD has been constructed having each of the angles at the base DB double of the remaining one."", ""ProofWordCount"" -> 458, ""GreekProof"" -> ""Ἐκκείσθω τις εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω κατὰ τὸ Γ σημεῖον, ὥστε τὸ ὑπὸ τῶν ΑΒ, ΒΓ περιεχόμενον ὀρθογώνιον ἴσον εἶναι τῷ ἀπὸ τῆς ΓΑ τετραγώνῳ: καὶ κέντρῳ τῷ Α καὶ διαστήματι τῷ ΑΒ κύκλος γεγράφθω ὁ ΒΔΕ, καὶ ἐνηρμόσθω εἰς τὸν ΒΔΕ κύκλον τῇ ΑΓ εὐθείᾳ μὴ μείζονι οὔσῃ τῆς τοῦ ΒΔΕ κύκλου διαμέτρου ἴση εὐθεῖα ἡ ΒΔ: καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ, καὶ περιγεγράφθω περὶ τὸ ΑΓΔ τρίγωνον κύκλος ὁ ΑΓΔ. καὶ ἐπεὶ τὸ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ, ἴση δὲ ἡ ΑΓ τῇ ΒΔ, τὸ ἄρα ὑπὸ τῶν ΑΒ, ΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΔ. καὶ ἐπεὶ κύκλου τοῦ ΑΓΔ εἴληπταί τι σημεῖον ἐκτὸς τὸ Β, καὶ ἀπὸ τοῦ Β πρὸς τὸν ΑΓΔ κύκλον προσπεπτώκασι δύο εὐθεῖαι αἱ ΒΑ, ΒΔ, καὶ ἡ μὲν αὐτῶν τέμνει, ἡ δὲ προσπίπτει, καί ἐστι τὸ ὑπὸ τῶν ΑΒ, ΒΓ ἴσον τῷ ἀπὸ τῆς ΒΔ, ἡ ΒΔ ἄρα ἐφάπτεται τοῦ ΑΓΔ κύκλου. ἐπεὶ οὖν ἐφάπτεται μὲν ἡ ΒΔ, ἀπὸ δὲ τῆς κατὰ τὸ Δ ἐπαφῆς διῆκται ἡ ΔΓ, ἡ ἄρα ὑπὸ ΒΔΓ γωνία ἴση ἐστὶ τῇ ἐν τῷ ἐναλλὰξ τοῦ κύκλου τμήματι γωνίᾳ τῇ ὑπὸ ΔΑΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ὑπὸ ΒΔΓ τῇ ὑπὸ ΔΑΓ, κοινὴ προσκείσθω ἡ ὑπὸ ΓΔΑ: ὅλη ἄρα ἡ ὑπὸ ΒΔΑ ἴση ἐστὶ δυσὶ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ. ἀλλὰ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ ἴση ἐστὶν ἡ ἐκτὸς ἡ ὑπὸ ΒΓΔ: καὶ ἡ ὑπὸ ΒΔΑ ἄρα ἴση ἐστὶ τῇ ὑπὸ ΒΓΔ. ἀλλὰ ἡ ὑπὸ ΒΔΑ τῇ ὑπὸ ΓΒΔ ἐστιν ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΑΔ τῇ ΑΒ ἐστιν ἴση: ὥστε καὶ ἡ ὑπὸ ΔΒΑ τῇ ὑπὸ ΒΓΔ ἐστιν ἴση. αἱ τρεῖς ἄρα αἱ ὑπὸ ΒΔΑ, ΔΒΑ, ΒΓΔ ἴσαι ἀλλήλαις εἰσίν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΒΓ γωνία τῇ ὑπὸ ΒΓΔ, ἴση ἐστὶ καὶ πλευρὰ ἡ ΒΔ πλευρᾷ τῇ ΔΓ. ἀλλὰ ἡ ΒΔ τῇ ΓΑ ὑπόκειται ἴση: καὶ ἡ ΓΑ ἄρα τῇ ΓΔ ἐστιν ἴση: ὥστε καὶ γωνία ἡ ὑπὸ ΓΔΑ γωνίᾳ τῇ ὑπὸ ΔΑΓ ἐστιν ἴση: αἱ ἄρα ὑπὸ ΓΔΑ, ΔΑΓ τῆς ὑπὸ ΔΑΓ εἰσι διπλασίους. ἴση δὲ ἡ ὑπὸ ΒΓΔ ταῖς ὑπὸ ΓΔΑ, ΔΑΓ: καὶ ἡ ὑπὸ ΒΓΔ ἄρα τῆς ὑπὸ ΓΑΔ ἐστι διπλῆ. ἴση δὲ ἡ ὑπὸ ΒΓΔ ἑκατέρᾳ τῶν ὑπὸ ΒΔΑ, ΔΒΑ: καὶ ἑκατέρα ἄρα τῶν ὑπὸ ΒΔΑ, ΔΒΑ τῆς ὑπὸ ΔΑΒ ἐστι διπλῆ. ἰσοσκελὲς ἄρα τρίγωνον συνέσταται τὸ ΑΒΔ ἔχον ἑκατέραν τῶν πρὸς τῇ ΔΒ βάσει γωνιῶν διπλασίονα τῆς λοιπῆς: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 396|>","<|""VertexLabel"" -> ""4.11"", ""Text"" -> ""In a given circle to inscribe an equilateral and equiangular pentagon."", ""TextWordCount"" -> 11, ""GreekText"" -> ""εἰς τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 9}, {""Book"" -> 3, ""Theorem"" -> 26}, {""Book"" -> 3, ""Theorem"" -> 27}, {""Book"" -> 3, ""Theorem"" -> 29}, {""Book"" -> 4, ""Theorem"" -> 2}, {""Book"" -> 4, ""Theorem"" -> 10}}, ""Proof"" -> ""Let ABCDE be the given circle; thus it is required to inscribe in the circle ABCDE an equilateral and equiangular pentagon. Let the isosceles triangle FGH be set out having each of the angles at G, H double of the angle at F; [IV. 10] let there be inscribed in the circle ABCDE the triangle ACD equiangular with the triangle FGH, so that the angle CAD is equal to the angle at F and the angles at G, H respectively equal to the angles ACD, CDA; [IV. 2] therefore each of the angles ACD, CDA is also double of the angle CAD. Now let the angles ACD, CDA be bisected respectively by the straight lines CE, DB [I. 9], and let AB, BC, DE, EA be joined. Then, since each of the angles ACD, CDA is double of the angle CAD, and they have been bisected by the straight lines CE, DB, therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another. But equal angles stand on equal circumferences; [III. 26] therefore the five circumferences AB, BC, CD, DE, EA are equal to one another. But equal circumferences are subtended by equal straight lines; [III. 29] therefore the five straight lines AB, BC, CD, DE, EA are equal to one another; therefore the pentagon ABCDE is equilateral. I say next that it is also equiangular. For, since the circumference AB is equal to the circumference DE, let BCD be added to each; therefore the whole circumference ABCD is equal to the whole circumference EDCB. And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is also equal to the angle AED. [III. 27] For the same reason each of the angles ABC, BCD, CDE is also equal to each of the angles BAE, AED; therefore the pentagon ABCDE is equiangular. But it was also proved equilateral; therefore in the given circle an equilateral and equiangular pentagon has been inscribed."", ""ProofWordCount"" -> 334, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ: δεῖ δὴ εἰς τὸν ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι. Ἐκκείσθω τρίγωνον ἰσοσκελὲς τὸ ΖΗΘ διπλασίονα ἔχον ἑκατέραν τῶν πρὸς τοῖς Η, Θ γωνιῶν τῆς πρὸς τῷ Ζ, καὶ ἐγγεγράφθω εἰς τὸν ΑΒΓΔΕ κύκλον τῷ ΖΗΘ τριγώνῳ ἰσογώνιον τρίγωνον τὸ ΑΓΔ, ὥστε τῇ μὲν πρὸς τῷ Ζ γωνίᾳ ἴσην εἶναι τὴν ὑπὸ ΓΑΔ, ἑκατέραν δὲ τῶν πρὸς τοῖς Η, Θ ἴσην ἑκατέρᾳ τῶν ὑπὸ ΑΓΔ, ΓΔΑ: καὶ ἑκατέρα ἄρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ τῆς ὑπὸ ΓΑΔ ἐστι διπλῆ. τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ δίχα ὑπὸ ἑκατέρας τῶν ΓΕ, ΔΒ εὐθειῶν, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ. ἐπεὶ οὖν ἑκατέρα τῶν ὑπὸ ΑΓΔ, ΓΔΑ γωνιῶν διπλασίων ἐστὶ τῆς ὑπὸ ΓΑΔ, καὶ τετμημέναι εἰσὶ δίχα ὑπὸ τῶν ΓΕ, ΔΒ εὐθειῶν, αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΔΑΓ, ΑΓΕ, ΕΓΔ, ΓΔΒ, ΒΔΑ ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν: αἱ πέντε ἄρα περιφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν. ὑπὸ δὲ τὰς ἴσας περιφερείας ἴσαι εὐθεῖαι ὑποτείνουσιν: αἱ πέντε ἄρα εὐθεῖαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ ἴσαι ἀλλήλαις εἰσίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἡ ΑΒ περιφέρεια τῇ ΔΕ περιφερείᾳ ἐστὶν ἴση, κοινὴ προσκείσθω ἡ ΒΓΔ: ὅλη ἄρα ἡ ΑΒΓΔ περιφέρεια ὅλῃ τῇ ΕΔΓΒ περιφερείᾳ ἐστὶν ἴση. καὶ βέβηκεν ἐπὶ μὲν τῆς ΑΒΓΔ περιφερείας γωνία ἡ ὑπὸ ΑΕΔ, ἐπὶ δὲ τῆς ΕΔΓΒ περιφερείας γωνία ἡ ὑπὸ ΒΑΕ: καὶ ἡ ὑπὸ ΒΑΕ ἄρα γωνία τῇ ὑπὸ ΑΕΔ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΕ γωνιῶν ἑκατέρᾳ τῶν ὑπὸ ΒΑΕ, ΑΕΔ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον. εἰς ἄρα τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 296|>","<|""VertexLabel"" -> ""4.12"", ""Text"" -> ""About a given circle to circumscribe an equilateral and equiangular pentagon."", ""TextWordCount"" -> 11, ""GreekText"" -> ""περὶ τὸν δοθέντα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον περιγράψαι."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 8}, {""Book"" -> 1, ""Theorem"" -> 26}, {""Book"" -> 1, ""Theorem"" -> 47}, {""Book"" -> 3, ""Theorem"" -> 1}, {""Book"" -> 3, ""Theorem"" -> 16}, {""Book"" -> 3, ""Theorem"" -> 18}, {""Book"" -> 3, ""Theorem"" -> 27}, {""Book"" -> 4, ""Theorem"" -> 11}}, ""Proof"" -> ""Let ABCDE be the given circle; thus it is required to circumscribe an equilateral and equiangular pentagon about the circle ABCDE. Let A, B, C, D, E be conceived to be the angular points of the inscribed pentagon, so that the circumferences AB, BC, CD, DE, EA are equal; [IV. 11] through A, B, C, D, E let GH, HK, KL, LM, MG be drawn touching the circle; [III. 16] let the centre F of the circle ABCDE be taken [III. 1], and let FB, FK, FC, FL, FD be joined. Then, since the straight line KL touches the circle ABCDE at C, and FC has been joined from the centre F to the point of contact at C, therefore FC is perpendicular to KL; [III. 18] therefore each of the angles at C is right. For the same reason the angles at the points B, D are also right. And, since the angle FCK is right, therefore the square on FK is equal to the squares on FC, CK. For the same reason [I. 47] the square on FK is also equal to the squares on FB, BK; so that the squares on FC, CK are equal to the squares on FB, BK, of which the square on FC is equal to the square on FB; therefore the square on CK which remains is equal to the square on BK. Therefore BK is equal to CK. And, since FB is equal to FC, and FK common, the two sides BF, FK are equal to the two sides CF, FK; and the base BK equal to the base CK; therefore the angle BFK is equal to the angle KFC, [I. 8]and the angle BKF to the angle FKC. Therefore the angle BFC is double of the angle KFC, and the angle BKC of the angle FKC. For the same reason the angle CFD is also double of the angle CFL, and the angle DLC of the angle FLC. Now, since the circumference BC is equal to CD, the angle BFC is also equal to the angle CFD. [III. 27] And the angle BFC is double of the angle KFC, and the angle DFC of the angle LFC; therefore the angle KFC is also equal to the angle LFC. But the angle FCK is also equal to the angle FCL; therefore FKC, FLC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them; therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle; [I. 26] therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC. And, since KC is equal to CL, therefore KL is double of KC. For the same reason it can be proved that HK is also double of BK. And BK is equal to KC; therefore HK is also equal to KL. Similarly each of the straight lines HG, GM, ML can also be proved equal to each of the straight lines HK, KL; therefore the pentagon GHKLM is equilateral. I say next that it is also equiangular. For, since the angle FKC is equal to the angle FLC, and the angle HKL was proved double of the angle FKC, and the angle KLM double of the angle FLC, therefore the angle HKL is also equal to the angle KLM. Similarly each of the angles KHG, HGM, GML can also be proved equal to each of the angles HKL, KLM; therefore the five angles GHK, HKL, KLM, LMG, MGH are equal to one another. Therefore the pentagon GHKLM is equiangular. And it was also proved equilateral; and it has been circumscribed about the circle ABCDE."", ""ProofWordCount"" -> 631, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕ: δεῖ δὴ περὶ τὸν ΑΒΓΔΕ κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον περιγράψαι. νενοήσθω τοῦ ἐγγεγραμμένου πενταγώνου τῶν γωνιῶν σημεῖα τὰ Α, Β, Γ, Δ, Ε, ὥστε ἴσας εἶναι τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ περιφερείας: καὶ διὰ τῶν Α, Β, Γ, Δ, Ε ἤχθωσαν τοῦ κύκλου ἐφαπτόμεναι αἱ ΗΘ, ΘΚ, ΚΛ, ΛΜ, ΜΗ, καὶ εἰλήφθω τοῦ ΑΒΓΔΕ κύκλου κέντρον τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΖΒ, ΖΚ, ΖΓ, ΖΛ, ΖΔ. καὶ ἐπεὶ ἡ μὲν ΚΛ εὐθεῖα ἐφάπτεται τοῦ ΑΒΓΔΕ κατὰ τὸ Γ, ἀπὸ δὲ τοῦ Ζ κέντρου ἐπὶ τὴν κατὰ τὸ Γ ἐπαφὴν ἐπέζευκται ἡ ΖΓ, ἡ ΖΓ ἄρα κάθετός ἐστιν ἐπὶ τὴν ΚΛ: ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν πρὸς τῷ Γ γωνιῶν. διὰ τὰ αὐτὰ δὴ καὶ αἱ πρὸς τοῖς Β, Δ σημείοις γωνίαι ὀρθαί εἰσιν. καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΖΓΚ γωνία, τὸ ἄρα ἀπὸ τῆς ΖΚ ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΖΓ, ΓΚ. διὰ τὰ αὐτὰ δὴ καὶ τοῖς ἀπὸ τῶν ΖΒ, ΒΚ ἴσον ἐστὶ τὸ ἀπὸ τῆς ΖΚ: ὥστε τὰ ἀπὸ τῶν ΖΓ, ΓΚ τοῖς ἀπὸ τῶν ΖΒ, ΒΚ ἐστιν ἴσα, ὧν τὸ ἀπὸ τῆς ΖΓ τῷ ἀπὸ τῆς ΖΒ ἐστιν ἴσον: λοιπὸν ἄρα τὸ ἀπὸ τῆς ΓΚ τῷ ἀπὸ τῆς ΒΚ ἐστιν ἴσον. ἴση ἄρα ἡ ΒΚ τῇ ΓΚ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΖΒ τῇ ΖΓ, καὶ κοινὴ ἡ ΖΚ, δύο δὴ αἱ ΒΖ, ΖΚ δυσὶ ταῖς ΓΖ, ΖΚ ἴσαι εἰσίν: καὶ βάσις ἡ ΒΚ βάσει τῇ ΓΚ ἐστιν ἴση: γωνία ἄρα ἡ μὲν ὑπὸ ΒΖΚ γωνίᾳ τῇ ὑπὸ ΚΖΓ ἐστιν ἴση: ἡ δὲ ὑπὸ ΒΚΖ τῇ ὑπὸ ΖΚΓ: διπλῆ ἄρα ἡ μὲν ὑπὸ ΒΖΓ τῆς ὑπὸ ΚΖΓ, ἡ δὲ ὑπὸ ΒΚΓ τῆς ὑπὸ ΖΚΓ. διὰ τὰ αὐτὰ δὴ καὶ ἡ μὲν ὑπὸ ΓΖΔ τῆς ὑπὸ ΓΖΛ ἐστι διπλῆ, ἡ δὲ ὑπὸ ΔΛΓ τῆς ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ περιφέρεια τῇ ΓΔ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΖΓ τῇ ὑπὸ ΓΖΔ. καί ἐστιν ἡ μὲν ὑπὸ ΒΖΓ τῆς ὑπὸ ΚΖΓ διπλῆ, ἡ δὲ ὑπὸ ΔΖΓ τῆς ὑπὸ ΛΖΓ: ἴση ἄρα καὶ ἡ ὑπὸ ΚΖΓ τῇ ὑπὸ ΛΖΓ: ἐστὶ δὲ καὶ ἡ ὑπὸ ΖΓΚ γωνία τῇ ὑπὸ ΖΓΛ ἴση. δύο δὴ τρίγωνά ἐστι τὰ ΖΚΓ, ΖΛΓ τὰς δύο γωνίας ταῖς δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει καὶ τὴν λοιπὴν γωνίαν τῇ λοιπῇ γωνίᾳ: ἴση ἄρα ἡ μὲν ΚΓ εὐθεῖα τῇ ΓΛ, ἡ δὲ ὑπὸ ΖΚΓ γωνία τῇ ὑπὸ ΖΛΓ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΚΓ τῇ ΓΛ, διπλῆ ἄρα ἡ ΚΛ τῆς ΚΓ. διὰ τὰ αὐτὰ δὴ δειχθήσεται καὶ ἡ ΘΚ τῆς ΒΚ διπλῆ. καί ἐστιν ἡ ΒΚ τῇ ΚΓ ἴση: καὶ ἡ ΘΚ ἄρα τῇ ΚΛ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται καὶ ἑκάστη τῶν ΘΗ, ΗΜ, ΜΛ ἑκατέρᾳ τῶν ΘΚ, ΚΛ ἴση: ἰσόπλευρον ἄρα ἐστὶ τὸ ΗΘΚΛΜ πεντάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΖΚΓ γωνία τῇ ὑπὸ ΖΛΓ, καὶ ἐδείχθη τῆς μὲν ὑπὸ ΖΚΓ διπλῆ ἡ ὑπὸ ΘΚΛ, τῆς δὲ ὑπὸ ΖΛΓ διπλῆ ἡ ὑπὸ ΚΛΜ, καὶ ἡ ὑπὸ ΘΚΛ ἄρα τῇ ὑπὸ ΚΛΜ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται καὶ ἑκάστη τῶν ὑπὸ ΚΘΗ, ΘΗΜ, ΗΜΛ ἑκατέρᾳ τῶν ὑπὸ ΘΚΛ, ΚΛΜ ἴση: αἱ πέντε ἄρα γωνίαι αἱ ὑπὸ ΗΘΚ, ΘΚΛ, ΚΛΜ, ΛΜΗ, ΜΗΘ ἴσαι ἀλλήλαις εἰσίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΗΘΚΛΜ πεντάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον, καὶ περιγέγραπται περὶ τὸν ΑΒΓΔΕ κύκλον. Περὶ τὸν δοθέντα ἄρα κύκλον πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον περιγέγραπται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 576|>","<|""VertexLabel"" -> ""4.13"", ""Text"" -> ""In a given pentagon, which is equilateral and equiangular, to inscribe a circle."", ""TextWordCount"" -> 13, ""GreekText"" -> ""εἰς τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλον ἐγγράψαι."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 26}, {""Book"" -> 3, ""Theorem"" -> 16}}, ""Proof"" -> ""Let ABCDE be the given equilateral and equiangular pentagon; thus it is required to inscribe a circle in the pentagon ABCDE. For let the angles BCD, CDE be bisected by the straight lines CF, DF respectively; and from the point F, at which the straight lines CF, DF meet one another, let the straight lines FB, FA, FE be joined. Then, since BC is equal to CD, and CF common, the two sides BC, CF are equal to the two sides DC, CF; and the angle BCF is equal to the angle DCF; therefore the base BF is equal to the base DF, and the triangle BCF is equal to the triangle DCF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4] Therefore the angle CBF is equal to the angle CDF. And, since the angle CDE is double of the angle CDF, and the angle CDE is equal to the angle ABC, while the angle CDF is equal to the angle CBF; therefore the angle CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle FBC; therefore the angle ABC has been bisected by the straight line BF. Similarly it can be proved that the angles BAE, AED have also been bisected by the straight lines FA, FE respectively. Now let FG, FH, FK, FL, FM be drawn from the point F perpendicular to the straight lines AB, BC, CD, DE, EA. Then, since the angle HCF is equal to the angle KCF, and the right angle FHC is also equal to the angle FKC, FHC, FKC are two triangles having two angles equal to two angles and one side equal to one side, namely FC which is common to them and subtends one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; [I. 26] therefore the perpendicular FH is equal to the perpendicular FK. Similarly it can be proved that each of the straight lines FL, FM, FG is also equal to each of the straight lines FH, FK; therefore the five straight lines FG, FH, FK, FL, FM are equal to one another. Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will pass also through the remaining points; and it will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points G, H, K, L, M are right. For, if it does not touch them. but cuts them, it will result that the straight line drawn at right angles to the diameter of the circle from its extremity falls within the circle: which was proved absurd. [III. 16] Therefore the circle described with centre F and distance one of the straight lines FG, FH, FK, FL, FM will not cut the straight lines AB, BC, CD, DE, EA; therefore it will touch them. Let it be described, as GHKLM. Therefore in the given pentagon, which is equilateral and equiangular, a circle has been inscribed."", ""ProofWordCount"" -> 519, ""GreekProof"" -> ""ἔστω τὸ δοθὲν πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον τὸ ΑΒΓΔΕ: δεῖ δὴ εἰς τὸ ΑΒΓΔΕ πεντάγωνον κύκλον ἐγγράψαι. τετμήσθω γὰρ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ εὐθειῶν: καὶ ἀπὸ τοῦ Ζ σημείου, καθ᾽ ὃ συμβάλλουσιν ἀλλήλαις αἱ ΓΖ, ΔΖ εὐθεῖαι, ἐπεζεύχθωσαν αἱ ΖΒ, ΖΑ, ΖΕ εὐθεῖαι. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΓΔ, κοινὴ δὲ ἡ ΓΖ, δύο δὴ αἱ ΒΓ, ΓΖ δυσὶ ταῖς ΔΓ, ΓΖ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΒΓΖ γωνίᾳ τῇ ὑπὸ ΔΓΖ ἐστιν ἴση: βάσις ἄρα ἡ ΒΖ βάσει τῇ ΔΖ ἐστιν ἴση, καὶ τὸ ΒΓΖ τρίγωνον τῷ ΔΓΖ τριγώνῳ ἐστιν ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν: ἴση ἄρα ἡ ὑπὸ ΓΒΖ γωνία τῇ ὑπὸ ΓΔΖ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ὑπὸ ΓΔΕ τῆς ὑπὸ ΓΔΖ, ἴση δὲ ἡ μὲν ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΒΓ, ἡ δὲ ὑπὸ ΓΔΖ τῇ ὑπὸ ΓΒΖ, καὶ ἡ ὑπὸ ΓΒΑ ἄρα τῆς ὑπὸ ΓΒΖ ἐστι διπλῆ: ἴση ἄρα ἡ ὑπὸ ΑΒΖ γωνία τῇ ὑπὸ ΖΒΓ: ἡ ἄρα ὑπὸ ΑΒΓ γωνία δίχα τέτμηται ὑπὸ τῆς ΒΖ εὐθείας. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκατέρα τῶν ὑπὸ ΒΑΕ, ΑΕΔ δίχα τέτμηται ὑπὸ ἑκατέρας τῶν ΖΑ, ΖΕ εὐθειῶν. ἤχθωσαν δὴ ἀπὸ τοῦ Ζ σημείου ἐπὶ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας κάθετοι αἱ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΘΓΖ γωνία τῇ ὑπὸ ΚΓΖ, ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΖΘΓ ὀρθῇ τῇ ὑπὸ ΖΚΓ ἴση, δύο δὴ τρίγωνά ἐστι τὰ ΖΘΓ, ΖΚΓ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν: καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει: ἴση ἄρα ἡ ΖΘ κάθετος τῇ ΖΚ καθέτῳ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΛ, ΖΜ, ΖΗ ἑκατέρᾳ τῶν ΖΘ, ΖΚ ἴση ἐστίν: αἱ πέντε ἄρα εὐθεῖαι αἱ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν Η, Θ, Κ, Λ, Μ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Η, Θ, Κ, Λ, Μ σημείοις γωνίας. εἰ γὰρ οὐκ ἐφάψεται αὐτῶν, ἀλλὰ τεμεῖ αὐτάς, συμβήσεται τὴν τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπ᾽ ἄκρας ἀγομένην ἐντὸς πίπτειν τοῦ κύκλου: ὅπερ ἄτοπον ἐδείχθη. οὐκ ἄρα ὁ κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν η, Θ, Κ, Λ, Μ σημείων γραφόμενος κύκλος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας: ἐφάψεται ἄρα αὐτῶν. γεγράφθω ὡς ὁ ΗΘΚΛΜ. εἰς ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλος ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 446|>","<|""VertexLabel"" -> ""4.14"", ""Text"" -> ""About a given pentagon, which is equilateral and equiangular, to circumscribe a circle."", ""TextWordCount"" -> 13, ""GreekText"" -> ""περὶ τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλον περιγράψαι."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 6}}, ""Proof"" -> ""Let ABCDE be the given pentagon, which is equilateral and equiangular; thus it is required to circumscribe a circle about the pentagon ABCDE. Let the angles BCD, CDE be bisected by the straight lines CF, DF respectively, and from the point F, at which the straight lines meet, let the straight lines FB, FA, FE be joined to the points B, A, E. Then in manner similar to the preceding it can be proved that the angles CBA, BAE, AED have also been bisected by the straight lines FB, FA, FE respectively. Now, since the angle BCD is equal to the angle CDE, and the angle FCD is half of the angle BCD, and the angle CDF half of the angle CDE, therefore the angle FCD is also equal to the angle CDF, so that the side FC is also equal to the side FD. [I. 6] Similarly it can be proved that each of the straight lines FB, FA, FE is also equal to each of the straight lines FC, FD; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another. Therefore the circle described with centre F and distance one of the straight lines FA, FB, FC, FD, FE will pass also through the remaining points, and will have been circumscribed. Let it be circumscribed, and let it be ABCDE. Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed."", ""ProofWordCount"" -> 242, ""GreekProof"" -> ""ἔστω τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, τὸ ΑΒΓΔΕ: δεῖ δὴ περὶ τὸ ΑΒΓΔΕ πεντάγωνον κύκλον περιγράψαι. τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ, καὶ ἀπὸ τοῦ Ζ σημείου, καθ᾽ ὃ συμβάλλουσιν αἱ εὐθεῖαι, ἐπὶ τὰ Β, Α, Ε σημεῖα ἐπεζεύχθωσαν εὐθεῖαι αἱ ΖΒ, ΖΑ, ΖΕ. ὁμοίως δὴ τῷ πρὸ τούτου δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΓΒΑ, ΒΑΕ, ΑΕΔ γωνιῶν δίχα τέτμηται ὑπὸ ἑκάστης τῶν ΖΒ, ΖΑ, ΖΕ εὐθειῶν. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΓΔ γωνία τῇ ὑπὸ ΓΔΕ, καί ἐστι τῆς μὲν ὑπὸ ΒΓΔ ἡμίσεια ἡ ὑπὸ ΖΓΔ, τῆς δὲ ὑπὸ ΓΔΕ ἡμίσεια ἡ ὑπὸ ΓΔΖ, καὶ ἡ ὑπὸ ΖΓΔ ἄρα τῇ ὑπὸ ΖΔΓ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΖΓ πλευρᾷ τῇ ΖΔ ἐστιν ἴση. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΒ, ΖΑ, ΖΕ ἑκατέρᾳ τῶν ΖΓ, ΖΔ ἐστιν ἴση: αἱ πέντε ἄρα εὐθεῖαι αἱ ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ ἴσαι ἀλλήλαις εἰσίν. ὁ ἄρα κέντρῳ τῷ Ζ καὶ διαστήματι ἑνὶ τῶν ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται περιγεγραμμένος. περιγεγράφθω καὶ ἔστω ὁ ΑΒΓΔΕ. περὶ ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλος περιγέγραπται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 205|>","<|""VertexLabel"" -> ""4.15"", ""Text"" -> ""In a given circle to inscribe an equilateral and equiangular hexagon."", ""TextWordCount"" -> 11, ""GreekText"" -> ""εἰς τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 15}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 3, ""Theorem"" -> 26}, {""Book"" -> 3, ""Theorem"" -> 27}, {""Book"" -> 3, ""Theorem"" -> 29}}, ""Proof"" -> ""Let ABCDEF be the given circle; thus it is required to inscribe an equilateral and equiangular hexagon in the circle ABCDEF. Let the diameter AD of the circle ABCDEF be drawn; let the centre G of the circle be taken, and with centre D and distance DG let the circle EGCH be described; let EG, CG be joined and carried through to the points B, F, and let AB, BC, CD, DE, EF, FA be joined. I say that the hexagon ABCDEF is equilateral and equiangular. For, since the point G is the centre of the circle ABCDEF, GE is equal to GD. Again, since the point D is the centre of the circle GCH, DE is equal to DG. But GE was proved equal to GD; therefore GE is also equal to ED; therefore the triangle EGD is equilateral; and therefore its three angles EGD, GDE, DEG are equal to one another, inasmuch as, in isosceles triangles, the angles at the base are equal to one another. [I. 5] And the three angles of the triangle are equal to two right angles; [I. 32] therefore the angle EGD is one-third of two right angles. Similarly, the angle DGC can also be proved to be onethird of two right angles. And, since the straight line CG standing on EB makes the adjacent angles EGC, CGB equal to two right angles, therefore the remaining angle CGB is also one-third of two right angles. Therefore the angles EGD, DGC, CGB are equal to one another; so that the angles vertical to them, the angles BGA, AGF, FGE are equal. [I. 15] Therefore the six angles EGD, DGC, CGB, BGA, AGF, FGE are equal to one another. But equal angles stand on equal circumferences; [III. 26] therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another. And equal circumferences are subtended by equal straight lines; [III. 29] therefore the six straight lines are equal to one another; therefore the hexagon ABCDEF is equilateral. I say next that it is also equiangular. For, since the circumference FA is equal to the circumference ED, let the circumference ABCD be added to each; therefore the whole FABCD is equal to the whole EDCBA; and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA; therefore the angle AFE is equal to the angle DEF. [III. 27] Similarly it can be proved that the remaining angles of the hexagon ABCDEF are also severally equal to each of the angles AFE, FED; therefore the hexagon ABCDEF is equiangular. But it was also proved equilateral; and it has been inscribed in the circle ABCDEF. Therefore in the given circle an equilateral and equiangular hexagon has been inscribed."", ""ProofWordCount"" -> 459, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕΖ: δεῖ δὴ εἰς τὸν ΑΒΓΔΕΖ κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι. ἤχθω τοῦ ΑΒΓΔΕΖ κύκλου διάμετρος ἡ ΑΔ, καὶ εἰλήφθω τὸ κέντρον τοῦ κύκλου τὸ Η, καὶ κέντρῳ μὲν τῷ Δ διαστήματι δὲ τῷ ΔΗ κύκλος γεγράφθω ὁ ΕΗΓΘ, καὶ ἐπιζευχθεῖσαι αἱ ΕΗ, ΓΗ διήχθωσαν ἐπὶ τὰ Β, Ζ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ: λέγω, ὅτι τὸ ΑΒΓΔΕΖ ἑξάγωνον ἰσόπλευρόν τέ ἐστι καὶ ἰσογώνιον. ἐπεὶ γὰρ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓΔΕΖ κύκλου, ἴση ἐστὶν ἡ ΗΕ τῇ ΗΔ. πάλιν, ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ τοῦ ΗΓΘ κύκλου, ἴση ἐστὶν ἡ ΔΕ τῇ ΔΗ. ἀλλ᾽ ἡ ΗΕ τῇ ΗΔ ἐδείχθη ἴση: καὶ ἡ ΗΕ ἄρα τῇ ΕΔ ἴση ἐστίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΕΗΔ τρίγωνον: καὶ αἱ τρεῖς ἄρα αὐτοῦ γωνίαι αἱ ὑπὸ ΕΗΔ, ΗΔΕ, ΔΕΗ ἴσαι ἀλλήλαις εἰσίν, ἐπειδήπερ τῶν ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει γωνίαι ἴσαι ἀλλήλαις εἰσίν: καί εἰσιν αἱ τρεῖς τοῦ τριγώνου γωνίαι δυσὶν ὀρθαῖς ἴσαι: ἡ ἄρα ὑπὸ ΕΗΔ γωνία τρίτον ἐστὶ δύο ὀρθῶν. ὁμοίως δὴ δειχθήσεται καὶ ἡ ὑπὸ ΔΗΓ τρίτον δύο ὀρθῶν. καὶ ἐπεὶ ἡ ΓΗ εὐθεῖα ἐπὶ τὴν ΕΒ σταθεῖσα τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΕΗΓ, ΓΗΒ δυσὶν ὀρθαῖς ἴσας ποιεῖ, καὶ λοιπὴ ἄρα ἡ ὑπὸ ΓΗΒ τρίτον ἐστὶ δύο ὀρθῶν: αἱ ἄρα ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ γωνίαι ἴσαι ἀλλήλαις εἰσίν: ὥστε καὶ αἱ κατὰ κορυφὴν αὐταῖς αἱ ὑπὸ ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι εἰσίν ταῖς ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ. αἱ ἓξ ἄρα γωνίαι αἱ ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ, ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι ἀλλήλαις εἰσίν. αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν: αἱ ἓξ ἄρα περιφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ ἴσαι ἀλλήλαις εἰσίν. ὑπὸ δὲ τὰς ἴσας περιφερείας αἱ ἴσαι εὐθεῖαι ὑποτείνουσιν: αἱ ἓξ ἄρα εὐθεῖαι ἴσαι ἀλλήλαις εἰσίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ ἑξάγωνον. λέγω δή, ὅτι καὶ ἰσογώνιον. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΖΑ περιφέρεια τῇ ΕΔ περιφερείᾳ, κοινὴ προσκείσθω ἡ ΑΒΓΔ περιφέρεια: ὅλη ἄρα ἡ ΖΑΒΓΔ ὅλῃ τῇ ΕΔΓΒΑ ἐστιν ἴση: καὶ βέβηκεν ἐπὶ μὲν τῆς ΖΑΒΓΔ περιφερείας ἡ ὑπὸ ΖΕΔ γωνία, ἐπὶ δὲ τῆς ΕΔΓΒΑ περιφερείας ἡ ὑπὸ ΑΖΕ γωνία: ἴση ἄρα ἡ ὑπὸ ΑΖΕ γωνία τῇ ὑπὸ ΔΕΖ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ λοιπαὶ γωνίαι τοῦ ΑΒΓΔΕΖ ἑξαγώνου κατὰ μίαν ἴσαι εἰσὶν ἑκατέρᾳ τῶν ὑπὸ ΑΖΕ, ΖΕΔ γωνιῶν: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ ἑξάγωνον. ἐδείχθη δὲ καὶ ἰσόπλευρον: καὶ ἐγγέγραπται εἰς τὸν ΑΒΓΔΕΖ κύκλον. εἰς ἄρα τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγέγραπται: ὅπερ ἔδει ποιῆσαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἡ τοῦ ἑξαγώνου πλευρὰ ἴση ἐστὶ τῇ ἐκ τοῦ κέντρου τοῦ κύκλου. ὁμοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν κατὰ τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν, περιγραφήσεται περὶ τὸν κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἀκολούθως τοῖς ἐπὶ τοῦ πενταγώνου εἰρημένοις. καὶ ἔτι διὰ τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου εἰρημένοις εἰς τὸ δοθὲν ἑξάγωνον κύκλον ἐγγράψομέν τε καὶ περιγράψομεν: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 483|>","<|""VertexLabel"" -> ""4.16"", ""Text"" -> ""In a given circle to inscribe a fifteen-angled figure which shall be both equilateral and equiangular."", ""TextWordCount"" -> 16, ""GreekText"" -> ""εἰς τὸν δοθέντα κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι."", ""GreekTextWordCount"" -> 10, ""References"" -> {{""Book"" -> 3, ""Theorem"" -> 30}}, ""Proof"" -> ""Let ABCD be the given circle; thus it is required to inscribe in the circle ABCD a fifteenangled figure which shall be both equilateral and equiangular. In the circle ABCD let there be inscribed a side AC of the equilateral triangle inscribed in it, and a side AB of an equilateral pentagon; therefore, of the equal segments of which there are fifteen in the circle ABCD, there will be five in the circumference ABC which is one-third of the circle, and there will be three in the circumference AB which is one-fifth of the circle; therefore in the remainder BC there will be two of the equal segments. Let BC be bisected at E; [III. 30] therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD. If therefore we join BE, EC and fit into the circle ABCD straight lines equal to them and in contiguity, a fifteen-angled figure which is both equilateral and equiangular will have been inscribed in it."", ""ProofWordCount"" -> 165, ""GreekProof"" -> ""ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ: δεῖ δὴ εἰς τὸν ΑΒΓΔ κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι. ἐγγεγράφθω εἰς τὸν ΑΒΓΔ κύκλον τριγώνου μὲν ἰσοπλεύρου τοῦ εἰς αὐτὸν ἐγγραφομένου πλευρὰ ἡ ΑΓ, πενταγώνου δὲ ἰσοπλεύρου ἡ ΑΒ: οἵων ἄρα ἐστὶν ὁ ΑΒΓΔ κύκλος ἴσων τμημάτων δεκαπέντε, τοιούτων ἡ μὲν ΑΒΓ περιφέρεια τρίτον οὖσα τοῦ κύκλου ἔσται πέντε, ἡ δὲ ΑΒ περιφέρεια πέμπτον οὖσα τοῦ κύκλου ἔσται τριῶν: λοιπὴ ἄρα ἡ ΒΓ τῶν ἴσων δύο. τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε: ἑκατέρα ἄρα τῶν ΒΕ, ΕΓ περιφερειῶν πεντεκαιδέκατόν ἐστι τοῦ ΑΒΓΔ κύκλου. ἐὰν ἄρα ἐπιζεύξαντες τὰς ΒΕ, ΕΓ ἴσας αὐταῖς κατὰ τὸ συνεχὲς εὐθείας ἐναρμόσωμεν εἰς τὸν ΑΒΓΔΕ κύκλον, ἔσται εἰς αὐτὸν ἐγγεγραμμένον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον: ὅπερ ἔδει ποιῆσαι. ὁμοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν κατὰ τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν, περιγραφήσεται περὶ τὸν κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον. ἔτι δὲ διὰ τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου δείξεων καὶ εἰς τὸ δοθὲν πεντεκαιδεκάγωνον κύκλον ἐγγράψομέν τε καὶ περιγράψομεν: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 171|>","<|""VertexLabel"" -> ""5.1"", ""Text"" -> ""If a magnitude be the same multiple of a magnitude that a part subtracted is of a part subtracted, the remainder will also be the same multiple of the remainder that the whole is of the whole."", ""TextWordCount"" -> 37, ""GreekText"" -> ""ἐὰν ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα τῶν πάντων."", ""GreekTextWordCount"" -> 26, ""References"" -> {}, ""Proof"" -> ""Let any number of magnitudes whatever AB, CD be respectively equimultiples of any magnitudes E, F equal in multitude; I say that, whatever multiple AB is of E, that multiple will AB, CD also be of E, F. For, since AB is the same multiple of E that CD is of F, as many magnitudes as there are in AB equal to E, so many also are there in CD equal to F. Let AB be divided into the magnitudes AG, GB equal to E, and CD into CH, HD equal to F; then the multitude of the magnitudes AG, GB will be equal to the multitude of the magnitudes CH, HD. Now, since AG is equal to E, and CH to F, therefore AG is equal to E, and AG, CH to E, F. For the same reason GB is equal to E, and GB, HD to E, F; therefore, as many magnitudes as there are in AB equal to E, so many also are there in AB, CD equal to E, F; therefore, whatever multiple AB is of E, that multiple will AB, CD also be of E, F."", ""ProofWordCount"" -> 191, ""GreekProof"" -> ""ἔστω ὁποσαοῦν μεγέθη τὰ ΑΒ, ΓΔ ὁποσωνοῦν μεγεθῶν τῶν Ε, Ζ ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον: λέγω, ὅτι ὁσαπλάσιόν ἐστι τὸ ΑΒ τοῦ Ε, τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ. ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ τὸ ΓΔ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μεγέθη ἴσα τῷ Ε, τοσαῦτα καὶ ἐν τῷ ΓΔ ἴσα τῷ Ζ. διῃρήσθω τὸ μὲν ΑΒ εἰς τὰ τῷ Ε μεγέθη ἴσα τὰ ΑΗ, ΗΒ, τὸ δὲ ΓΔ εἰς τὰ τῷ Ζ ἴσα τὰ ΓΘ, ΘΔ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει τῶν ΓΘ, ΘΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΗ τῷ Ε, τὸ δὲ ΓΘ τῷ Ζ, ἴσον ἄρα τὸ ΑΗ τῷ Ε, καὶ τὰ ΑΗ, ΓΘ τοῖς Ε, Ζ. διὰ τὰ αὐτὰ δὴ ἴσον ἐστὶ τὸ ΗΒ τῷ ε, καὶ τὰ ΗΒ, ΘΔ τοῖς Ε, Ζ: ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Ε, τοσαῦτα καὶ ἐν τοῖς ΑΒ, ΓΔ ἴσα τοῖς Ε, Ζ: ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΒ τοῦ Ε, τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ. ἐὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα τῶν πάντων: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 213|>","<|""VertexLabel"" -> ""5.2"", ""Text"" -> ""If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, the sum of the first and fifth will also be the same multiple of the second that the sum of the third and sixth is of the fourth."", ""TextWordCount"" -> 64, ""GreekText"" -> ""ἐὰν πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολλαπλάσιον καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ πέμπτον δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τετάρτου."", ""GreekTextWordCount"" -> 33, ""References"" -> {}, ""Proof"" -> ""Let a first magnitude, AB, be the same multiple of a second, C, that a third, DE, is of a fourth, F, and let a fifth, BG, also be the same multiple of the second, C, that a sixth, EH, is of the fourth F; I say that the sum of the first and fifth, AG, will be the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F. For, since AB is the same multiple of C that DE is of F, therefore, as many magnitudes as there are in AB equal to C, so many also are there in DE equal to F. For the same reason also, as many as there are in BG equal to C, so many are there also in EH equal to F; therefore, as many as there are in the whole AG equal to C, so many also are there in the whole DH equal to F. Therefore, whatever multiple AG is of C, that multiple also is DH of F. Therefore the sum of the first and fifth, AG, is the same multiple of the second, C, that the sum of the third and sixth, DH, is of the fourth, F."", ""ProofWordCount"" -> 210, ""GreekProof"" -> ""πρῶτον γὰρ τὸ ΑΒ δευτέρου τοῦ Γ ἰσάκις ἔστω πολλαπλάσιον καὶ τρίτον τὸ ΔΕ τετάρτου τοῦ Ζ, ἔστω δὲ καὶ πέμπτον τὸ ΒΗ δευτέρου τοῦ Γ ἰσάκις πολλαπλάσιον καὶ ἕκτον τὸ ΕΘ τετάρτου τοῦ Ζ: λέγω, ὅτι καὶ συντεθὲν πρῶτον καὶ πέμπτον τὸ ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ. ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ τὸ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΔΕ ἴσα τῷ Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὅσα ἐστὶν ἐν τῷ ΒΗ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΕΘ ἴσα τῷ Ζ: ὅσα ἄρα ἐστὶν ἐν ὅλῳ τῷ ΑΗ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν ὅλῳ τῷ ΔΘ ἴσα τῷ Ζ: ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΗ τοῦ Γ, τοσαυταπλάσιον ἔσται καὶ τὸ ΔΘ τοῦ Ζ. καὶ συντεθὲν ἄρα πρῶτον καὶ πέμπτον τὸ ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ. ἐὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολλαπλάσιον καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ πέμπτον δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τετάρτου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 204|>","<|""VertexLabel"" -> ""5.3"", ""Text"" -> ""If a first magnitude be the same multiple of a second that a third is of a fourth, and if equimultiples be taken of the first and third, then also ex aequali the magnitudes taken will be equimultiples respectively, the one of the second and the other of the fourth."", ""TextWordCount"" -> 50, ""GreekText"" -> ""ἐὰν πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ληφθῇ δὲ ἰσάκις πολλαπλάσια τοῦ τε πρώτου καὶ τρίτου, καὶ δι᾽ ἴσου τῶν ληφθέντων ἑκάτερον ἑκατέρου ἰσάκις ἔσται πολλαπλάσιον τὸ μὲν τοῦ δευτέρου τὸ δὲ τοῦ τετάρτου."", ""GreekTextWordCount"" -> 37, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 2}}, ""Proof"" -> ""Let a first magnitude A be the same multiple of a second B that a third C is of a fourth D, and let equimultiples EF, GH be taken of A, C; I say that EF is the same multiple of B that GH is of D. For, since EF is the same multiple of A that GH is of C, therefore, as many magnitudes as there are in EF equal to A, so many also are there in GH equal to C. Let EF be divided into the magnitudes EK, KF equal to A, and GH into the magnitudes GL, LH equal to C; then the multitude of the magnitudes EK, KF will be equal to the multitude of the magnitudes GL, LH. And, since A is the same multiple of B that C is of D, while EK is equal to A, and GL to C, therefore EK is the same multiple of B that GL is of D. For the same reason KF is the same multiple of B that LH is of D. Since, then, a first magnitude EK is the same multiple of a second B that a third GL is of a fourth D, and a fifth KF is also the same multiple of the second B that a sixth LH is of the fourth D, therefore the sum of the first and fifth, EF, is also the same multiple of the second B that the sum of the third and sixth, GH, is of the fourth D. [V. 2]"", ""ProofWordCount"" -> 256, ""GreekProof"" -> ""πρῶτον γὰρ τὸ Α δευτέρου τοῦ Β ἰσάκις ἔστω πολλαπλάσιον καὶ τρίτον τὸ Γ τετάρτου τοῦ Δ, καὶ εἰλήφθω τῶν Α, Γ ἰσάκις πολλαπλάσια τὰ ΕΖ, ΗΘ: λέγω, ὅτι ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΕΖ τοῦ Β καὶ τὸ ΗΘ τοῦ Δ. ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΕΖ τοῦ Α καὶ τὸ ΗΘ τοῦ Γ, ὅσα ἄρα ἐστὶν ἐν τῷ ΕΖ ἴσα τῷ Α, τοσαῦτα καὶ ἐν τῷ ΗΘ ἴσα τῷ Γ. διῃρήσθω τὸ μὲν ΕΖ εἰς τὰ τῷ Α μεγέθη ἴσα τὰ ΕΚ, ΚΖ, τὸ δὲ ΗΘ εἰς τὰ τῷ Γ ἴσα τὰ ΗΛ, ΛΘ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΕΚ, ΚΖ τῷ πλήθει τῶν ΗΛ, ΛΘ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Α τοῦ Β καὶ τὸ Γ τοῦ Δ, ἴσον δὲ τὸ μὲν ΕΚ τῷ Α, τὸ δὲ ΗΛ τῷ Γ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΕΚ τοῦ Β καὶ τὸ ΗΛ τοῦ Δ. διὰ τὰ αὐτὰ δὴ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΚΖ τοῦ Β καὶ τὸ ΛΘ τοῦ Δ. ἐπεὶ οὖν πρῶτον τὸ ΕΚ δευτέρου τοῦ Β ἰσάκις ἐστὶ πολλαπλάσιον καὶ τρίτον τὸ ΗΛ τετάρτου τοῦ Δ, ἔστι δὲ καὶ πέμπτον τὸ ΚΖ δευτέρου τοῦ Β ἰσάκις πολλαπλάσιον καὶ ἕκτον τὸ ΛΘ τετάρτου τοῦ Δ, καὶ συντεθὲν ἄρα πρῶτον καὶ πέμπτον τὸ ΕΖ δευτέρου τοῦ Β ἰσάκις ἐστὶ πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΗΘ τετάρτου τοῦ Δ. ἐὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ληφθῇ δὲ τοῦ πρώτου καὶ τρίτου ἰσάκις πολλαπλάσια, καὶ δι᾽ ἴσου τῶν ληφθέντων ἑκάτερον ἑκατέρου ἰσάκις ἔσται πολλαπλάσιον τὸ μὲν τοῦ δευτέρου τὸ δὲ τοῦ τετάρτου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 263|>","<|""VertexLabel"" -> ""5.4"", ""Text"" -> ""If a first magnitude have to a second the same ratio as a third to a fourth, any equimultiples whatever of the first and third will also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order."", ""TextWordCount"" -> 45, ""GreekText"" -> ""ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, καὶ τὰ ἰσάκις πολλαπλάσια τοῦ τε πρώτου καὶ τρίτου πρὸς τὰ ἰσάκις πολλαπλάσια τοῦ δευτέρου καὶ τετάρτου καθ᾽ ὁποιονοῦν πολλαπλασιασμὸν τὸν αὐτὸν ἕξει λόγον ληφθέντα κατάλληλα."", ""GreekTextWordCount"" -> 39, ""References"" -> {{""Book"" -> 5, ""Definition"" -> 5}, {""Book"" -> 5, ""Theorem"" -> 3}}, ""Proof"" -> ""For let a first magnitude A have to a second B the same ratio as a third C to a fourth D; and let equimultiples E, F be taken of A, C, and G, H other, chance, equimultiples of B, D; I say that, as E is to G, so is F to H. For let equimultiples K, L be taken of E, F, and other, chance, equimultiples M, N of G, H. Since E is the same multiple of A that F is of C, and equimultiples K, L of E, F have been taken, therefore K is the same multiple of A that L is of C. [V. 3] For the same reason M is the same multiple of B that N is of D. And, since, as A is to B, so is C to D, and of A, C equimultiples K, L have been taken, and of B, D other, chance, equimultiples M, N, therefore, if K is in excess of M, L also is in excess of N, if it is equal, equal, and if less, less. [V. Def. 5] And K, L are equimultiples of E, F, and M, N other, chance, equimultiples of G, H; therefore, as E is to G, so is F to H. [V. Def. 5]"", ""ProofWordCount"" -> 216, ""GreekProof"" -> ""πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, καὶ εἰλήφθω τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Ε, Ζ, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Η, Θ: λέγω, ὅτι ἐστὶν ὡς τὸ Ε πρὸς τὸ Η, οὕτως τὸ Ζ πρὸς τὸ Θ. εἰλήφθω γὰρ τῶν μὲν Ε, Ζ ἰσάκις πολλαπλάσια τὰ Κ, Λ, τῶν δὲ Η, Θ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν. Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ μὲν Ε τοῦ Α, τὸ δὲ Ζ τοῦ Γ, καὶ εἴληπται τῶν Ε, Ζ ἰσάκις πολλαπλάσια τὰ Κ, Λ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ Κ τοῦ α καὶ τὸ Λ τοῦ Γ. διὰ τὰ αὐτὰ δὴ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Μ τοῦ Β καὶ τὸ Ν τοῦ Λ. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Κ, Λ, τῶν δὲ Β, Δ ἄλλα ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Κ τοῦ Μ, ὑπερέχει καὶ τὸ Λ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Κ, Λ τῶν Ε, Ζ ἰσάκις πολλαπλάσια, τὰ δὲ Μ, Ν τῶν Η, Θ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Ε πρὸς τὸ Η, οὕτως τὸ Ζ πρὸς τὸ Θ. ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, καὶ τὰ ἰσάκις πολλαπλάσια τοῦ τε πρώτου καὶ τρίτου πρὸς τὰ ἰσάκις πολλαπλάσια τοῦ δευτέρου καὶ τετάρτου τὸν αὐτὸν ἕξει λόγον καθ᾽ ὁποιονοῦν πολλαπλασιασμὸν ληφθέντα κατάλληλα: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 274|>","<|""VertexLabel"" -> ""5.5"", ""Text"" -> ""If a magnitude be the same multiple of a magnitude that a part subtracted is of a part subtracted, the remainder will also be the same multiple of the remainder that the whole is of the whole."", ""TextWordCount"" -> 37, ""GreekText"" -> ""ἐὰν μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, ὅπερ ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι τὸ ὅλον τοῦ ὅλου."", ""GreekTextWordCount"" -> 23, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 1}}, ""Proof"" -> ""For let the magnitude AB be the same multiple of the magnitude CD that the part AE subtracted is of the part CF subtracted; I say that the remainder EB is also the same multiple of the remainder FD that the whole AB is of the whole CD. For, whatever multiple AE is of CF, let EB be made that multiple of CG. Then, since AE is the same multiple of CF that EB is of GC, therefore AE is the same multiple of CF that AB is of GF. [V. 1] But, by the assumption, AE is the same multiple of CF that AB is of CD. Therefore AB is the same multiple of each of the magnitudes GF, CD; therefore GF is equal to CD. Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD. And, since AE is the same multiple of CF that EB is of GC, and GC is equal to DF,therefore AE is the same multiple of CF that EB is of FD. But, by hypothesis, AE is the same multiple of CF that AB is of CD; therefore EB is the same multiple of FD that AB is of CD. That is, the remainder EB will be the same multiple ofthe remainder FD that the whole AB is of the whole CD."", ""ProofWordCount"" -> 226, ""GreekProof"" -> ""μέγεθος γὰρ τὸ ΑΒ μεγέθους τοῦ ΓΔ ἰσάκις ἔστω πολλαπλάσιον, ὅπερ ἀφαιρεθὲν τὸ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ: λέγω, ὅτι καὶ λοιπὸν τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. ὁσαπλάσιον γάρ ἐστι τὸ ΑΕ τοῦ ΓΖ, τοσαυταπλάσιον γεγονέτω καὶ τὸ ΕΒ τοῦ ΓΗ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΗΓ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΗΖ. κεῖται δὲ ἰσάκις πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ. ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΒ ἑκατέρου τῶν ΗΖ, ΓΔ: ἴσον ἄρα τὸ ΗΖ τῷ ΓΔ. κοινὸν ἀφῃρήσθω τὸ ΓΖ: λοιπὸν ἄρα τὸ ΗΓ λοιπῷ τῷ ΖΔ ἴσον ἐστίν. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΗΓ, ἴσον δὲ τὸ ΗΓ τῷ ΔΖ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΖΔ. ἰσάκις δὲ ὑπόκειται πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΕΒ τοῦ ΖΔ καὶ τὸ ΑΒ τοῦ ΓΔ. καὶ λοιπὸν ἄρα τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. ἐὰν ἄρα μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, ὅπερ ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι καὶ τὸ ὅλον τοῦ ὅλου: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 226|>","<|""VertexLabel"" -> ""5.6"", ""Text"" -> ""If two magnitudes be equimultiples of two magnitudes, and any magnitudes subtracted from them be equimultiples of the same, the remainders also are either equal to the same or equimultiples of them."", ""TextWordCount"" -> 32, ""GreekText"" -> ""ἐὰν δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολλαπλάσια, καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολλαπλάσια, καὶ τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν πολλαπλάσια."", ""GreekTextWordCount"" -> 28, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 2}}, ""Proof"" -> ""For let two magnitudes AB, CD be equimultiples of two magnitudes E, F, and let AG, CH subtracted from them be equimultiples of the same two E, F; I say that the remainders also, GB, HD, are either equal to E, F or equimultiples of them. For, first, let GB be equal to E; I say that HD is also equal to F. For let CK be made equal to F. Since AG is the same multiple of E that CH is of F, while GB is equal to E and KC to F, therefore AB is the same multiple of E that KH is of F. [V. 2] But, by hypothesis, AB is the same multiple of E that CD is of F; therefore KH is the same multiple of F that CD is of F. Since then each of the magnitudes KH, CD is the same multiple of F, therefore KH is equal to CD. Let CH be subtracted from each; therefore the remainder KC is equal to the remainder HD. But F is equal to KC; therefore HD is also equal to F. Hence, if GB is equal to E, HD is also equal to F. Similarly we can prove that, even if GB be a multiple of E, HD is also the same multiple of F."", ""ProofWordCount"" -> 220, ""GreekProof"" -> ""δύο γὰρ μεγέθη τὰ ΑΒ, ΓΔ δύο μεγεθῶν τῶν Ε, Ζ ἰσάκις ἔστω πολλαπλάσια, καὶ ἀφαιρεθέντα τὰ ΑΗ, ΓΘ τῶν αὐτῶν τῶν Ε, Ζ ἰσάκις ἔστω πολλαπλάσια: λέγω, ὅτι καὶ λοιπὰ τὰ ΗΒ, ΘΔ τοῖς Ε, Ζ ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν πολλαπλάσια. ἔστω γὰρ πρότερον τὸ ΗΒ τῷ Ε ἴσον. λέγω, ὅτι καὶ τὸ ΘΔ τῷ Ζ ἴσον ἐστίν. κείσθω γὰρ τῷ Ζ ἴσον τὸ ΓΚ. ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΗ τοῦ Ε καὶ τὸ ΓΘ τοῦ Ζ, ἴσον δὲ τὸ μὲν ΗΒ τῷ Ε, τὸ δὲ ΚΓ τῷ Ζ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ τὸ ΚΘ τοῦ Ζ. ἰσάκις δὲ ὑπόκειται πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ τὸ ΓΔ τοῦ Ζ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΚΘ τοῦ Ζ καὶ τὸ ΓΔ τοῦ Ζ. ἐπεὶ οὖν ἑκάτερον τῶν ΚΘ, ΓΔ τοῦ Ζ ἰσάκις ἐστὶ πολλαπλάσιον, ἴσον ἄρα ἐστὶ τὸ ΚΘ τῷ ΓΔ. κοινὸν ἀφῃρήσθω τὸ ΓΘ: λοιπὸν ἄρα τὸ ΚΓ λοιπῷ τῷ ΘΔ ἴσον ἐστίν. ἀλλὰ τὸ Ζ τῷ ΚΓ ἐστιν ἴσον: καὶ τὸ ΘΔ ἄρα τῷ Ζ ἴσον ἐστίν. ὥστε εἰ τὸ ΗΒ τῷ Ε ἴσον ἐστίν, καὶ τὸ ΘΔ ἴσον ἔσται τῷ Ζ. ὁμοίως δὴ δείξομεν, ὅτι, κἂν πολλαπλάσιον ᾖ τὸ ΗΒ τοῦ Ε, τοσαυταπλάσιον ἔσται καὶ τὸ ΘΔ τοῦ Ζ. ἐὰν ἄρα δύο μεγέθη δύο μεγεθῶν ἰσάκις ᾖ πολλαπλάσια, καὶ ἀφαιρεθέντα τινὰ τῶν αὐτῶν ἰσάκις ᾖ πολλαπλάσια, καὶ τὰ λοιπὰ τοῖς αὐτοῖς ἤτοι ἴσα ἐστὶν ἢ ἰσάκις αὐτῶν πολλαπλάσια: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 243|>","<|""VertexLabel"" -> ""5.7"", ""Text"" -> ""Equal magnitudes have to the same the same ratio, as also has the same to equal magnitudes."", ""TextWordCount"" -> 17, ""GreekText"" -> ""τὰ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ αὐτὸ πρὸς τὰ ἴσα."", ""GreekTextWordCount"" -> 15, ""References"" -> {{""Book"" -> 5, ""Definition"" -> 5}}, ""Proof"" -> ""Let A, B be equal magnitudes and C any other, chance, magnitude; I say that each of the magnitudes A, B has the same ratio to C, and C has the same ratio to each of the magnitudes A, B. For let equimultiples D, E of A, B be taken, and of C another, chance, multiple F. Then, since D is the same multiple of A that E is of B, while A is equal to B, therefore D is equal to E. But F is another, chance, magnitude. If therefore D is in excess of F, E is also in excess of F, if equal to it, equal; and, if less, less. And D, E are equimultiples of A, B, while F is another, chance, multiple of C; therefore, as A is to C, so is B to C. [V. Def. 5] I say next that C also has the same ratio to each of the magnitudes A, B. For, with the same construction, we can prove similarly that D is equal to E; and F is some other magnitude. If therefore F is in excess of D, it is also in excess of E, if equal, equal; and, if less, less. And F is a multiple of C, while D, E are other, chance, equimultiples of A, B; therefore, as C is to A, so is C to B. [V. Def. 5]"", ""ProofWordCount"" -> 234, ""GreekProof"" -> ""ἔστω ἴσα μεγέθη τὰ Α, Β, ἄλλο δέ τι, ὃ ἔτυχεν, μέγεθος τὸ Γ: λέγω, ὅτι ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν ἔχει λόγον, καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β. εἰλήφθω γὰρ τῶν μὲν Α, Β ἰσάκις πολλαπλάσια τὰ Δ, Ε, τοῦ δὲ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον τὸ Ζ. ἐπεὶ οὖν ἰσάκις ἐστὶ πολλαπλάσιον τὸ Δ τοῦ Α καὶ τὸ Ε τοῦ Β, ἴσον δὲ τὸ Α τῷ Β, ἴσον ἄρα καὶ τὸ Δ τῷ Ε. ἄλλο δέ, ὃ ἔτυχεν, τὸ Ζ. εἰ ἄρα ὑπερέχει τὸ Δ τοῦ Ζ, ὑπερέχει καὶ τὸ Ε τοῦ Ζ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Δ, Ε τῶν Α, Β ἰσάκις πολλαπλάσια, τὸ δὲ Ζ τοῦ Γ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον: ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ Γ. λέγω δή, ὅτι καὶ τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν ἔχει λόγον. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἴσον ἐστὶ τὸ Δ τῷ Ε: ἄλλο δέ τι τὸ Ζ: εἰ ἄρα ὑπερέχει τὸ Ζ τοῦ Δ, ὑπερέχει καὶ τοῦ Ε, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὸ μὲν Ζ τοῦ Γ πολλαπλάσιον, τὰ δὲ Δ, Ε τῶν Α, Β ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Α, οὕτως τὸ Γ πρὸς τὸ Β. τὰ ἴσα ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον καὶ τὸ αὐτὸ πρὸς τὰ ἴσα. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν μεγέθη τινὰ ἀνάλογον ᾖ, καὶ ἀνάπαλιν ἀνάλογον ἔσται. ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 263|>","<|""VertexLabel"" -> ""5.8"", ""Text"" -> ""Of unequal magnitudes, the greater has to the same a greater ratio than the less has; and the same has to the less a greater ratio than it has to the greater."", ""TextWordCount"" -> 32, ""GreekText"" -> ""τῶν ἀνίσων μεγεθῶν τὸ μεῖζον πρὸς τὸ αὐτὸ μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον."", ""GreekTextWordCount"" -> 14, ""References"" -> {{""Book"" -> 5, ""Definition"" -> 4}, {""Book"" -> 5, ""Definition"" -> 7}, {""Book"" -> 5, ""Theorem"" -> 1}}, ""Proof"" -> ""Let AB, C be unequal magnitudes, and let AB be greater; let D be another, chance, magnitude; I say that AB has to D a greater ratio than C has to D, and D has to C a greater ratio than it has to AB. For, since AB is greater than C, let BE be made equal to C; then the less of the magnitudes AE, EB, if multiplied, will sometime be greater than D. [V. Def. 4] [Case I.] First, let AE be less than EB; let AE be multiplied, and let FG be a multiple of it which is greater than D; then, whatever multiple FG is of AE, let GH be made the same multiple of EB and K of C; and let L be taken double of D, M triple of it, and successive multiples increasing by one, until what is taken is a multiple of D and the first that is greater than K. Let it be taken, and let it be N which is quadruple of D and the first multiple of it that is greather than K. Then, since K is less than N first, therefore K is not less than M. And, since FG is the same multiple of AE that GH is of EB, therefore FG is the same multiple of AE that FH is of AB. [V. 1] But FG is the same multiple of AE that K is of C; therefore FH is the same multiple of AB that K is of C; therefore FH, K are equimultiples of AB, C. Again, since GH is the same multiple of EB that K is of C, and EB is equal to C, therefore GH is equal to K. But K is not less than M; therefore neither is GH less than M. And FG is greater than D; therefore the whole FH is greater than D, M together. But D, M together are equal to N, inasmuch as M is triple of D, and M, D together are quadruple of D, while N is also quadruple of D; whence M, D together are equal to N. But FH is greater than M, D; therefore FH is in excess of N, while K is not in excess of N. And FH, K are equimultiples of AB, C, while N is another, chance, multiple of D; therefore AB has to D a greater ratio than C has to D. [V. Def. 7] I say next, that D also has to C a greater ratio than D has to AB. For, with the same construction, we can prove similarly that N is in excess of K, while N is not in excess of FH. And N is a multiple of D, while FH, K are other, chance, equimultiples of AB, C; therefore D has to C a greater ratio than D has to AB. [V. Def. 7] [Case 2.] Again, let AE be greater than EB. Then the less, EB, if multiplied, will sometime be greater than D. [V. Def. 4] Let it be multiplied, and let GH be a multiple of EB and greater than D; and, whatever multiple GH is of EB, let FG be made the same multiple of AE, and K of C. Then we can prove similarly that FH, K are equimultiples of AB, C; and, similarly, let N be taken a multiple of D but the first that is greater than FG, so that FG is again not less than M. But GH is greater than D; therefore the whole FH is in excess of D, M, that is, of N. Now K is not in excess of N, inasmuch as FG also, which is greater than GH, that is, than K, is not in excess of N. And in the same manner, by following the above argument, we complete the demonstration."", ""ProofWordCount"" -> 645, ""GreekProof"" -> ""καὶ τὸ αὐτὸ πρὸς τὸ ἔλαττον μείζονα λόγον ἔχει ἤπερ πρὸς τὸ μεῖζον. ἔστω ἄνισα μεγέθη τὰ ΑΒ, Γ, καὶ ἔστω μεῖζον τὸ ΑΒ, ἄλλο δέ, ὃ ἔτυχεν, τὸ Δ: λέγω, ὅτι τὸ ΑΒ πρὸς τὸ Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Δ, καὶ τὸ Δ πρὸς τὸ Γ μείζονα λόγον ἔχει ἤπερ πρὸς τὸ ΑΒ. ἐπεὶ γὰρ μεῖζόν ἐστι τὸ ΑΒ τοῦ Γ, κείσθω τῷ Γ ἴσον τὸ ΒΕ: τὸ δὴ ἔλασσον τῶν ΑΕ, ΕΒ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ Δ μεῖζον. ἔστω πρότερον τὸ ΑΕ ἔλαττον τοῦ ΕΒ, καὶ πεπολλαπλασιάσθω τὸ ΑΕ, καὶ ἔστω αὐτοῦ πολλαπλάσιον τὸ ΖΗ μεῖζον ὂν τοῦ δ, καὶ ὁσαπλάσιόν ἐστι τὸ ΖΗ τοῦ ΑΕ, τοσαυταπλάσιον γεγονέτω καὶ τὸ μὲν ΗΘ τοῦ ΕΒ τὸ δὲ Κ τοῦ Γ: καὶ εἰλήφθω τοῦ Δ διπλάσιον μὲν τὸ Λ, τριπλάσιον δὲ τὸ Μ, καὶ ἑξῆς ἑνὶ πλεῖον, ἕως ἂν τὸ λαμβανόμενον πολλαπλάσιον μὲν γένηται τοῦ Δ, πρώτως δὲ μεῖζον τοῦ Κ. εἰλήφθω, καὶ ἔστω τὸ Ν τετραπλάσιον μὲν τοῦ Δ, πρώτως δὲ μεῖζον τοῦ Κ. ἐπεὶ οὖν τὸ Κ τοῦ Ν πρώτως ἐστὶν ἔλαττον, τὸ Κ ἄρα τοῦ Μ οὔκ ἐστιν ἔλαττον. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ ΗΘ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ ΖΘ τοῦ ΑΒ. ἰσάκις δέ ἐστι πολλαπλάσιον τὸ ΖΗ τοῦ ΑΕ καὶ τὸ Κ τοῦ Γ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΖΘ τοῦ ΑΒ καὶ τὸ Κ τοῦ Γ. τὰ ΖΘ, Κ ἄρα τῶν ΑΒ, Γ ἰσάκις ἐστὶ πολλαπλάσια. πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΕΒ καὶ τὸ Κ τοῦ Γ, ἴσον δὲ τὸ ΕΒ τῷ Γ, ἴσον ἄρα καὶ τὸ ΗΘ τῷ Κ: τὸ δὲ Κ τοῦ Μ οὔκ ἐστιν ἔλαττον: οὐδ᾽ ἄρα τὸ ΗΘ τοῦ μ ἔλαττόν ἐστιν. μεῖζον δὲ τὸ ΖΗ τοῦ Δ: ὅλον ἄρα τὸ ΖΘ συναμφοτέρων τῶν Δ, Μ μεῖζόν ἐστιν. ἀλλὰ συναμφότερα τὰ Δ, Μ τῷ Ν ἐστιν ἴσα, ἐπειδήπερ τὸ Μ τοῦ Δ τριπλάσιόν ἐστιν, συναμφότερα δὲ τὰ Μ, Δ τοῦ Δ ἐστι τετραπλάσια, ἔστι δὲ καὶ τὸ Ν τοῦ Δ τετραπλάσιον: συναμφότερα ἄρα τὰ Μ, Δ τῷ Ν ἴσα ἐστίν. ἀλλὰ τὸ ΖΘ τῶν Μ, Δ μεῖζόν ἐστιν: τὸ ΖΘ ἄρα τοῦ Ν ὑπερέχει: τὸ δὲ Κ τοῦ Ν οὐχ ὑπερέχει. καί ἐστι τὰ μὲν ΖΘ, Κ τῶν ΑΒ, Γ ἰσάκις πολλαπλάσια, τὸ δὲ Ν τοῦ Δ ἄλλο, ὃ ἔτυχεν, πολλαπλάσιον: τὸ ΑΒ ἄρα πρὸς τὸ Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Δ. λέγω δή, ὅτι καὶ τὸ Δ πρὸς τὸ Γ μείζονα λόγον ἔχει ἤπερ τὸ Δ πρὸς τὸ ΑΒ. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι τὸ μὲν Ν τοῦ Κ ὑπερέχει, τὸ δὲ Ν τοῦ ΖΘ οὐχ ὑπερέχει. καί ἐστι τὸ μὲν Ν τοῦ Δ πολλαπλάσιον, τὰ δὲ ΖΘ, Κ τῶν ΑΒ, Γ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: τὸ Δ ἄρα πρὸς τὸ Γ μείζονα λόγον ἔχει ἤπερ τὸ Δ πρὸς τὸ ΑΒ. ἀλλὰ δὴ τὸ ΑΕ τοῦ ΕΒ μεῖζον ἔστω. τὸ δὴ ἔλαττον τὸ ΕΒ πολλαπλασιαζόμενον ἔσται ποτὲ τοῦ Δ μεῖζον. πεπολλαπλασιάσθω, καὶ ἔστω τὸ ΗΘ πολλαπλάσιον μὲν τοῦ ΕΒ, μεῖζον δὲ τοῦ Δ: καὶ ὁσαπλάσιόν ἐστι τὸ ΗΘ τοῦ ΕΒ, τοσαυταπλάσιον γεγονέτω καὶ τὸ μὲν ΖΗ τοῦ ΑΕ, τὸ δὲ Κ τοῦ Γ. ὁμοίως δὴ δείξομεν, ὅτι τὰ ΖΘ, Κ τῶν ΑΒ, Γ ἰσάκις ἐστὶ πολλαπλάσια: καὶ εἰλήφθω ὁμοίως τὸ Ν πολλαπλάσιον μὲν τοῦ Δ, πρώτως δὲ μεῖζον τοῦ ΖΗ: ὥστε πάλιν τὸ ΖΗ τοῦ Μ οὔκ ἐστιν ἔλασσον. μεῖζον δὲ τὸ ΗΘ τοῦ Δ: ὅλον ἄρα τὸ ΖΘ τῶν Δ, Μ, τουτέστι τοῦ Ν, ὑπερέχει. τὸ δὲ Κ τοῦ Ν οὐχ ὑπερέχει, ἐπειδήπερ καὶ τὸ ΖΗ μεῖζον ὂν τοῦ ΗΘ, τουτέστι τοῦ Κ, τοῦ Ν οὐχ ὑπερέχει. καὶ ὡσαύτως κατακολουθοῦντες τοῖς ἐπάνω περαίνομεν τὴν ἀπόδειξιν. τῶν ἄρα ἀνίσων μεγεθῶν τὸ μεῖζον πρὸς τὸ αὐτὸ μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον: καὶ τὸ αὐτὸ πρὸς τὸ ἔλαττον μείζονα λόγον ἔχει ἤπερ πρὸς τὸ μεῖζον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 647|>","<|""VertexLabel"" -> ""5.9"", ""Text"" -> ""Magnitudes which have the same ratio to the same are equal to one another; and magnitudes to which the same has the same ratio are equal."", ""TextWordCount"" -> 26, ""GreekText"" -> ""τὰ πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχοντα λόγον ἴσα ἀλλήλοις ἐστίν : καὶ πρὸς ἃ τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον, ἐκεῖνα ἴσα ἐστίν."", ""GreekTextWordCount"" -> 23, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 8}}, ""Proof"" -> ""For let each of the magnitudes A, B have the same ratio to C; I say that A is equal to B. For, otherwise, each of the magnitudes A, B would not have had the same ratio to C; [V. 8] but it has; therefore A is equal to B. Again, let C have the same ratio to each of the magnitudes A, B; I say that A is equal to B. For, otherwise, C would not have had the same ratio to each of the magnitudes A, B; [V. 8] but it has; therefore A is equal to B."", ""ProofWordCount"" -> 100, ""GreekProof"" -> ""ἐχέτω γὰρ ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν λόγον: λέγω, ὅτι ἴσον ἐστὶ τὸ Α τῷ Β. εἰ γὰρ μή, οὐκ ἂν ἑκάτερον τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν εἶχε λόγον: ἔχει δέ: ἴσον ἄρα ἐστὶ τὸ Α τῷ Β. ἐχέτω δὴ πάλιν τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν λόγον: λέγω, ὅτι ἴσον ἐστὶ τὸ Α τῷ Β. εἰ γὰρ μή, οὐκ ἂν τὸ Γ πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν εἶχε λόγον: ἔχει δέ: ἴσον ἄρα ἐστὶ τὸ Α τῷ Β. τὰ ἄρα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχοντα λόγον ἴσα ἀλλήλοις ἐστίν: καὶ πρὸς ἃ τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον, ἐκεῖνα ἴσα ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 118|>","<|""VertexLabel"" -> ""5.10"", ""Text"" -> ""Of magnitudes which have a ratio to the same, that which has a greater ratio is greater; and that to which the same has a greater ratio is less."", ""TextWordCount"" -> 29, ""GreekText"" -> ""τῶν πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον ἐκεῖνο μεῖζόν ἐστιν: πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλαττόν ἐστιν."", ""GreekTextWordCount"" -> 24, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 7}, {""Book"" -> 5, ""Theorem"" -> 8}}, ""Proof"" -> ""For let A have to C a greater ratio than B has to C; I say that A is greater than B. For, if not, A is either equal to B or less. Now A is not equal to B; for in that case each of the magnitudes A, B would have had the same ratio to C; [V. 7] but they have not; therefore A is not equal to B. Nor again is A less than B; for in that case A would have had to C a less ratio than B has to C; [V. 8] but it has not; therefore A is not less than B. But it was proved not to be equal either; therefore A is greater than B. Again, let C have to B a greater ratio than C has to A; I say that B is less than A. For, if not, it is either equal or greater. Now B is not equal to A; for in that case C would have had the same ratio to each of the magnitudes A, B; [V. 7] but it has not; therefore A is not equal to B. Nor again is B greater than A; for in that case C would have had to B a less ratio than it has to A; [V. 8] but it has not; therefore B is not greater than A. But it was proved that it is not equal either; therefore B is less than A."", ""ProofWordCount"" -> 247, ""GreekProof"" -> ""ἐχέτω γὰρ τὸ Α πρὸς τὸ Γ μείζονα λόγον ἤπερ τὸ Β πρὸς τὸ Γ: λέγω, ὅτι μεῖζόν ἐστι τὸ Α τοῦ Β. εἰ γὰρ μή, ἤτοι ἴσον ἐστὶ τὸ Α τῷ Β ἢ ἔλασσον. ἴσον μὲν οὖν οὔκ ἐστι τὸ Α τῷ Β: ἑκάτερον γὰρ ἂν τῶν Α, Β πρὸς τὸ Γ τὸν αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ: οὐκ ἄρα ἴσον ἐστὶ τὸ Α τῷ Β. οὐδὲ μὴν ἔλασσόν ἐστι τὸ Α τοῦ Β: τὸ Α γὰρ ἂν πρὸς τὸ Γ ἐλάσσονα λόγον εἶχεν ἤπερ τὸ Β πρὸς τὸ Γ. οὐκ ἔχει δέ: οὐκ ἄρα ἔλασσόν ἐστι τὸ Α τοῦ Β. ἐδείχθη δὲ οὐδὲ ἴσον: μεῖζον ἄρα ἐστὶ τὸ Α τοῦ Β. ἐχέτω δὴ πάλιν τὸ Γ πρὸς τὸ Β μείζονα λόγον ἤπερ τὸ Γ πρὸς τὸ Α: λέγω, ὅτι ἔλασσόν ἐστι τὸ Β τοῦ Α. εἰ γὰρ μή, ἤτοι ἴσον ἐστὶν ἢ μεῖζον. ἴσον μὲν οὖν οὔκ ἐστι τὸ Β τῷ Α: τὸ Γ γὰρ ἂν πρὸς ἑκάτερον τῶν Α, Β τὸν αὐτὸν εἶχε λόγον. οὐκ ἔχει δέ: οὐκ ἄρα ἴσον ἐστὶ τὸ Α τῷ Β. οὐδὲ μὴν μεῖζόν ἐστι τὸ Β τοῦ Α: τὸ Γ γὰρ ἂν πρὸς τὸ Β ἐλάσσονα λόγον εἶχεν ἤπερ πρὸς τὸ Α. οὐκ ἔχει δέ: οὐκ ἄρα μεῖζόν ἐστι τὸ Β τοῦ Α. ἐδείχθη δέ, ὅτι οὐδὲ ἴσον: ἔλαττον ἄρα ἐστὶ τὸ Β τοῦ Α. τῶν ἄρα πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον μεῖζόν ἐστιν: καὶ πρὸς ὃ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλαττόν ἐστιν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 251|>","<|""VertexLabel"" -> ""5.11"", ""Text"" -> ""Ratios which are the same with the same ratio are also the same with one another."", ""TextWordCount"" -> 16, ""GreekText"" -> ""οἱ τῷ αὐτῷ λόγῳ οἱ αὐτοὶ καὶ ἀλλήλοις εἰσὶν οἱ αὐτοί."", ""GreekTextWordCount"" -> 11, ""References"" -> {}, ""Proof"" -> ""For, as A is to B, so let C be to D, and, as C is to D, so let E be to F; I say that, as A is to B, so is E to F. For of A, C, E let equimultiples G, H, K be taken, and of B, D, F other, chance, equimultiples L, M, N. Then since, as A is to B, so is C to D, and of A, C equimultiples G, H have been taken, and of B, D other, chance, equimultiples L, M, therefore, if G is in excess of L, H is also in excess of M, if equal, equal, and if less, less. Again, since, as C is to D, so is E to F, and of C, E equimultiples H, K have been taken, and of D, F other, chance, equimultiples M, N, therefore, if H is in excess of M, K is also in excess of N, if equal, equal, and if less, less. But we saw that, if H was in excess of M, G was also in excess of L; if equal, equal; and if less, less; so that, in addition, if G is in excess of L, K is also in excess of N, if equal, equal, and if less, less. And G, K are equimultiples of A, E, while L, N are other, chance, equimultiples of B, F; therefore, as A is to B, so is E to F."", ""ProofWordCount"" -> 245, ""GreekProof"" -> ""ἔστωσαν γὰρ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, ὡς δὲ τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ: λέγω, ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. εἰλήφθω γὰρ τῶν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, Κ, τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Η, Θ, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, εἰ ἄρα ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Θ τοῦ Μ, καὶ εἰ ἴσον ἐστίν, ἴσον, καὶ εἰ ἐλλείπει, ἐλλείπει. πάλιν, ἐπεί ἐστιν ὡς τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ, καὶ εἴληπται τῶν Γ, Ε ἰσάκις πολλαπλάσια τὰ Θ, Κ, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Θ τοῦ Μ, ὑπερέχει καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. ἀλλὰ εἰ ὑπερεῖχε τὸ Θ τοῦ Μ, ὑπερεῖχε καὶ τὸ Η τοῦ Λ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον: ὥστε καὶ εἰ ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Η, Κ τῶν Α, Ε ἰσάκις πολλαπλάσια, τὰ δὲ Λ, Ν τῶν Β, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. οἱ ἄρα τῷ αὐτῷ λόγῳ οἱ αὐτοὶ καὶ ἀλλήλοις εἰσὶν οἱ αὐτοί: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 282|>","<|""VertexLabel"" -> ""5.12"", ""Text"" -> ""If any number of magnitudes be proportional, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents."", ""TextWordCount"" -> 28, ""GreekText"" -> ""ἐὰν ᾖ ὁποσαοῦν μεγέθη ἀνάλογον, ἔσται ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα."", ""GreekTextWordCount"" -> 22, ""References"" -> {{""Book"" -> 5, ""Definition"" -> 5}, {""Book"" -> 5, ""Theorem"" -> 1}}, ""Proof"" -> ""Let any number of magnitudes A, B, C, D, E, F be proportional, so that, as A is to B, so is C to D and E to F; I say that, as A is to B, so are A, C, E to B, D, F. For of A, C, E let equimultiples G, H, K be taken, and of B, D, F other, chance, equimultiples L, M, N. Then since, as A is to B, so is C to D, and E to F, and of A, C, E equimultiples G, H, K have been taken, and of B, D, F other, chance, equimultiples L, M, N, therefore, if G is in excess of L, H is also in excess of M, and K of N, if equal, equal, and if less, less; so that, in addition, if G is in excess of L, then G, H, K are in excess of L, M, N, if equal, equal, and if less, less. Now G and G, H, K are equimultiples of A and A, C, E, since, if any number of magnitudes whatever are respectively equimultiples of any magnitudes equal in multitude, whatever multiple one of the magnitudes is of one, that multiple also will all be of all. [V. 1] For the same reason L and L, M, N are also equimultiples of B and B, D, F; therefore, as A is to B, so are A, C, E to B, D, F. [V. Def. 5]"", ""ProofWordCount"" -> 248, ""GreekProof"" -> ""ἔστωσαν ὁποσαοῦν μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, ε, Ζ, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ τὸ Ε πρὸς τὸ Ζ: λέγω, ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὰ Α, Γ, Ε πρὸς τὰ Β, Δ, Ζ. εἰλήφθω γὰρ τῶν μὲν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, Κ, τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ τὸ Ε πρὸς τὸ Ζ, καὶ εἴληπται τῶν μὲν Α, Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, Κ τῶν δὲ Β, Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν, εἰ ἄρα ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Θ τοῦ Μ, καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. ὥστε καὶ εἰ ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὰ Η, Θ, Κ τῶν Λ, Μ, Ν, καὶ εἰ ἴσον, ἴσα, καὶ εἰ ἔλαττον, ἐλάττονα. καί ἐστι τὸ μὲν Η καὶ τὰ Η, Θ, Κ τοῦ Α καὶ τῶν Α, Γ, Ε ἰσάκις πολλαπλάσια, ἐπειδήπερ ἐὰν ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα τῶν πάντων. διὰ τὰ αὐτὰ δὴ καὶ τὸ Λ καὶ τὰ Λ, Μ, Ν τοῦ Β καὶ τῶν β, Δ, Ζ ἰσάκις ἐστὶ πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὰ Α, Γ, Ε πρὸς τὰ Β, Δ, Ζ. ἐὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ἀνάλογον, ἔσται ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 287|>","<|""VertexLabel"" -> ""5.13"", ""Text"" -> ""If a first magnitude have to a second the same ratio as a third to a fourth, and the third have to the fourth a greater ratio than a fifth has to a sixth, the first will also have to the second a greater ratio than the fifth to the sixth."", ""TextWordCount"" -> 51, ""GreekText"" -> ""ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα λόγον ἔχῃ ἢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον μείζονα λόγον ἕξει ἢ πέμπτον πρὸς ἕκτον."", ""GreekTextWordCount"" -> 34, ""References"" -> {{""Book"" -> 5, ""Definition"" -> 5}, {""Book"" -> 5, ""Definition"" -> 7}}, ""Proof"" -> ""For let a first magnitude A have to a second B the same ratio as a third C has to a fourth D, and let the third C have to the fourth D a greater ratio than a fifth E has to a sixth F; I say that the first A will also have to the second B a greater ratio than the fifth E to the sixth F. For, since there are some equimultiples of C, E, and of D, F other, chance, equimultiples, such that the multiple of C is in excess of the multiple of D, while the multiple of E is not in excess of the multiple of F, [V. Def. 7] let them be taken, and let G, H be equimultiples of C, E, and K, L other, chance, equimultiples of D, F, so that G is in excess of K, but H is not in excess of L; and, whatever multiple G is of C, let M be also that multiple of A, and, whatever multiple K is of D, let N be also that multiple of B. Now, since, as A is to B, so is C to D, and of A, C equimultiples M, G have been taken, and of B, D other, chance, equimultiples N, K, therefore, if M is in excess of N, G is also in excess of K, if equal, equal, and if less, less. [V. Def. 5] But G is in excess of K; therefore M is also in excess of N. But H is not in excess of L; and M, H are equimultiples of A, E, and N, L other, chance, equimultiples of B, F; therefore A has to B a greater ratio than E has to F. [V. Def. 7]"", ""ProofWordCount"" -> 296, ""GreekProof"" -> ""πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, τρίτον δὲ τὸ Γ πρὸς τέταρτον τὸ Δ μείζονα λόγον ἐχέτω ἢ πέμπτον τὸ Ε πρὸς ἕκτον τὸ Ζ. λέγω, ὅτι καὶ πρῶτον τὸ Α πρὸς δεύτερον τὸ Β μείζονα λόγον ἕξει ἤπερ πέμπτον τὸ Ε πρὸς ἕκτον τὸ Ζ. ἐπεὶ γὰρ ἔστι τινὰ τῶν μὲν Γ, Ε ἰσάκις πολλαπλάσια, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια, καὶ τὸ μὲν τοῦ Γ πολλαπλάσιον τοῦ τοῦ Δ πολλαπλασίου ὑπερέχει, τὸ δὲ τοῦ Ε πολλαπλάσιον τοῦ τοῦ Ζ πολλαπλασίου οὐχ ὑπερέχει, εἰλήφθω, καὶ ἔστω τῶν μὲν Γ, Ε ἰσάκις πολλαπλάσια τὰ Η, Θ, τῶν δὲ Δ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Κ, Λ, ὥστε τὸ μὲν Η τοῦ Κ ὑπερέχειν, τὸ δὲ Θ τοῦ Λ μὴ ὑπερέχειν: καὶ ὁσαπλάσιον μέν ἐστι τὸ Η τοῦ Γ, τοσαυταπλάσιον ἔστω καὶ τὸ Μ τοῦ Α, ὁσαπλάσιον δὲ τὸ Κ τοῦ Δ, τοσαυταπλάσιον ἔστω καὶ τὸ Ν τοῦ Β. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ, καὶ εἴληπται τῶν μὲν Α, Γ ἰσάκις πολλαπλάσια τὰ Μ, Η, τῶν δὲ Β, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Ν, Κ, εἰ ἄρα ὑπερέχει τὸ Μ τοῦ Ν, ὑπερέχει καὶ τὸ Η τοῦ Κ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. ὑπερέχει δὲ τὸ Η τοῦ Κ: ὑπερέχει ἄρα καὶ τὸ Μ τοῦ Ν. τὸ δὲ Θ τοῦ Λ οὐχ ὑπερέχει: καί ἐστι τὰ μὲν Μ, Θ τῶν Α, Ε ἰσάκις πολλαπλάσια, τὰ δὲ Ν, Λ τῶν Β, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: τὸ ἄρα Α πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Ε πρὸς τὸ Ζ. ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τρίτον δὲ πρὸς τέταρτον μείζονα λόγον ἔχῃ ἢ πέμπτον πρὸς ἕκτον, καὶ πρῶτον πρὸς δεύτερον μείζονα λόγον ἕξει ἢ πέμπτον πρὸς ἕκτον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 322|>","<|""VertexLabel"" -> ""5.14"", ""Text"" -> ""If a first magnitude have to a second the same ratio as a third has to a fourth, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less."", ""TextWordCount"" -> 42, ""GreekText"" -> ""ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον."", ""GreekTextWordCount"" -> 32, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 8}, {""Book"" -> 5, ""Theorem"" -> 10}, {""Book"" -> 5, ""Theorem"" -> 13}}, ""Proof"" -> ""For let a first magnitude A have the same ratio to a second B as a third C has to a fourth D; and let A be greater than C; I say that B is also greater than D. For, since A is greater than C, and B is another, chance, magnitude, therefore A has to B a greater ratio than C has to B. [V. 8] But, as A is to B, so is C to D; therefore C has also to D a greater ratio than C has to B. [V. 13] But that to which the same has a greater ratio is less; [V. 10] therefore D is less than B; so that B is greater than D. Similarly we can prove that, if A be equal to C, B will also be equal to D; and, if A be less than C, B will also be less than D."", ""ProofWordCount"" -> 153, ""GreekProof"" -> ""πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, μεῖζον δὲ ἔστω τὸ Α τοῦ Γ: λέγω, ὅτι καὶ τὸ Β τοῦ Δ μεῖζόν ἐστιν. ἐπεὶ γὰρ τὸ Α τοῦ Γ μεῖζόν ἐστιν, ἄλλο δέ, ὃ ἔτυχεν, μέγεθος τὸ Β, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ: καὶ τὸ Γ ἄρα πρὸς τὸ Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλασσόν ἐστιν: ἔλασσον ἄρα τὸ Δ τοῦ Β: ὥστε μεῖζόν ἐστι τὸ Β τοῦ Δ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Β τῷ Δ, κἂν ἔλασσον ᾖ τὸ Α τοῦ Γ, ἔλασσον ἔσται καὶ τὸ Β τοῦ Δ. ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 187|>","<|""VertexLabel"" -> ""5.15"", ""Text"" -> ""Parts have the same ratio as the same multiples of them taken in corresponding order."", ""TextWordCount"" -> 15, ""GreekText"" -> ""τὰ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον ληφθέντα κατάλληλα."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 7}, {""Book"" -> 5, ""Theorem"" -> 12}}, ""Proof"" -> ""For let AB be the same multiple of C that DE is of F; I say that, as C is to F, so is AB to DE. For, since AB is the same multiple of C that DE is of F, as many magnitudes as there are in AB equal to C, so many are there also in DE equal to F. Let AB be divided into the magnitudes AG, GH, HB equal to C, and DE into the magnitudes DK, KL, LE equal to F; then the multitude of the magnitudes AG, GH, HB will be equal to the multitude of the magnitudes DK, KL, LE. And, since AG. GH, HB are equal to one another, and DK, KL, LE are also equal to one another, therefore, as AG is to DK, so is GH to KL, and HB to LE. [V. 7] Therefore, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [V. 12] therefore, as AG is to DK, so is AB to DE. But AG is equal to C and DK to F; therefore, as C is to F, so is AB to DE."", ""ProofWordCount"" -> 200, ""GreekProof"" -> ""ἔστω γὰρ ἰσάκις πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ τὸ ΔΕ τοῦ Ζ: λέγω, ὅτι ἐστὶν ὡς τὸ Γ πρὸς τὸ Ζ, οὕτως τὸ ΑΒ πρὸς τὸ ΔΕ. ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ τὸ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μεγέθη ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΔΕ ἴσα τῷ Ζ. διῃρήσθω τὸ μὲν ΑΒ εἰς τὰ τῷ Γ ἴσα τὰ ΑΗ, ΗΘ, ΘΒ, τὸ δὲ ΔΕ εἰς τὰ τῷ Ζ ἴσα τὰ ΔΚ, ΚΛ, ΛΕ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΘ, ΘΒ τῷ πλήθει τῶν ΔΚ, ΚΛ, ΛΕ. καὶ ἐπεὶ ἴσα ἐστὶ τὰ ΑΗ, ΗΘ, ΘΒ ἀλλήλοις, ἔστι δὲ καὶ τὰ ΔΚ, ΚΛ, ΛΕ ἴσα ἀλλήλοις, ἔστιν ἄρα ὡς τὸ ΑΗ πρὸς τὸ ΔΚ, οὕτως τὸ ΗΘ πρὸς τὸ ΚΛ, καὶ τὸ ΘΒ πρὸς τὸ ΛΕ. ἔσται ἄρα καὶ ὡς ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ἔστιν ἄρα ὡς τὸ ΑΗ πρὸς τὸ ΔΚ, οὕτως τὸ ΑΒ πρὸς τὸ ΔΕ. ἴσον δὲ τὸ μὲν ΑΗ τῷ Γ, τὸ δὲ ΔΚ τῷ Ζ: ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ζ οὕτως τὸ ΑΒ πρὸς τὸ ΔΕ. τὰ ἄρα μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον ληφθέντα κατάλληλα: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 212|>","<|""VertexLabel"" -> ""5.16"", ""Text"" -> ""If four magnitudes be proportional, they will also be proportional alternately."", ""TextWordCount"" -> 11, ""GreekText"" -> ""ἐὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, καὶ ἐναλλὰξ ἀνάλογον ἔσται."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 5, ""Definition"" -> 5}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 5, ""Theorem"" -> 14}, {""Book"" -> 5, ""Theorem"" -> 15}}, ""Proof"" -> ""Let A, B, C, D be four proportional magnitudes, so that, as A is to B, so is C to D; I say that they will also be so alternately, that is, as A is to C, so is B to D. For of A, B let equimultiples E, F be taken, and of C, D other, chance, equimultiples G, H. Then, since E is the same multiple of A that F is of B, and parts have the same ratio as the same multiples of them, [V. 15] therefore, as A is to B, so is E to F. But as A is to B, so is C to D; therefore also, as C is to D, so is E to F. [V. 11] Again, since G, H are equimultiples of C, D, therefore, as C is to D, so is G to H. [V. 15] But, as C is to D, so is E to F; therefore also, as E is to F, so is G to H. [V. 11] But, if four magnitudes be proportional, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less. [V. 14] Therefore, if E is in excess of G, F is also in excess of H, if equal, equal, and if less, less. Now E, F are equimultiples of A, B, and G, H other, chance, equimultiples of C, D; therefore, as A is to C, so is B to D. [V. Def. 5]"", ""ProofWordCount"" -> 257, ""GreekProof"" -> ""ἔστω τέσσαρα μεγέθη ἀνάλογον τὰ Α, Β, Γ, Δ, ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ: λέγω, ὅτι καὶ ἐναλλὰξ ἀνάλογον ἔσται, ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ Δ. εἰλήφθω γὰρ τῶν μὲν Α, Β ἰσάκις πολλαπλάσια τὰ Ε, Ζ, τῶν δὲ Γ, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Η, Θ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ Ε τοῦ Α καὶ τὸ Ζ τοῦ Β, τὰ δὲ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ: καὶ ὡς ἄρα τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ. πάλιν, ἐπεὶ τὰ Η, Θ τῶν Γ, Δ ἰσάκις ἐστὶ πολλαπλάσια, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Η πρὸς τὸ Θ. ὡς δὲ τὸ Γ πρὸς τὸ Δ, οὕτως τὸ Ε πρὸς τὸ Ζ: καὶ ὡς ἄρα τὸ Ε πρὸς τὸ Ζ, οὕτως τὸ Η πρὸς τὸ Θ. ἐὰν δὲ τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. εἰ ἄρα ὑπερέχει τὸ Ε τοῦ Η, ὑπερέχει καὶ τὸ Ζ τοῦ Θ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Ε, Ζ τῶν Α, Β ἰσάκις πολλαπλάσια, τὰ δὲ Η, Θ τῶν Γ, Δ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Β πρὸς τὸ Δ. ἐὰν ἄρα τέσσαρα μεγέθη ἀνάλογον ᾖ, καὶ ἐναλλὰξ ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 276|>","<|""VertexLabel"" -> ""5.17"", ""Text"" -> ""If magnitudes be proportional componendo, they will also be proportional separando."", ""TextWordCount"" -> 11, ""GreekText"" -> ""ἐὰν συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 1}, {""Book"" -> 5, ""Theorem"" -> 2}}, ""Proof"" -> ""Let AB, BE, CD, DF be magnitudes proportional componendo, so that, as AB is to BE, so is CD to DF; I say that they will also be proportional separando, that is, as AE is to EB, so is CF to DF. For of AE, EB, CF, FD let equimultiples GH, HK, LM, MN be taken, and of EB, FD other, chance, equimultiples, KO, NP. Then, since GH is the same multiple of AE that HK is of EB, therefore GH is the same multiple of AE that GK is of AB. [V. 1] But GH is the same multiple of AE that LM is of CF; therefore GK is the same multiple of AB that LM is of CF. Again, since LM is the same multiple of CF that MN is of FD, therefore LM is the same multiple of CF that LN is of CD. [V. 1] But LM was the same multiple of CF that GK is of AB; therefore GK is the same multiple of AB that LN is of CD. Therefore GK, LN are equimultiples of AB, CD. Again, since HK is the same multiple of EB that MN is of FD, and KO is also the same multiple of EB that NP is of FD, therefore the sum HO is also the same multiple of EB that MP is of FD. [V. 2] And, since, as AB is to BE, so is CD to DF, and of AB, CD equimultiples GK, LN have been taken, and of EB, FD equimultiples HO, MP, therefore, if GK is in excess of HO, LN is also in excess of MP, if equal, equal, and if less, less. Let GK be in excess of HO; then, if HK be subtracted from each, GH is also in excess of KO. But we saw that, if GK was in excess of HO, LN was also in excess of MP; therefore LN is also in excess of MP, and, if MN be subtracted from each, LM is also in excess of NP; so that, if GH is in excess of KO, LM is also in excess of NP. Similarly we can prove that, if GH be equal to KO, LM will also be equal to NP, and if less, less. And GH, LM are equimultiples of AE, CF, while KO, NP are other, chance, equimultiples of EB, FD; therefore, as AE is to EB, so is CF to FD."", ""ProofWordCount"" -> 408, ""GreekProof"" -> ""ἔστω συγκείμενα μεγέθη ἀνάλογον τὰ ΑΒ, ΒΕ, ΓΔ, ΔΖ, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ: λέγω, ὅτι καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΔΖ. εἰλήφθω γὰρ τῶν μὲν ΑΕ, ΕΒ, ΓΖ, ΖΔ ἰσάκις πολλαπλάσια τὰ ΗΘ, ΘΚ, ΛΜ, ΜΝ, τῶν δὲ ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ ΚΞ, ΝΠ. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΘΚ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΗΚ τοῦ ΑΒ. ἰσάκις δέ ἐστι πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΛΜ τοῦ ΓΖ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΜ τοῦ ΓΖ. πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΜΝ τοῦ ΖΔ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΛΝ τοῦ ΓΔ. ἰσάκις δὲ ἦν πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΗΚ τοῦ ΑΒ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΝ τοῦ ΓΔ. τὰ ΗΚ, ΛΝ ἄρα τῶν ΑΒ, ΓΔ ἰσάκις ἐστὶ πολλαπλάσια. πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΘΚ τοῦ ΕΒ καὶ τὸ ΜΝ τοῦ ΖΔ, ἔστι δὲ καὶ τὸ ΚΞ τοῦ ΕΒ ἰσάκις πολλαπλάσιον καὶ τὸ ΝΠ τοῦ ΖΔ, καὶ συντεθὲν τὸ ΘΞ τοῦ ΕΒ ἰσάκις ἐστὶ πολλαπλάσιον καὶ τὸ ΜΠ τοῦ ΖΔ. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, καὶ εἴληπται τῶν μὲν ΑΒ, ΓΔ ἰσάκις πολλαπλάσια τὰ ΗΚ, ΛΝ, τῶν δὲ ΕΒ, ΖΔ ἰσάκις πολλαπλάσια τὰ ΘΞ, ΜΠ, εἰ ἄρα ὑπερέχει τὸ ΗΚ τοῦ ΘΞ, ὑπερέχει καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. ὑπερεχέτω δὴ τὸ ΗΚ τοῦ ΘΞ, καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΘΚ ὑπερέχει ἄρα καὶ τὸ ΗΘ τοῦ ΚΞ. ἀλλὰ εἰ ὑπερεῖχε τὸ ΗΚ τοῦ ΘΞ, ὑπερεῖχε καὶ τὸ ΛΝ τοῦ ΜΠ: ὑπερέχει ἄρα καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΜΝ ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ: ὥστε εἰ ὑπερέχει τὸ ΗΘ τοῦ ΚΞ, ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ ΗΘ τῷ ΚΞ, ἴσον ἔσται καὶ τὸ ΛΜ τῷ ΝΠ, κἂν ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν ΗΘ, ΛΜ τῶν ΑΕ, ΓΖ ἰσάκις πολλαπλάσια, τὰ δὲ ΚΞ, ΝΠ τῶν ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. ἐὰν ἄρα συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 413|>","<|""VertexLabel"" -> ""5.18"", ""Text"" -> ""If magnitudes be proportional separando, they will also be proportional componendo."", ""TextWordCount"" -> 11, ""GreekText"" -> ""ἐὰν διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα ἀνάλογον ἔσται."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 5, ""Theorem"" -> 14}, {""Book"" -> 5, ""Theorem"" -> 17}}, ""Proof"" -> ""Let AE, EB, CF, FD be magnitudes proportional separando, so that, as AE is to EB, so is CF to FD; I say that they will also be proportional componendo, that is, as AB is to BE, so is CD to FD. For, if CD be not to DF as AB to BE, then, as AB is to BE, so will CD be either to some magnitude less than DF or to a greater. First, let it be in that ratio to a less magnitude DG. Then, since, as AB is to BE, so is CD to DG, they are magnitudes proportional componendo; so that they will also be proportional separando. [V. 17] Therefore, as AE is to EB, so is CG to GD. But also, by hypothesis, as AE is to EB, so is CF to FD. Therefore also, as CG is to GD, so is CF to FD. [V. 11] But the first CG is greater than the third CF; therefore the second GD is also greater than the fourth FD. [V. 14] But it is also less: which is impossible. Therefore, as AB is to BE, so is not CD to a less magnitude than FD. Similarly we can prove that neither is it in that ratio to a greater; it is therefore in that ratio to FD itself."", ""ProofWordCount"" -> 222, ""GreekProof"" -> ""ἔστω διῃρημένα μεγέθη ἀνάλογον τὰ ΑΕ, ΕΒ, ΓΖ, ΖΔ, ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ: λέγω, ὅτι καὶ συντεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΖΔ. εἰ γὰρ μή ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, ἔσται ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ ἤτοι πρὸς ἔλασσόν τι τοῦ ΔΖ ἢ πρὸς μεῖζον. ἔστω πρότερον πρὸς ἔλασσον τὸ ΔΗ. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΗ, συγκείμενα μεγέθη ἀνάλογόν ἐστιν: ὥστε καὶ διαιρεθέντα ἀνάλογον ἔσται. ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΗ πρὸς τὸ ΗΔ. ὑπόκειται δὲ καὶ ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. καὶ ὡς ἄρα τὸ ΓΗ πρὸς τὸ ΗΔ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. μεῖζον δὲ τὸ πρῶτον τὸ ΓΗ τοῦ τρίτου τοῦ ΓΖ: μεῖζον ἄρα καὶ τὸ δεύτερον τὸ ΗΔ τοῦ τετάρτου τοῦ ΖΔ. ἀλλὰ καὶ ἔλαττον: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα ἐστὶν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς ἔλασσον τοῦ ΖΔ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ πρὸς μεῖζον: πρὸς αὐτὸ ἄρα. ἐὰν ἄρα διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 213|>","<|""VertexLabel"" -> ""5.19"", ""Text"" -> ""If, as a whole is to a whole, so is a part subtracted to a part subtracted, the remainder will also be to the remainder as whole to whole."", ""TextWordCount"" -> 29, ""GreekText"" -> ""ἐὰν ᾖ ὡς ὅλον πρὸς ὅλον, οὕτως ἀφαιρεθὲν πρὸς ἀφαιρεθέν, καὶ τὸ λοιπὸν πρὸς τὸ λοιπὸν ἔσται ὡς ὅλον πρὸς ὅλον."", ""GreekTextWordCount"" -> 21, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 5, ""Theorem"" -> 16}, {""Book"" -> 5, ""Theorem"" -> 17}}, ""Proof"" -> ""For, as the whole AB is to the whole CD, so let the part AE subtracted be to the part CF subtracted; I say that the remainder EB will also be to the remainder FD as the whole AB to the whole CD. For since, as AB is to CD, so is AE to CF, alternately also, as BA is to AE, so is DC to CF. [V. 16] And, since the magnitudes are proportional componendo, they will also be proportional separando, [V. 17] that is, as BE is to EA, so is DF to CF, and, alternately, as BE is to DF, so is EA to FC. [V. 16] But, as AE is to CF, so by hypothesis is the whole AB to the whole CD. Therefore also the remainder EB will be to the remainder FD as the whole AB is to the whole CD. [V. 11]"", ""ProofWordCount"" -> 149, ""GreekProof"" -> ""ἔστω γὰρ ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ, οὕτως ἀφαιρεθὲν τὸ ΑΕ πρὸς ἀφαιρεθὲν τὸ ΓΖ: λέγω, ὅτι καὶ λοιπὸν τὸ ΕΒ πρὸς λοιπὸν τὸ ΖΔ ἔσται ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. ἐπεὶ γάρ ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΑΕ πρὸς τὸ ΓΖ, καὶ ἐναλλὰξ ὡς τὸ ΒΑ πρὸς τὸ ΑΕ, οὕτως τὸ ΔΓ πρὸς τὸ ΓΖ. καὶ ἐπεὶ συγκείμενα μεγέθη ἀνάλογόν ἐστιν, καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΒΕ πρὸς τὸ ΕΑ, οὕτως τὸ ΔΖ πρὸς τὸ ΓΖ: καὶ ἐναλλάξ, ὡς τὸ ΒΕ πρὸς τὸ ΔΖ, οὕτως τὸ ΕΑ πρὸς τὸ ΖΓ. ὡς δὲ τὸ ΑΕ πρὸς τὸ ΓΖ, οὕτως ὑπόκειται ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. καὶ λοιπὸν ἄρα τὸ ΕΒ πρὸς λοιπὸν τὸ ΖΔ ἔσται ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. ἐὰν ἄρα ᾖ ὡς ὅλον πρὸς ὅλον, οὕτως ἀφαιρεθὲν πρὸς ἀφαιρεθέν, καὶ τὸ λοιπὸν πρὸς τὸ λοιπὸν ἔσται ὡς ὅλον πρὸς ὅλον ὅπερ ἔδει δεῖξαι. Καὶ ἐπεὶ ἐδείχθη ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΕΒ πρὸς τὸ ΖΔ, καὶ ἐναλλὰξ ὡς τὸ ΑΒ πρὸς τὸ ΒΕ οὕτως τὸ ΓΔ πρὸς τὸ ΖΔ, συγκείμενα ἄρα μεγέθη ἀνάλογόν ἐστιν: ἐδείχθη δὲ ὡς τὸ ΒΑ πρὸς τὸ ΑΕ, οὕτως τὸ ΔΓ πρὸς τὸ ΓΖ: καί ἐστιν ἀναστρέψαντι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ ἀναστρέψαντι ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 230|>","<|""VertexLabel"" -> ""5.20"", ""Text"" -> ""If there be three magnitudes, and others equal to them in multitude, which taken two and two are in the same ratio, and if ex aequali the first be greater than the third, the fourth will also be greater than the sixth; if equal, equal; and, if less, less."", ""TextWordCount"" -> 49, ""GreekText"" -> ""ἐὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, δι᾽ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον."", ""GreekTextWordCount"" -> 40, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 8}, {""Book"" -> 5, ""Theorem"" -> 10}, {""Book"" -> 5, ""Theorem"" -> 13}}, ""Proof"" -> ""Let there be three magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two are in the same ratio, so that, as A is to B, so is D to E, and as B is to C, so is E to F; and let A be greater than C ex aequali; I say that D will also be greater than F; if A is equal to C, equal; and, if less, less. For, since A is greater than C, and B is some other magnitude, and the greater has to the same a greater ratio than the less has, [V. 8] therefore A has to B a greater ratio than C has to B. But, as A is to B, so is D to E, and, as C is to B, inversely, so is F to E; therefore D has also to E a greater ratio than F has to E. [V. 13] But, of magnitudes which have a ratio to the same, that which has a greater ratio is greater; [V. 10] therefore D is greater than F. Similarly we can prove that, if A be equal to C, D will also be equal to F; and if less, less."", ""ProofWordCount"" -> 210, ""GreekProof"" -> ""ἔστω τρία μεγέθη τὰ Α, Β, Γ, καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, ὡς δὲ τὸ Β πρὸς τὸ Γ, οὕτως τὸ Ε πρὸς τὸ Ζ, δι᾽ ἴσου δὲ μεῖζον ἔστω τὸ Α τοῦ Γ: λέγω, ὅτι καὶ τὸ Δ τοῦ Ζ μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. ἐπεὶ γὰρ μεῖζόν ἐστι τὸ Α τοῦ Γ, ἄλλο δέ τι τὸ Β, τὸ δὲ μεῖζον πρὸς τὸ αὐτὸ μείζονα λόγον ἔχει ἤπερ τὸ ἔλαττον, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. ἀλλ᾽ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, ὡς δὲ τὸ Γ πρὸς τὸ Β, ἀνάπαλιν οὕτως τὸ Ζ πρὸς τὸ Ε: καὶ τὸ Δ ἄρα πρὸς τὸ Ε μείζονα λόγον ἔχει ἤπερ τὸ Ζ πρὸς τὸ Ε. τῶν δὲ πρὸς τὸ αὐτὸ λόγον ἐχόντων τὸ μείζονα λόγον ἔχον μεῖζόν ἐστιν. μεῖζον ἄρα τὸ Δ τοῦ Ζ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Δ τῷ Ζ, κἂν ἔλαττον, ἔλαττον. ἐὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, δι᾽ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 243|>","<|""VertexLabel"" -> ""5.21"", ""Text"" -> ""If there be three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, then, if ex aequali the first magnitude is greater than the third, the fourth will also be greater than the sixth; if equal, equal; and if less, less."", ""TextWordCount"" -> 58, ""GreekText"" -> ""ἐὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, δι᾽ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον."", ""GreekTextWordCount"" -> 46, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 8}, {""Book"" -> 5, ""Theorem"" -> 10}, {""Book"" -> 5, ""Theorem"" -> 13}}, ""Proof"" -> ""Let there be three magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two are in the same ratio, and let the proportion of them be perturbed, so that, as A is to B, so is E to F, and, as B is to C, so is D to E, and let A be greater than C ex aequali; I say that D will also be greater than F; if A is equal to C, equal; and if less, less. For, since A is greater than C, and B is some other magnitude, therefore A has to B a greater ratio than C has to B. [V. 8] But, as A is to B, so is E to F, and, as C is to B, inversely, so is E to D. Therefore also E has to F a greater ratio than E has to D. [V. 13] But that to which the same has a greater ratio is less; [V. 10] therefore F is less than D; therefore D is greater than F. Similarly we can prove that, if A be equal to C, D will also be equal to F; and if less, less."", ""ProofWordCount"" -> 204, ""GreekProof"" -> ""ἔστω τρία μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ἔστω δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς δὲ τὸ Β πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ε, δι᾽ ἴσου δὲ τὸ Α τοῦ Γ μεῖζον ἔστω: λέγω, ὅτι καὶ τὸ Δ τοῦ Ζ μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. ἐπεὶ γὰρ μεῖζόν ἐστι τὸ Α τοῦ Γ, ἄλλο δέ τι τὸ Β, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. ἀλλ᾽ ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς δὲ τὸ Γ πρὸς τὸ Β, ἀνάπαλιν οὕτως τὸ Ε πρὸς τὸ Δ. καὶ τὸ Ε ἄρα πρὸς τὸ Ζ μείζονα λόγον ἔχει ἤπερ τὸ Ε πρὸς τὸ Δ. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλασσόν ἐστιν: ἔλασσον ἄρα ἐστὶ τὸ Ζ τοῦ Δ: μεῖζον ἄρα ἐστὶ τὸ Δ τοῦ Ζ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Δ τῷ Ζ, κἂν ἔλαττον, ἔλαττον. ἐὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, δι᾽ ἴσου δὲ τὸ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ τέταρτον τοῦ ἕκτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 250|>","<|""VertexLabel"" -> ""5.22"", ""Text"" -> ""If there be any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, they will also be in the same ratio ex aequali."", ""TextWordCount"" -> 36, ""GreekText"" -> ""ἐὰν ᾖ ὁποσαοῦν μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται."", ""GreekTextWordCount"" -> 26, ""References"" -> {{""Book"" -> 5, ""Definition"" -> 5}, {""Book"" -> 5, ""Theorem"" -> 4}, {""Book"" -> 5, ""Theorem"" -> 20}}, ""Proof"" -> ""Let there be any number of magnitudes A, B, C, and others D, E, F equal to them in multitude, which taken two and two together are in the same ratio, so that, as A is to B, so is D to E, and, as B is to C, so is E to F; I say that they will also be in the same ratio ex aequali, . For of A, D let equimultiples G, H be taken, and of B, E other, chance, equimultiples K, L; and, further, of C, F other, chance, equimultiples M, N. Then, since, as A is to B, so is D to E, and of A, D equimultiples G, H have been taken, and of B, E other, chance, equimultiples K, L, therefore, as G is to K, so is H to L. [V. 4] For the same reason also, as K is to M, so is L to N. Since, then, there are three magnitudes G, K, M, and others H, L, N equal to them in multitude, which taken two and two together are in the same ratio, therefore, ex aequali, if G is in excess of M, H is also in excess of N; if equal, equal; and if less, less. [V. 20] And G, H are equimultiples of A, D, and M, N other, chance, equimultiples of C, F. Therefore, as A is to C, so is D to F. [V. Def. 5]"", ""ProofWordCount"" -> 254, ""GreekProof"" -> ""ἔστω ὁποσαοῦν μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Δ, Ε, Ζ, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, ὡς δὲ τὸ Β πρὸς τὸ Γ, οὕτως τὸ Ε πρὸς τὸ Ζ: λέγω, ὅτι καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται. εἰλήφθω γὰρ τῶν μὲν Α, Δ ἰσάκις πολλαπλάσια τὰ η, Θ, τῶν δὲ Β, Ε ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Κ, Λ, καὶ ἔτι τῶν Γ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Μ, Ν. καὶ ἐπεί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Δ πρὸς τὸ Ε, καὶ εἴληπται τῶν μὲν Α, Δ ἰσάκις πολλαπλάσια τὰ η, Θ, τῶν δὲ Β, Ε ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Κ, Λ, ἔστιν ἄρα ὡς τὸ Η πρὸς τὸ Κ, οὕτως τὸ Θ πρὸς τὸ Λ. διὰ τὰ αὐτὰ δὴ καὶ ὡς τὸ Κ πρὸς τὸ Μ, οὕτως τὸ Λ πρὸς τὸ Ν. ἐπεὶ οὖν τρία μεγέθη ἐστὶ τὰ Η, Κ, Μ, καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Θ, Λ, Ν, σύνδυο λαμβανόμενα καὶ ἐν τῷ αὐτῷ λόγῳ, δι᾽ ἴσου ἄρα, εἰ ὑπερέχει τὸ Η τοῦ Μ, ὑπερέχει καὶ τὸ Θ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Η, Θ τῶν Α, Δ ἰσάκις πολλαπλάσια, τὰ δὲ Μ, Ν τῶν Γ, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια. ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ζ. ἐὰν ἄρα ᾖ ὁποσαοῦν μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος, σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 281|>","<|""VertexLabel"" -> ""5.23"", ""Text"" -> ""If there be three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, they will also be in the same ratio ex aequali."", ""TextWordCount"" -> 40, ""GreekText"" -> ""ἐὰν ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται."", ""GreekTextWordCount"" -> 31, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 5, ""Theorem"" -> 15}, {""Book"" -> 5, ""Theorem"" -> 16}, {""Book"" -> 5, ""Theorem"" -> 21}}, ""Proof"" -> ""Let there be three magnitudes A, B, C, and others equal to them in multitude, which, taken two and two together, are in the same proportion, namely D, E, F; and let the proportion of them be perturbed, so that, as A is to B, so is E to F, and, as B is to C, so is D to E; I say that, as A is to C, so is D to F. Of A, B, D let equimultiples G, H, K be taken, and of C, E, F other, chance, equimultiples L, M, N. Then, since G, H are equimultiples of A, B, and parts have the same ratio as the same multiples of them, [V. 15] therefore, as A is to B, so is G to H. For the same reason also, as E is to F, so is M to N. And, as A is to B, so is E to F; therefore also, as G is to H, so is M to N. [V. 11] Next, since, as B is to C, so is D to E, alternately, also, as B is to D, so is C to E. [V. 16] And, since H, K are equimultiples of B, D, and parts have the same ratio as their equimultiples, therefore, as B is to D, so is H to K. [V. 15] But, as B is to D, so is C to E; therefore also, as H is to K, so is C to E. [V. 11] Again, since L, M are equimultiples of C, E, therefore, as C is to E, so is L to M. [V. 15] But, as C is to E, so is H to K; therefore also, as H is to K, so is L to M, [V. 11]and, alternately, as H is to L, so is K to M. [V. 16] But it was also proved that, as G is to H, so is M to N. Since, then, there are three magnitudes G, H, L, and others equal to them in multitude K, M, N, which taken two and two together are in the same ratio, and the proportion of them is perturbed, therefore, ex aequali, if G is in excess of L, K is also in excess of N; if equal, equal; and if less, less. [V. 21] And G, K are equimultiples of A, D, and L, N of C, F. Therefore, as A is to C, so is D to F."", ""ProofWordCount"" -> 417, ""GreekProof"" -> ""ἔστω τρία μεγέθη τὰ Α, Β, Γ καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ τὰ Δ, Ε, Ζ, ἔστω δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, ὡς μὲν τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ, ὡς δὲ τὸ Β πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ε: λέγω, ὅτι ἐστὶν ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ζ. εἰλήφθω τῶν μὲν Α, Β, Δ ἰσάκις πολλαπλάσια τὰ Η, Θ, Κ, τῶν δὲ Γ, Ε, Ζ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ Λ, Μ, Ν. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσια τὰ Η, Θ τῶν Α, Β, τὰ δὲ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Η πρὸς τὸ Θ. διὰ τὰ αὐτὰ δὴ καὶ ὡς τὸ Ε πρὸς τὸ Ζ, οὕτως τὸ Μ πρὸς τὸ Ν: καί ἐστιν ὡς τὸ Α πρὸς τὸ Β, οὕτως τὸ Ε πρὸς τὸ Ζ: καὶ ὡς ἄρα τὸ Η πρὸς τὸ Θ, οὕτως τὸ Μ πρὸς τὸ Ν. καὶ ἐπεί ἐστιν ὡς τὸ Β πρὸς τὸ Γ, οὕτως τὸ δ πρὸς τὸ Ε, καὶ ἐναλλὰξ ὡς τὸ Β πρὸς τὸ Δ, οὕτως τὸ Γ πρὸς τὸ Ε. καὶ ἐπεὶ τὰ Θ, Κ τῶν Β, Δ ἰσάκις ἐστὶ πολλαπλάσια, τὰ δὲ μέρη τοῖς ἰσάκις πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ Β πρὸς τὸ Δ, οὕτως τὸ Θ πρὸς τὸ Κ. ἀλλ᾽ ὡς τὸ Β πρὸς τὸ Δ, οὕτως τὸ Γ πρὸς τὸ Ε: καὶ ὡς ἄρα τὸ Θ πρὸς τὸ Κ, οὕτως τὸ Γ πρὸς τὸ Ε. πάλιν, ἐπεὶ τὰ Λ, Μ τῶν Γ, Ε ἰσάκις ἐστι πολλαπλάσια, ἔστιν ἄρα ὡς τὸ Γ πρὸς τὸ Ε, οὕτως τὸ Λ πρὸς τὸ Μ. ἀλλ᾽ ὡς τὸ Γ πρὸς τὸ Ε, οὕτως τὸ Θ πρὸς τὸ Κ: καὶ ὡς ἄρα τὸ Θ πρὸς τὸ Κ, οὕτως τὸ Λ πρὸς τὸ Μ, καὶ ἐναλλὰξ ὡς τὸ Θ πρὸς τὸ Λ, τὸ Κ πρὸς τὸ Μ. ἐδείχθη δὲ καὶ ὡς τὸ Η πρὸς τὸ Θ, οὕτως τὸ Μ πρὸς τὸ Ν. ἐπεὶ οὖν τρία μεγέθη ἐστὶ τὰ Η, Θ, Λ, καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος τὰ Κ, Μ, Ν σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, καί ἐστιν αὐτῶν τεταραγμένη ἡ ἀναλογία, δι᾽ ἴσου ἄρα, εἰ ὑπερέχει τὸ Η τοῦ Λ, ὑπερέχει καὶ τὸ Κ τοῦ Ν, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν Η, Κ τῶν Α, Δ ἰσάκις πολλαπλάσια, τὰ δὲ Λ, Ν τῶν Γ, Ζ. ἔστιν ἄρα ὡς τὸ Α πρὸς τὸ Γ, οὕτως τὸ Δ πρὸς τὸ Ζ. ἐὰν ἄρα ᾖ τρία μεγέθη καὶ ἄλλα αὐτοῖς ἴσα τὸ πλῆθος σύνδυο λαμβανόμενα ἐν τῷ αὐτῷ λόγῳ, ᾖ δὲ τεταραγμένη αὐτῶν ἡ ἀναλογία, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 471|>","<|""VertexLabel"" -> ""5.24"", ""Text"" -> ""If a first magnitude have to a second the same ratio as a third has to a fourth, and also a fifth have to the second the same ratio as a sixth to the fourth, the first and fifth added together will have to the second the same ratio as the third and sixth have to the fourth."", ""TextWordCount"" -> 58, ""GreekText"" -> ""ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, ἔχῃ δὲ καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν λόγον καὶ ἕκτον πρὸς τέταρτον, καὶ συντεθὲν πρῶτον καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν ἕξει λόγον καὶ τρίτον καὶ ἕκτον πρὸς τέταρτον."", ""GreekTextWordCount"" -> 42, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 18}, {""Book"" -> 5, ""Theorem"" -> 22}}, ""Proof"" -> ""Let a first magnitude AB have to a second C the same ratio as a third DE has to a fourth F; and let also a fifth BG have to the second C the same ratio as a sixth EH has to the fourth F; I say that the first and fifth added together, AG, will have to the second C the same ratio as the third and sixth, DH, has to the fourth F. For since, as BG is to C, so is EH to F, inversely, as C is to BG, so is F to EH. Since, then, as AB is to C, so is DE to F, and, as C is to BG, so is F to EH, therefore, ex aequali, as AB is to BG, so is DE to EH. [V. 22] And, since the magnitudes are proportional separando, they will also be proportional componendo; [V. 18] therefore, as AG is to GB, so is DH to HE. But also, as BG is to C, so is EH to F; therefore, ex aequali, as AG is to C, so is DH to F. [V. 22]"", ""ProofWordCount"" -> 189, ""GreekProof"" -> ""πρῶτον γὰρ τὸ ΑΒ πρὸς δεύτερον τὸ Γ τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ ΔΕ πρὸς τέταρτον τὸ Ζ, ἐχέτω δὲ καὶ πέμπτον τὸ ΒΗ πρὸς δεύτερον τὸ Γ τὸν αὐτὸν λόγον καὶ ἕκτον τὸ ΕΘ πρὸς τέταρτον τὸ Ζ: λέγω, ὅτι καὶ συντεθὲν πρῶτον καὶ πέμπτον τὸ ΑΗ πρὸς δεύτερον τὸ Γ τὸν αὐτὸν ἕξει λόγον, καὶ τρίτον καὶ ἕκτον τὸ ΔΘ πρὸς τέταρτον τὸ Ζ. ἐπεὶ γάρ ἐστιν ὡς τὸ ΒΗ πρὸς τὸ Γ, οὕτως τὸ ΕΘ πρὸς τὸ Ζ, ἀνάπαλιν ἄρα ὡς τὸ Γ πρὸς τὸ ΒΗ, οὕτως τὸ Ζ πρὸς τὸ ΕΘ. ἐπεὶ οὖν ἐστιν ὡς τὸ ΑΒ πρὸς τὸ Γ, οὕτως τὸ ΔΕ πρὸς τὸ Ζ, ὡς δὲ τὸ Γ πρὸς τὸ ΒΗ, οὕτως τὸ Ζ πρὸς τὸ ΕΘ, δι᾽ ἴσου ἄρα ἐστὶν ὡς τὸ ΑΒ πρὸς τὸ ΒΗ, οὕτως τὸ ΔΕ πρὸς τὸ ΕΘ. καὶ ἐπεὶ διῃρημένα μεγέθη ἀνάλογόν ἐστιν, καὶ συντεθέντα ἀνάλογον ἔσται: ἔστιν ἄρα ὡς τὸ ΑΗ πρὸς τὸ ΗΒ, οὕτως τὸ ΔΘ πρὸς τὸ ΘΕ. ἔστι δὲ καὶ ὡς τὸ ΒΗ πρὸς τὸ Γ, οὕτως τὸ ΕΘ πρὸς τὸ Ζ: δι᾽ ἴσου ἄρα ἐστὶν ὡς τὸ ΑΗ πρὸς τὸ Γ, οὕτως τὸ ΔΘ πρὸς τὸ Ζ. ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, ἔχῃ δὲ καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν λόγον καὶ ἕκτον πρὸς τέταρτον, καὶ συντεθὲν πρῶτον καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν ἕξει λόγον καὶ τρίτον καὶ ἕκτον πρὸς τέταρτον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 244|>","<|""VertexLabel"" -> ""5.25"", ""Text"" -> ""If four magnitudes be proportional, the greatest and the least are greater than the remaining two."", ""TextWordCount"" -> 16, ""GreekText"" -> ""ἐὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ μέγιστον αὐτῶν καὶ τὸ ἐλάχιστον δύο τῶν λοιπῶν μείζονά ἐστιν."", ""GreekTextWordCount"" -> 16, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 19}}, ""Proof"" -> ""Let the four magnitudes AB, CD, E, F be proportional so that, as AB is to CD, so is E to F, and let AB be the greatest of them and F the least; I say that AB, F are greater than CD, E. For let AG be made equal to E, and CH equal to F. Since, as AB is to CD, so is E to F, and E is equal to AG, and F to CH, therefore, as AB is to CD, so is AG to CH. And since, as the whole AB is to the whole CD, so is the part AG subtracted to the part CH subtracted, the remainder GB will also be to the remainder HD as the whole AB is to the whole CD. [V. 19] But AB is greater than CD; therefore GB is also greater than HD. And, since AG is equal to E, and CH to F, therefore AG, F are equal to CH, E. And if, GB, HD being unequal, and GB greater, AG, F be added to GB and CH, E be added to HD, it follows that AB, F are greater than CD, E."", ""ProofWordCount"" -> 196, ""GreekProof"" -> ""ἔστω τέσσαρα μεγέθη ἀνάλογον τὰ ΑΒ, ΓΔ, Ε, Ζ, ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ Ε πρὸς τὸ Ζ, ἔστω δὲ μέγιστον μὲν αὐτῶν τὸ ΑΒ, ἐλάχιστον δὲ τὸ Ζ: λέγω, ὅτι τὰ ΑΒ, Ζ τῶν ΓΔ, Ε μείζονά ἐστιν. κείσθω γὰρ τῷ μὲν Ε ἴσον τὸ ΑΗ, τῷ δὲ Ζ ἴσον τὸ ΓΘ. ἐπεὶ οὖν ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ Ε πρὸς τὸ Ζ, ἴσον δὲ τὸ μὲν Ε τῷ ΑΗ, τὸ δὲ Ζ τῷ ΓΘ, ἔστιν ἄρα ὡς τὸ ΑΒ πρὸς τὸ ΓΔ, οὕτως τὸ ΑΗ πρὸς τὸ ΓΘ. καὶ ἐπεί ἐστιν ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ, οὕτως ἀφαιρεθὲν τὸ ΑΗ πρὸς ἀφαιρεθὲν τὸ ΓΘ, καὶ λοιπὸν ἄρα τὸ ΗΒ πρὸς λοιπὸν τὸ ΘΔ ἔσται ὡς ὅλον τὸ ΑΒ πρὸς ὅλον τὸ ΓΔ. μεῖζον δὲ τὸ ΑΒ τοῦ ΓΔ: μεῖζον ἄρα καὶ τὸ ΗΒ τοῦ ΘΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΗ τῷ Ε, τὸ δὲ ΓΘ τῷ Ζ, τὰ ἄρα ΑΗ, Ζ ἴσα ἐστὶ τοῖς ΓΘ, ε. καὶ ἐπεὶ ἐὰν ἀνίσοις ἴσα προστεθῇ, τὰ ὅλα ἄνισά ἐστιν, ἐὰν ἄρα τῶν ΗΒ, ΘΔ ἀνίσων ὄντων καὶ μείζονος τοῦ ΗΒ τῷ μὲν ΗΒ προστεθῇ τὰ ΑΗ, Ζ, τῷ δὲ ΘΔ προστεθῇ τὰ ΓΘ, Ε, συνάγεται τὰ ΑΒ, Ζ μείζονα τῶν ΓΔ, Ε. ἐὰν ἄρα τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ μέγιστον αὐτῶν καὶ τὸ ἐλάχιστον δύο τῶν λοιπῶν μείζονά ἐστιν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 233|>","<|""VertexLabel"" -> ""6.1"", ""Text"" -> ""Triangles and parallelograms which are under the same height are to one another as their bases."", ""TextWordCount"" -> 16, ""GreekText"" -> ""τὰ τρίγωνα καὶ τὰ παραλληλόγραμμα, τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις."", ""GreekTextWordCount"" -> 17, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 38}, {""Book"" -> 1, ""Theorem"" -> 41}, {""Book"" -> 5, ""Definition"" -> 5}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 5, ""Theorem"" -> 15}}, ""Proof"" -> ""Let ABC, ACD be triangles and EC, CF parallelograms under the same height;I say that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF. For let BD be produced in both directions to the points H, L and let [any number of straight lines] BG, GH bemade equal to the base BC, and any number of straight lines DK, KL equal to the base CD; let AG, AH, AK, AL be joined. Then, since CB, BG, GH are equal to one another, the triangles ABC, AGB, AHG are also equal to one another. [I. 38] Therefore, whatever multiple the base HC is of the base BC, that multiple also is the triangle AHC of the triangle ABC. For the same reason,whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD; and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38] if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ACL, and, if less, less. Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD, equimultiples have been taken of the base BC and thetriangle ABC, namely the base HC and the triangle AHC, and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC; and it has been proved that, if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ALC; if equal, equal; and, if less, less. Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. [V. Def. 5] Next, since the parallelogram EC is double of the triangleABC, [I. 41] and the parallelogram FC is double of the triangle ACD, while parts have the same ratio as the same multiples of them, [V. 15] therefore, as the triangle ABC is to the triangle ACD, so isthe parallelogram EC to the parallelogram FC. Since, then, it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD, and, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF,therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC. [V. 11]"", ""ProofWordCount"" -> 436, ""GreekProof"" -> ""ἔστω τρίγωνα μὲν τὰ ΑΒΓ, ΑΓΔ, παραλληλόγραμμα δὲ τὰ ΕΓ, ΓΖ ὑπὸ τὸ αὐτὸ ὕψος τὸ ΑΓ: λέγω, ὅτι ἐστὶν ὡς ἡ ΒΓ βάσις πρὸς τὴν ΓΔ βάσιν, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, καὶ τὸ ΕΓ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον. Ἐκβεβλήσθω γὰρ ἡ ΒΔ ἐφ᾽ ἑκάτερα τὰ μέρη ἐπὶ τὰ Θ, Λ σημεῖα, καὶ κείσθωσαν τῇ μὲν ΒΓ βάσει ἴσαι ὁσαιδηποτοῦν αἱ ΒΗ, ΗΘ, τῇ δὲ ΓΔ βάσει ἴσαι ὁσαιδηποτοῦν αἱ ΔΚ, ΚΛ, καὶ ἐπεζεύχθωσαν αἱ ΑΗ, ΑΘ, ΑΚ, ΑΛ. καὶ ἐπεὶ ἴσαι εἰσὶν αἱ ΓΒ, ΒΗ, ΗΘ ἀλλήλαις, ἴσα ἐστὶ καὶ τὰ ΑΘΗ, ΑΗΒ, ΑΒΓ τρίγωνα ἀλλήλοις. ὁσαπλασίων ἄρα ἐστὶν ἡ ΘΓ βάσις τῆς ΒΓ βάσεως, τοσαυταπλάσιόν ἐστι καὶ τὸ ΑΘΓ τρίγωνον τοῦ ΑΒΓ τριγώνου. διὰ τὰ αὐτὰ δὴ ὁσαπλασίων ἐστὶν ἡ ΛΓ βάσις τῆς ΓΔ βάσεως, τοσαυταπλάσιόν ἐστι καὶ τὸ ΑΛΓ τρίγωνον τοῦ ΑΓΔ τριγώνου: καὶ εἰ ἴση ἐστὶν ἡ ΘΓ βάσις τῇ ΓΛ βάσει, ἴσον ἐστὶ καὶ τὸ ΑΘΓ τρίγωνον τῷ ΑΓΛ τριγώνῳ, καὶ εἰ ὑπερέχει ἡ ΘΓ βάσις τῆς ΓΛ βάσεως, ὑπερέχει καὶ τὸ ΑΘΓ τρίγωνον τοῦ ΑΓΛ τριγώνου, καὶ εἰ ἐλάσσων, ἔλασσον. τεσσάρων δὴ ὄντων μεγεθῶν δύο μὲν βάσεων τῶν ΒΓ, ΓΔ, δύο δὲ τριγώνων τῶν ΑΒΓ, ΑΓΔ εἴληπται ἰσάκις πολλαπλάσια τῆς μὲν ΒΓ βάσεως καὶ τοῦ ΑΒΓ τριγώνου ἥ τε ΘΓ βάσις καὶ τὸ ΑΘΓ τρίγωνον, τῆς δὲ ΓΔ βάσεως καὶ τοῦ ΑΔΓ τριγώνου ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια ἥ τε ΛΓ βάσις καὶ τὸ ΑΛΓ τρίγωνον: καὶ δέδεικται, ὅτι, εἰ ὑπερέχει ἡ ΘΓ βάσις τῆς ΓΛ βάσεως, ὑπερέχει καὶ τὸ ΑΘΓ τρίγωνον τοῦ ΑΛΓ τριγώνου, καὶ εἰ ἴση, ἴσον, καὶ εἰ ἐλάσσων, ἔλασσον: ἔστιν ἄρα ὡς ἡ ΒΓ βάσις πρὸς τὴν ΓΔ βάσιν, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον. καὶ ἐπεὶ τοῦ μὲν ΑΒΓ τριγώνου διπλάσιόν ἐστι τὸ ΕΓ παραλληλόγραμμον, τοῦ δὲ ΑΓΔ τριγώνου διπλάσιόν ἐστι τὸ ΖΓ παραλληλόγραμμον, τὰ δὲ μέρη τοῖς ὡσαύτως πολλαπλασίοις τὸν αὐτὸν ἔχει λόγον, ἔστιν ἄρα ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, οὕτως τὸ ΕΓ παραλληλόγραμμον πρὸς τὸ ΖΓ παραλληλόγραμμον. ἐπεὶ οὖν ἐδείχθη, ὡς μὲν ἡ ΒΓ βάσις πρὸς τὴν ΓΔ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, ὡς δὲ τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΓΔ τρίγωνον, οὕτως τὸ ΕΓ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον, καὶ ὡς ἄρα ἡ ΒΓ βάσις πρὸς τὴν ΓΔ βάσιν, οὕτως τὸ ΕΓ παραλληλόγραμμον πρὸς τὸ ΖΓ παραλληλόγραμμον. τὰ ἄρα τρίγωνα καὶ τὰ παραλληλόγραμμα τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα πρὸς ἄλληλά ἐστιν ὡς αἱ βάσεις: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 406|>","<|""VertexLabel"" -> ""6.2"", ""Text"" -> ""If a straight line be drawn parallel to one of the sides of a triangle, it will cut the sides of the triangle proportionally; and, if the sides of the triangle be cut proportionally, the line joining the points of section will be parallel to the remaining side of the triangle."", ""TextWordCount"" -> 51, ""GreekText"" -> ""ἐὰν τριγώνου παρὰ μίαν τῶν πλευρῶν ἀχθῇ τις εὐθεῖα, ἀνάλογον τεμεῖ τὰς τοῦ τριγώνου πλευράς: καὶ ἐὰν αἱ τοῦ τριγώνου πλευραὶ ἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς τομὰς ἐπιζευγνυμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ τριγώνου πλευράν."", ""GreekTextWordCount"" -> 36, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 38}, {""Book"" -> 1, ""Theorem"" -> 39}, {""Book"" -> 5, ""Theorem"" -> 7}, {""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 1}}, ""Proof"" -> ""For let DE be drawn parallel to BC, one of the sides of the triangle ABC; I say that, as BD is to DA, so is CE to EA. For let BE, CD be joined. Therefore the triangle BDE is equal to the triangle CDE; for they are on the same base DE and in the same parallels DE, BC. [I. 38] And the triangle ADE is another area. But equals have the same ratio to the same; [V. 7] therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. But, as the triangle BDE is to ADE, so is BD to DA; for, being under the same height, the perpendicular drawn from E to AB, they are to one another as their bases. [VI. 1] For the same reason also, as the triangle CDE is to ADE, so is CE to EA. Therefore also, as BD is to DA, so is CE to EA. [V. 11] Again, let the sides AB, AC of the triangle ABC be cut proportionally, so that, as BD is to DA, so is CE to EA; and let DE be joined. I say that DE is parallel to BC. For, with the same construction, since, as BD is to DA, so is CE to EA, but, as BD is to DA, so is the triangle BDE to the triangle ADE, and, as CE is to EA, so is the triangle CDE to the triangle ADE, [VI. 1] therefore also, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. [V. 11] Therefore each of the triangles BDE, CDE has the same ratio to ADE. Therefore the triangle BDE is equal to the triangle CDE; [V. 9] and they are on the same base DE. But equal triangles which are on the same base are also in the same parallels. [I. 39] Therefore DE is parallel to BC."", ""ProofWordCount"" -> 330, ""GreekProof"" -> ""τριγώνου γὰρ τοῦ ΑΒΓ παράλληλος μιᾷ τῶν πλευρῶν τῇ ΒΓ ἤχθω ἡ ΔΕ: λέγω, ὅτι ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ. ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΔ. ἴσον ἄρα ἐστὶ ΒΔΕ τρίγωνον τῷ ΓΔΕ τριγώνῳ: ἐπὶ γὰρ τῆς αὐτῆς βάσεώς ἐστι τῆς ΔΕ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΔΕ, ΒΓ: ἄλλο δέ τι τὸ ΑΔΕ τρίγωνον. τὰ δὲ ἴσα πρὸς τὸ αὐτὸ τὸν αὐτὸν ἔχει λόγον: ἔστιν ἄρα ὡς τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, οὕτως τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον. ἀλλ᾽ ὡς μὲν τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ, οὕτως ἡ ΒΔ πρὸς τὴν ΔΑ: ὑπὸ γὰρ τὸ αὐτὸ ὕψος ὄντα τὴν ἀπὸ τοῦ Ε ἐπὶ τὴν ΑΒ κάθετον ἀγομένην πρὸς ἄλληλά εἰσιν ὡς αἱ βάσεις. διὰ τὰ αὐτὰ δὴ ὡς τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ: καὶ ὡς ἄρα ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ. ἀλλὰ δὴ αἱ τοῦ ΑΒΓ τριγώνου πλευραὶ αἱ ΑΒ, ΑΓ ἀνάλογον τετμήσθωσαν, ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ, καὶ ἐπεζεύχθω ἡ ΔΕ: λέγω, ὅτι παράλληλός ἐστιν ἡ ΔΕ τῇ ΒΓ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΑ, ἀλλ᾽ ὡς μὲν ἡ ΒΔ πρὸς τὴν ΔΑ, οὕτως τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, ὡς δὲ ἡ ΓΕ πρὸς τὴν ΕΑ, οὕτως τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, καὶ ὡς ἄρα τὸ ΒΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον, οὕτως τὸ ΓΔΕ τρίγωνον πρὸς τὸ ΑΔΕ τρίγωνον. ἑκάτερον ἄρα τῶν ΒΔΕ, ΓΔΕ τριγώνων πρὸς τὸ ΑΔΕ τὸν αὐτὸν ἔχει λόγον. ἴσον ἄρα ἐστὶ τὸ ΒΔΕ τρίγωνον τῷ ΓΔΕ τριγώνῳ: καί εἰσιν ἐπὶ τῆς αὐτῆς βάσεως τῆς ΔΕ. τὰ δὲ ἴσα τρίγωνα καὶ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. παράλληλος ἄρα ἐστὶν ἡ ΔΕ τῇ ΒΓ. ἐὰν ἄρα τριγώνου παρὰ μίαν τῶν πλευρῶν ἀχθῇ τις εὐθεῖα, ἀνάλογον τεμεῖ τὰς τοῦ τριγώνου πλευράς: καὶ ἐὰν αἱ τοῦ τριγώνου πλευραὶ ἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς τομὰς ἐπιζευγνυμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ τριγώνου πλευράν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 356|>","<|""VertexLabel"" -> ""6.3"", ""Text"" -> ""If an angle of a triangle be bisected and the straight line cutting the angle cut the base also, the segments of the base will have the same ratio as the remaining sides of the triangle; and, if the segments of the base have the same ratio as the remaining sides of the triangle, the straight line joined from the vertex to the point of section will bisect the angle of the triangle."", ""TextWordCount"" -> 73, ""GreekText"" -> ""ἐὰν τριγώνου ἡ γωνία δίχα τμηθῇ, ἡ δὲ τέμνουσα τὴν γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς: καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τεμεῖ τὴν τοῦ τριγώνου γωνίαν."", ""GreekTextWordCount"" -> 59, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 6}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 2}}, ""Proof"" -> ""Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; I say that, as BD is to CD, so is BA to AC. For let CE be drawn through C parallel to DA, and let BA be carried through and meet it at E. Then, since the straight line AC falls upon the parallels AD, EC, the angle ACE is equal to the angle CAD. [I. 29] But the angle CAD is by hypothesis equal to the angle BAD; therefore the angle BAD is also equal to the angle ACE. Again, since the straight line BAE falls upon the parallels AD, EC, the exterior angle BAD is equal to the interior angle AEC. [I. 29] But the angle ACE was also proved equal to the angle BAD; therefore the angle ACE is also equal to the angle AEC, so that the side AE is also equal to the side AC. [I. 6] And, since AD has been drawn parallel to EC, one of the sides of the triangle BCE, therefore, proportionally, as BD is to DC, so is BA to AE. But AE is equal to AC; [VI. 2] therefore, as BD is to DC, so is BA to AC. Again, let BA be to AC as BD to DC, and let AD be joined; I say that the angle BAC has been bisected by the straight line A.D. For, with the same construction, since, as BD is to DC, so is BA to AC, and also, as BD is to DC, so is BA to AE: for AD has been drawn parallel to EC, one of the sides of the triangle BCE: [VI. 2] therefore also, as BA is to AC, so is BA to AE. [V. 11] Therefore AC is equal to AE, [V. 9] so that the angle AEC is also equal to the angle ACE. [I. 5] But the angle AEC is equal to the exterior angle BAD, [I. 29] and the angle ACE is equal to the alternate angle CAD; [id.]therefore the angle BAD is also equal to the angle CAD. Therefore the angle BAC has been bisected by the straight line AD."", ""ProofWordCount"" -> 366, ""GreekProof"" -> ""ἔστω τρίγωνον τὸ ΑΒΓ, καὶ τετμήσθω ἡ ὑπὸ ΒΑΓ γωνία δίχα ὑπὸ τῆς ΑΔ εὐθείας: λέγω, ὅτι ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΓΔ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ. ἤχθω γὰρ διὰ τοῦ Γ τῇ ΔΑ παράλληλος ἡ ΓΕ καὶ διαχθεῖσα ἡ ΒΑ συμπιπτέτω αὐτῇ κατὰ τὸ Ε. καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΔ, ΕΓ εὐθεῖα ἐνέπεσεν ἡ ΑΓ, ἡ ἄρα ὑπὸ ΑΓΕ γωνία ἴση ἐστὶ τῇ ὑπὸ ΓΑΔ. ἀλλ᾽ ἡ ὑπὸ ΓΑΔ τῇ ὑπὸ ΒΑΔ ὑπόκειται ἴση: καὶ ἡ ὑπὸ ΒΑΔ ἄρα τῇ ὑπὸ ΑΓΕ ἐστιν ἴση. πάλιν, ἐπεὶ εἰς παραλλήλους τὰς ΑΔ, ΕΓ εὐθεῖα ἐνέπεσεν ἡ ΒΑΕ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΑΔ ἴση ἐστὶ τῇ ἐντὸς τῇ ὑπὸ ΑΕΓ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΔ ἴση: καὶ ἡ ὑπὸ ΑΓΕ ἄρα γωνία τῇ ὑπὸ ΑΕΓ ἐστιν ἴση: ὥστε καὶ πλευρὰ ἡ ΑΕ πλευρᾷ τῇ ΑΓ ἐστιν ἴση. καὶ ἐπεὶ τριγώνου τοῦ ΒΓΕ παρὰ μίαν τῶν πλευρῶν τὴν ΕΓ ἦκται ἡ ΑΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΕ. ἴση δὲ ἡ ΑΕ τῇ ΑΓ: ὡς ἄρα ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ. ἀλλὰ δὴ ἔστω ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ, καὶ ἐπεζεύχθω ἡ ΑΔ: λέγω, ὅτι δίχα τέτμηται ἡ ὑπὸ ΒΑΓ γωνία ὑπὸ τῆς ΑΔ εὐθείας. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεί ἐστιν ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΓ, ἀλλὰ καὶ ὡς ἡ ΒΔ πρὸς τὴν ΔΓ, οὕτως ἐστὶν ἡ ΒΑ πρὸς τὴν ΑΕ: τριγώνου γὰρ τοῦ ΒΓΕ παρὰ μίαν τὴν ΕΓ ἦκται ἡ ΑΔ: καὶ ὡς ἄρα ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΕ. ἴση ἄρα ἡ ΑΓ τῇ ΑΕ: ὥστε καὶ γωνία ἡ ὑπὸ ΑΕΓ τῇ ὑπὸ ΑΓΕ ἐστιν ἴση. ἀλλ᾽ ἡ μὲν ὑπὸ ΑΕΓ τῇ ἐκτὸς τῇ ὑπὸ ΒΑΔ ἐστιν ἴση, ἡ δὲ ὑπὸ ΑΓΕ τῇ ἐναλλὰξ τῇ ὑπὸ ΓΑΔ ἐστιν ἴση: καὶ ἡ ὑπὸ ΒΑΔ ἄρα τῇ ὑπὸ ΓΑΔ ἐστιν ἴση. ἡ ἄρα ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ὑπὸ τῆς ΑΔ εὐθείας. ἐὰν ἄρα τριγώνου ἡ γωνία δίχα τμηθῇ, ἡ δὲ τέμνουσα τὴν γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς: καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τέμνει τὴν τοῦ τριγώνου γωνίαν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 407|>","<|""VertexLabel"" -> ""6.4"", ""Text"" -> ""In equiangular triangles the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles."", ""TextWordCount"" -> 21, ""GreekText"" -> ""τῶν ἰσογωνίων τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι."", ""GreekTextWordCount"" -> 20, ""References"" -> {{""Postulate"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 17}, {""Book"" -> 1, ""Theorem"" -> 28}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 5, ""Theorem"" -> 16}, {""Book"" -> 5, ""Theorem"" -> 22}, {""Book"" -> 6, ""Theorem"" -> 2}}, ""Proof"" -> ""Let ABC, DCE be equiangular triangles having the angle ABC equal to the angle DCE, the angle BAC to the angle CDE, and further the angle ACB to the angle CED; I say that in the triangles ABC, DCE the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles. For let BC be placed in a straight line with CE. Then, since the angles ABC, ACB are less than two right angles, [I. 17] and the angle ACB is equal to the angle DEC, therefore the angles ABC, DEC are less than two right angles; therefore BA, ED, when produced, will meet. [I. Post. 5] Let them be produced and meet at F. Now, since the angle DCE is equal to the angle ABC, BF is parallel to CD. [I. 28] Again, since the angle ACB is equal to the angle DEC, AC is parallel to FE. [I. 28] Therefore FACD is a parallelogram; therefore FA is equal to DC, and AC to FD. [I. 34] And, since AC has been drawn parallel to FE, one side of the triangle FBE, therefore, as BA is to AF, so is BC to CE. [VI. 2] But AF is equal to CD; therefore, as BA is to CD, so is BC to CE, and alternately, as AB is to BC, so is DC to CE. [V. 16] Again, since CD is parallel to BF, therefore, as BC is to CE, so is FD to DE. [VI. 2] But FD is equal to AC; therefore, as BC is to CE, so is AC to DE, and alternately, as BC is to CA, so is CE to ED. [V. 16] Since then it was proved that, as AB is to BC, so is DC to CE, and, as BC is to CA, so is CE to ED; therefore, ex aequali, as BA is to AC, so is CD to DE. [V. 22]"", ""ProofWordCount"" -> 326, ""GreekProof"" -> ""ἔστω ἰσογώνια τρίγωνα τὰ ΑΒΓ, ΔΓΕ ἴσην ἔχοντα τὴν μὲν ὑπὸ ΑΒΓ γωνίαν τῇ ὑπὸ ΔΓΕ, τὴν δὲ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΕ καὶ ἔτι τὴν ὑπὸ ΑΓΒ τῇ ὑπὸ ΓΕΔ: λέγω, ὅτι τῶν ΑΒΓ, ΔΓΕ τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι. κείσθω γὰρ ἐπ᾽ εὐθείας ἡ ΒΓ τῇ ΓΕ. καὶ ἐπεὶ αἱ ὑπὸ ΑΒΓ, ΑΓΒ γωνίαι δύο ὀρθῶν ἐλάττονές εἰσιν, ἴση δὲ ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΕΓ, αἱ ἄρα ὑπὸ ΑΒΓ, ΔΕΓ δύο ὀρθῶν ἐλάττονές εἰσιν: αἱ ΒΑ, ΕΔ ἄρα ἐκβαλλόμεναι συμπεσοῦνται. ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν κατὰ τὸ Ζ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΓΕ γωνία τῇ ὑπὸ ΑΒΓ, παράλληλός ἐστιν ἡ ΒΖ τῇ ΓΔ. πάλιν, ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΕΓ, παράλληλός ἐστιν ἡ ΑΓ τῇ ΖΕ. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΑΓΔ: ἴση ἄρα ἡ μὲν ΖΑ τῇ ΔΓ, ἡ δὲ ΑΓ τῇ ΖΔ. καὶ ἐπεὶ τριγώνου τοῦ ΖΒΕ παρὰ μίαν τὴν ΖΕ ἦκται ἡ ΑΓ, ἔστιν ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΖ, οὕτως ἡ ΒΓ πρὸς τὴν ΓΕ. ἴση δὲ ἡ ΑΖ τῇ ΓΔ: ὡς ἄρα ἡ ΒΑ πρὸς τὴν ΓΔ, οὕτως ἡ ΒΓ πρὸς τὴν ΓΕ, καὶ ἐναλλὰξ ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΓ πρὸς τὴν ΓΕ. πάλιν, ἐπεὶ παράλληλός ἐστιν ἡ ΓΔ τῇ ΒΖ, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΕ, οὕτως ἡ ΖΔ πρὸς τὴν ΔΕ. ἴση δὲ ἡ ΖΔ τῇ ΑΓ: ὡς ἄρα ἡ ΒΓ πρὸς τὴν ΓΕ, οὕτως ἡ ΑΓ πρὸς τὴν ΔΕ, καὶ ἐναλλὰξ ὡς ἡ ΒΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΔ. ἐπεὶ οὖν ἐδείχθη ὡς μὲν ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΓ πρὸς τὴν ΓΕ, ὡς δὲ ἡ ΒΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΓΕ πρὸς τὴν ΕΔ, δι᾽ ἴσου ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΓΔ πρὸς τὴν ΔΕ. τῶν ἄρα ἰσογωνίων τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 336|>","<|""VertexLabel"" -> ""6.5"", ""Text"" -> ""If two triangles have their sides proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend."", ""TextWordCount"" -> 23, ""GreekText"" -> ""ἐὰν δύο τρίγωνα τὰς πλευρὰς ἀνάλογον ἔχῃ, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν."", ""GreekTextWordCount"" -> 23, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 8}, {""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 4}}, ""Proof"" -> ""Let ABC, DEF be two triangles having their sides proportional, so that, as AB is to BC, so is DE to EF, as BC is to CA, so is EF to FD, and further, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and they will have those angles equal which the corresponding sides subtend, namely the angle ABC to the angle DEF, the angle BCA to the angle EFD, and further the angle BAC to the angle EDF. For on the straight line EF, and at the points E, F on it, let there be constructed the angle FEG equal to the angle ABC, and the angle EFG equal to the angle ACB; [I. 23] therefore the remaining angle at A is equal to the remaining angle at G. [I. 32] Therefore the triangle ABC is equiangular with the triangle GEF. Therefore in the triangles ABC, GEF the sides about the equal angles are proportional, and those are corresponding sides which subtend the equal angles; [VI. 4] therefore, as AB is to BC, so is GE to EF. But, as AB is to BC, so by hypothesis is DE to EF; therefore, as DE is to EF, so is GE to EF. [V. 11] Therefore each of the straight lines DE, GE has the same ratio to EF; therefore DE is equal to GE. [V. 9] For the same reason DF is also equal to GF. Since then DE is equal to EG, and EF is common, the two sides DE, EF are equal to the two sides GE, EF; and the base DF is equal to the base FG; therefore the angle DEF is equal to the angle GEF, [I. 8] and the triangle DEF is equal to the triangle GEF, and the remaining angles are equal to the remaining angles, namely those which the equal sides subtend. [I. 4] Therefore the angle DFE is also equal to the angle GFE, and the angle EDF to the angle EGF. And, since the angle FED is equal to the angle GEF, while the angle GEF is equal to the angle ABC, therefore the angle ABC is also equal to the angle DEF. For the same reason the angle ACB is also equal to the angle DFE, and further, the angle at A to the angle at D; therefore the triangle ABC is equiangular with the triangle DEF."", ""ProofWordCount"" -> 411, ""GreekProof"" -> ""ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς πλευρὰς ἀνάλογον ἔχοντα, ὡς μὲν τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, ὡς δὲ τὴν ΒΓ πρὸς τὴν ΓΑ, οὕτως τὴν ΕΖ πρὸς τὴν ΖΔ, καὶ ἔτι ὡς τὴν ΒΑ πρὸς τὴν ΑΓ, οὕτως τὴν ΕΔ πρὸς τὴν ΔΖ. λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ καὶ ἴσας ἕξουσι τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν, τὴν μὲν ὑπὸ ΑΒΓ τῇ ὑπὸ ΔΕΖ, τὴν δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΕΖΔ καὶ ἔτι τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ. συνεστάτω γὰρ πρὸς τῇ ΕΖ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Ε, Ζ τῇ μὲν ὑπὸ ΑΒΓ γωνίᾳ ἴση ἡ ὑπὸ ΖΕΗ, τῇ δὲ ὑπὸ ΑΓΒ ἴση ἡ ὑπὸ ΕΖΗ: λοιπὴ ἄρα ἡ πρὸς τῷ Α λοιπῇ τῇ πρὸς τῷ Η ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΕΗΖ τριγώνῳ. τῶν ἄρα ΑΒΓ, ΕΗΖ τριγώνων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας καὶ ὁμόλογοι αἱ ὑπὸ τὰς ἴσας γωνίας ὑποτείνουσαι: ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΖ. ἀλλ᾽ ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ὑπόκειται ἡ ΔΕ πρὸς τὴν ΕΖ: ὡς ἄρα ἡ ΔΕ πρὸς τὴν ΕΖ, οὕτως ἡ ΗΕ πρὸς τὴν ΕΖ. ἑκατέρα ἄρα τῶν ΔΕ, ΗΕ πρὸς τὴν ΕΖ τὸν αὐτὸν ἔχει λόγον: ἴση ἄρα ἐστὶν ἡ ΔΕ τῇ ΗΕ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΔΖ τῇ ΗΖ ἐστιν ἴση. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΕ τῇ ΕΗ, κοινὴ δὲ ἡ ΕΖ, δύο δὴ αἱ ΔΕ, ΕΖ δυσὶ ταῖς ΗΕ, ΕΖ ἴσαι εἰσίν: καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΗ ἐστιν ἴση: γωνία ἄρα ἡ ὑπὸ ΔΕΖ γωνίᾳ τῇ ὑπὸ ΗΕΖ ἐστιν ἴση, καὶ τὸ ΔΕΖ τρίγωνον τῷ ΗΕΖ τριγώνῳ ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν. ἴση ἄρα ἐστὶ καὶ ἡ μὲν ὑπὸ ΔΖΕ γωνία τῇ ὑπὸ ΗΖΕ, ἡ δὲ ὑπὸ ΕΔΖ τῇ ὑπὸ ΕΗΖ. καὶ ἐπεὶ ἡ μὲν ὑπὸ ΖΕΔ τῇ ὑπὸ ΗΕΖ ἐστιν ἴση, ἀλλ᾽ ἡ ὑπὸ ΗΕΖ τῇ ὑπὸ ΑΒΓ, καὶ ἡ ὑπὸ ΑΒΓ ἄρα γωνία τῇ ὑπὸ ΔΕΖ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση, καὶ ἔτι ἡ πρὸς τῷ Α τῇ πρὸς τῷ Δ: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἐὰν ἄρα δύο τρίγωνα τὰς πλευρὰς ἀνάλογον ἔχῃ, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 415|>","<|""VertexLabel"" -> ""6.6"", ""Text"" -> ""If two triangles have one angle equal to one angle and the sides about the equal angles proportional, the triangles will be equiangular and will have those angles equal which the corresponding sides subtend."", ""TextWordCount"" -> 34, ""GreekText"" -> ""ἐὰν δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν."", ""GreekTextWordCount"" -> 33, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 4}, {""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 4}}, ""Proof"" -> ""Let ABC, DEF be two triangles having one angle BAC equal to one angle EDF and the sides about the equal angles proportional, so that, as BA is to AC, so is ED to DF; I say that the triangle ABC is equiangular with the triangle DEF, and will have the angle ABC equal to the angle DEF, and the angle ACB to the angle DFE. For on the straight line DF, and at the points D, F on it, let there be constructed the angle FDG equal to either of the angles BAC, EDF, and the angle DFG equal to the angle ACB; [I. 23] therefore the remaining angle at B is equal to the remaining angle at G. [I. 32] Therefore the triangle ABC is equiangular with the triangle DGF. Therefore, proportionally, as BA is to AC, so is GD to DF. [VI. 4] But, by hypothesis, as BA is to AC, so also is ED to DF; therefore also, as ED is to DF, so is GD to DF. [V. 11] Therefore ED is equal to DG; [V. 9] and DF is common; therefore the two sides ED, DF are equal to the two sides GD, DF; and the angle EDF is equal to the angle GDF; therefore the base EF is equal to the base GF, and the triangle DEF is equal to the triangle DGF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend. [I. 4] Therefore the angle DFG is equal to the angle DFE, and the angle DGF to the angle DEF. But the angle DFG is equal to the angle ACB; therefore the angle ACB is also equal to the angle DFE. And, by hypothesis, the angle BAC is also equal to the angle EDF; therefore the remaining angle at B is also equal to the remaining angle at E; [I. 32] therefore the triangle ABC is equiangular with the triangle DEF."", ""ProofWordCount"" -> 329, ""GreekProof"" -> ""ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ μίαν γωνίαν τὴν ὑπὸ ΒΑΓ μιᾷ γωνίᾳ τῇ ὑπὸ ΕΔΖ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΒΑ πρὸς τὴν ΑΓ, οὕτως τὴν ΕΔ πρὸς τὴν ΔΖ: λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ καὶ ἴσην ἕξει τὴν ὑπὸ ΑΒΓ γωνίαν τῇ ὑπὸ ΔΕΖ, τὴν δὲ ὑπὸ ΑΓΒ τῇ ὑπὸ ΔΖΕ. συνεστάτω γὰρ πρὸς τῇ ΔΖ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Δ, Ζ ὁποτέρᾳ μὲν τῶν ὑπὸ ΒΑΓ, ΕΔΖ ἴση ἡ ὑπὸ ΖΔΗ, τῇ δὲ ὑπὸ ΑΓΒ ἴση ἡ ὑπὸ ΔΖΗ: λοιπὴ ἄρα ἡ πρὸς τῷ Β γωνία λοιπῇ τῇ πρὸς τῷ Η ἴση ἐστίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΗΖ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΗΔ πρὸς τὴν ΔΖ. ὑπόκειται δὲ καὶ ὡς ἡ ΒΑ πρὸς τὴν ΑΓ, οὕτως ἡ ΕΔ πρὸς τὴν ΔΖ: καὶ ὡς ἄρα ἡ ΕΔ πρὸς τὴν ΔΖ, οὕτως ἡ ΗΔ πρὸς τὴν ΔΖ. ἴση ἄρα ἡ ΕΔ τῇ ΔΗ: καὶ κοινὴ ἡ ΔΖ: δύο δὴ αἱ ΕΔ, ΔΖ δυσὶ ταῖς ΗΔ, ΔΖ ἴσαι εἰσίν: καὶ γωνία ἡ ὑπὸ ΕΔΖ γωνίᾳ τῇ ὑπὸ ΗΔΖ ἐστιν ἴση: βάσις ἄρα ἡ ΕΖ βάσει τῇ ΗΖ ἐστιν ἴση, καὶ τὸ ΔΕΖ τρίγωνον τῷ ΗΔΖ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφ᾽ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν. ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΔΖΗ τῇ ὑπὸ ΔΖΕ, ἡ δὲ ὑπὸ ΔΗΖ τῇ ὑπὸ ΔΕΖ. ἀλλ᾽ ἡ ὑπὸ ΔΖΗ τῇ ὑπὸ ΑΓΒ ἐστιν ἴση: καὶ ἡ ὑπὸ ΑΓΒ ἄρα τῇ ὑπὸ ΔΖΕ ἐστιν ἴση. ὑπόκειται δὲ καὶ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ ἴση: καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Β λοιπῇ τῇ πρὸς τῷ Ε ἴση ἐστίν: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἐὰν ἄρα δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, ὑφ᾽ ἃς αἱ ὁμόλογοι πλευραὶ ὑποτείνουσιν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 337|>","<|""VertexLabel"" -> ""6.7"", ""Text"" -> ""If two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional."", ""TextWordCount"" -> 48, ""GreekText"" -> ""ἐὰν δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ ἄλλας γωνίας τὰς πλευρὰς ἀνάλογον, τῶν δὲ λοιπῶν ἑκατέραν ἅμα ἤτοι ἐλάσσονα ἢ μὴ ἐλάσσονα ὀρθῆς, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, περὶ ἃς ἀνάλογόν εἰσιν αἱ πλευραί."", ""GreekTextWordCount"" -> 42, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 5}, {""Book"" -> 1, ""Theorem"" -> 13}, {""Book"" -> 1, ""Theorem"" -> 17}, {""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 4}}, ""Proof"" -> ""Let ABC, DEF be two triangles having one angle equal to one angle, the angle BAC to the angle EDF, the sides about other angles ABC, DEF proportional, so that, as AB is to BC, so is DE to EF, and, first, each of the remaining angles at C, F less than a right angle; I say that the triangle ABC is equiangular with the triangle DEF, the angle ABC will be equal to the angle DEF, and the remaining angle, namely the angle at C, equal to the remaining angle, the angle at F. For, if the angle ABC is unequal to the angle DEF, one of them is greater. Let the angle ABC be greater; and on the straight line AB, and at the point B on it, let the angle ABG be constructed equal to the angle DEF. [I. 23] Then, since the angle A is equal to D, and the angle ABG to the angle DEF, therefore the remaining angle AGB is equal to the remaining angle DFE. [I. 32] Therefore the triangle ABG is equiangular with the triangle DEF. Therefore, as AB is to BG, so is DE to EF [VI. 4] But, as DE is to EF, so by hypothesis is AB to BC; therefore AB has the same ratio to each of the straight lines BC, BG; [V. 11] therefore BC is equal to BG, [V. 9]so that the angle at C is also equal to the angle BGC. [I. 5] But, by hypothesis, the angle at C is less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle. [I. 13] And it was proved equal to the angle at F; therefore the angle at F is also greater than a right angle. But it is by hypothesis less than a right angle: which is absurd. Therefore the angle ABC is not unequal to the angle DEF; therefore it is equal to it. But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32] Therefore the triangle ABC is equiangular with the triangle DEF. But, again, let each of the angles at C, F be supposed not less than a right angle; I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF. For, with the same construction, we can prove similarly that BC is equal to BG; so that the angle at C is also equal to the angle BGC. [I. 5] But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle. Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [I. 17] Therefore, once more, the angle ABC is not unequal to the angle DEF; therefore it is equal to it. But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32] Therefore the triangle ABC is equiangular with the triangle DEF."", ""ProofWordCount"" -> 546, ""GreekProof"" -> ""ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχοντα τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΕΔΖ, περὶ δὲ ἄλλας γωνίας τὰς ὑπὸ ΑΒΓ, ΔΕΖ τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, τῶν δὲ λοιπῶν τῶν πρὸς τοῖς Γ, Ζ πρότερον ἑκατέραν ἅμα ἐλάσσονα ὀρθῆς: λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ, καὶ ἴση ἔσται ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ, καὶ λοιπὴ δηλονότι ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση. εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ὑπὸ ΑΒΓ. καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Β τῇ ὑπὸ ΔΕΖ γωνίᾳ ἴση ἡ ὑπὸ ΑΒΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν Α γωνία τῇ Δ, ἡ δὲ ὑπὸ ΑΒΗ τῇ ὑπὸ ΔΕΖ, λοιπὴ ἄρα ἡ ὑπὸ ΑΗΒ λοιπῇ τῇ ὑπὸ ΔΖΕ ἐστιν ἴση. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΒΗ, οὕτως ἡ ΔΕ πρὸς τὴν ΕΖ. ὡς δὲ ἡ ΔΕ πρὸς τὴν ΕΖ, οὕτως ὑπόκειται ἡ ΑΒ πρὸς τὴν ΒΓ: ἡ ΑΒ ἄρα πρὸς ἑκατέραν τῶν ΒΓ, ΒΗ τὸν αὐτὸν ἔχει λόγον: ἴση ἄρα ἡ ΒΓ τῇ ΒΗ. ὥστε καὶ γωνία ἡ πρὸς τῷ Γ γωνίᾳ τῇ ὑπὸ ΒΗΓ ἐστιν ἴση. ἐλάττων δὲ ὀρθῆς ὑπόκειται ἡ πρὸς τῷ Γ: ἐλάττων ἄρα ἐστὶν ὀρθῆς καὶ ἡ ὑπὸ ΒΗΓ: ὥστε ἡ ἐφεξῆς αὐτῇ γωνία ἡ ὑπὸ ΑΗΒ μείζων ἐστὶν ὀρθῆς. καὶ ἐδείχθη ἴση οὖσα τῇ πρὸς τῷ Ζ: καὶ ἡ πρὸς τῷ Ζ ἄρα μείζων ἐστὶν ὀρθῆς. ὑπόκειται δὲ ἐλάσσων ὀρθῆς: ὅπερ ἐστὶν ἄτοπον. οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ: ἴση ἄρα. ἔστι δὲ καὶ ἡ πρὸς τῷ Α ἴση τῇ πρὸς τῷ Δ: καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση ἐστίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἀλλὰ δὴ πάλιν ὑποκείσθω ἑκατέρα τῶν πρὸς τοῖς Γ, Ζ μὴ ἐλάσσων ὀρθῆς: λέγω πάλιν, ὅτι καὶ οὕτως ἐστὶν ἰσογώνιον τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἴση ἐστὶν ἡ ΒΓ τῇ ΒΗ: ὥστε καὶ γωνία ἡ πρὸς τῷ Γ τῇ ὑπὸ ΒΗΓ ἴση ἐστίν. οὐκ ἐλάττων δὲ ὀρθῆς ἡ πρὸς τῷ Γ: οὐκ ἐλάττων ἄρα ὀρθῆς οὐδὲ ἡ ὑπὸ ΒΗΓ. τριγώνου δὴ τοῦ ΒΗΓ αἱ δύο γωνίαι δύο ὀρθῶν οὔκ εἰσιν ἐλάττονες: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα πάλιν ἄνισός ἐστιν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΕΖ: ἴση ἄρα. ἔστι δὲ καὶ ἡ πρὸς τῷ Α τῇ πρὸς τῷ Δ ἴση: λοιπὴ ἄρα ἡ πρὸς τῷ Γ λοιπῇ τῇ πρὸς τῷ Ζ ἴση ἐστίν. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. ἐὰν ἄρα δύο τρίγωνα μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχῃ, περὶ δὲ ἄλλας γωνίας τὰς πλευρὰς ἀνάλογον, τῶν δὲ λοιπῶν ἑκατέραν ἅμα ἐλάττονα ἢ μὴ ἐλάττονα ὀρθῆς, ἰσογώνια ἔσται τὰ τρίγωνα καὶ ἴσας ἕξει τὰς γωνίας, περὶ ἃς ἀνάλογόν εἰσιν αἱ πλευραί: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 498|>","<|""VertexLabel"" -> ""6.8"", ""Text"" -> ""If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another."", ""TextWordCount"" -> 31, ""GreekText"" -> ""ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, τὰ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις."", ""GreekTextWordCount"" -> 25, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 6, ""Definition"" -> 1}, {""Book"" -> 6, ""Theorem"" -> 4}}, ""Proof"" -> ""Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC; I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another. For, since the angle BAC is equal to the angle ADB, for each is right, and the angle at B is common to the two triangles ABC and ABD, therefore the remaining angle ACB is equal to the remaining angle BAD; [I. 32] therefore the triangle ABC is equiangular with the triangle ABD. Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD, so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD, and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4] Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional. Therefore the triangle ABC is similar to the triangle ABD. [VI. Def. 1] Similarly we can prove that the triangle ABC is also similar to the triangle ADC; therefore each of the triangles ABD, ADC is similar to the whole ABC. I say next that the triangles ABD, ADC are also similar to one another. For, since the right angle BDA is equal to the right angle ADC, and moreover the angle BAD was also proved equal to the angle at C, therefore the remaining angle at B is also equal to the remaining angle DAC; [I. 32] therefore the triangle ABD is equiangular with the triangle ADC. Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B, and so also is BA to AC, these sides subtending the right angles; [VI. 4] therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1]"", ""ProofWordCount"" -> 384, ""GreekProof"" -> ""ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν, καὶ ἤχθω ἀπὸ τοῦ Α ἐπὶ τὴν ΒΓ κάθετος ἡ ΑΔ: λέγω, ὅτι ὅμοιόν ἐστιν ἑκάτερον τῶν ΑΒΔ, ΑΔΓ τριγώνων ὅλῳ τῷ ΑΒΓ καὶ ἔτι ἀλλήλοις. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΑΔΒ: ὀρθὴ γὰρ ἑκατέρα: καὶ κοινὴ τῶν δύο τριγώνων τοῦ τε ΑΒΓ καὶ τοῦ ΑΒΔ ἡ πρὸς τῷ Β, λοιπὴ ἄρα ἡ ὑπὸ ΑΓΒ λοιπῇ τῇ ὑπὸ ΒΑΔ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΒΔ τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΒΓ ὑποτείνουσα τὴν ὀρθὴν τοῦ ΑΒΓ τριγώνου πρὸς τὴν ΒΑ ὑποτείνουσαν τὴν ὀρθὴν τοῦ ΑΒΔ τριγώνου, οὕτως αὐτὴ ἡ ΑΒ ὑποτείνουσα τὴν πρὸς τῷ Γ γωνίαν τοῦ ΑΒΓ τριγώνου πρὸς τὴν ΒΔ ὑποτείνουσαν τὴν ἴσην τὴν ὑπὸ ΒΑΔ τοῦ ΑΒΔ τριγώνου, καὶ ἔτι ἡ ΑΓ πρὸς τὴν ΑΔ ὑποτείνουσαν τὴν πρὸς τῷ Β γωνίαν κοινὴν τῶν δύο τριγώνων. τὸ ΑΒΓ ἄρα τρίγωνον τῷ ΑΒΔ τριγώνῳ ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. ὅμοιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΒΔ τριγώνῳ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τῷ ΑΔΓ τριγώνῳ ὅμοιόν ἐστι τὸ ΑΒΓ τρίγωνον: ἑκάτερον ἄρα τῶν ΑΒΔ, ΑΔΓ τριγώνων ὅμοιόν ἐστιν ὅλῳ τῷ ΑΒΓ. λέγω δή, ὅτι καὶ ἀλλήλοις ἐστὶν ὅμοια τὰ ΑΒΔ, ΑΔΓ τρίγωνα. ἐπεὶ γὰρ ὀρθὴ ἡ ὑπὸ ΒΔΑ ὀρθῇ τῇ ὑπὸ ΑΔΓ ἐστιν ἴση, ἀλλὰ μὴν καὶ ἡ ὑπὸ ΒΑΔ τῇ πρὸς τῷ Γ ἐδείχθη ἴση, καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Β λοιπῇ τῇ ὑπὸ ΔΑΓ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΑΔΓ τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΒΔ τοῦ ΑΒΔ τριγώνου ὑποτείνουσα τὴν ὑπὸ ΒΑΔ πρὸς τὴν ΔΑ τοῦ ΑΔΓ τριγώνου ὑποτείνουσαν τὴν πρὸς τῷ Γ ἴσην τῇ ὑπὸ ΒΑΔ, οὕτως αὐτὴ ἡ ΑΔ τοῦ ΑΒΔ τριγώνου ὑποτείνουσα τὴν πρὸς τῷ Β γωνίαν πρὸς τὴν ΔΓ ὑποτείνουσαν τὴν ὑπὸ ΔΑΓ τοῦ ΑΔΓ τριγώνου ἴσην τῇ πρὸς τῷ Β, καὶ ἔτι ἡ ΒΑ πρὸς τὴν ΑΓ ὑποτείνουσαι τὰς ὀρθάς: ὅμοιον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΑΔΓ τριγώνῳ. ἐὰν ἄρα ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, τὰ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, ἡ ἀχθεῖσα τῶν τῆς βάσεως τμημάτων μέση ἀνάλογόν ἐστιν: ὅπερ ἔδει δεῖξαι καὶ ἔτι τῆς βάσεως καὶ ἑνὸς ὁποιουοῦν τῶν τμημάτων ἡ πρὸς τῷ τμήματι πλευρὰ μέση ἀνάλογόν ἐστιν."", ""GreekProofWordCount"" -> 406|>","<|""VertexLabel"" -> ""6.9"", ""Text"" -> ""From a given straight line to cut off a prescribed part."", ""TextWordCount"" -> 11, ""GreekText"" -> ""τῆς δοθείσης εὐθείας τὸ προσταχθὲν μέρος ἀφελεῖν."", ""GreekTextWordCount"" -> 7, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 6, ""Theorem"" -> 2}}, ""Proof"" -> ""Let AB be the given straight line; thus it is required to cut off from AB a prescribed part. Let the third part be that prescribed. Let a straight line AC be drawn through from A containing with AB any angle; let a point D be taken at random on AC, and let DE, EC be made equal to AD. [I. 3] Let BC be joined, and through D let DF be drawn parallel to it. [I. 31] Then, since FD has been drawn parallel to BC, one of the sides of the triangle ABC, therefore, proportionally, as CD is to DA, so is BF to FA. [VI. 2] But CD is double of DA; therefore BF is also double of FA; therefore BA is triple of AF. Therefore from the given straight line AB the prescribed third part AF has been cut off."", ""ProofWordCount"" -> 144, ""GreekProof"" -> ""ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ: δεῖ δὴ τῆς ΑΒ τὸ προσταχθὲν μέρος ἀφελεῖν. Ἐπιτετάχθω δὴ τὸ τρίτον. καὶ διήχθω τις ἀπὸ τοῦ α εὐθεῖα ἡ ΑΓ γωνίαν περιέχουσα μετὰ τῆς ΑΒ τυχοῦσαν: καὶ εἰλήφθω τυχὸν σημεῖον ἐπὶ τῆς ΑΓ τὸ Δ, καὶ κείσθωσαν τῇ ΑΔ ἴσαι αἱ ΔΕ, ΕΓ. καὶ ἐπεζεύχθω ἡ ΒΓ, καὶ διὰ τοῦ Δ παράλληλος αὐτῇ ἤχθω ἡ ΔΖ. ἐπεὶ οὖν τριγώνου τοῦ ΑΒΓ παρὰ μίαν τῶν πλευρῶν τὴν ΒΓ ἦκται ἡ ΖΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΒΖ πρὸς τὴν ΖΑ. διπλῆ δὲ ἡ ΓΔ τῆς ΔΑ: διπλῆ ἄρα καὶ ἡ ΒΖ τῆς ΖΑ: τριπλῆ ἄρα ἡ ΒΑ τῆς ΑΖ. τῆς ἄρα δοθείσης εὐθείας τῆς ΑΒ τὸ ἐπιταχθὲν τρίτον μέρος ἀφῄρηται τὸ ΑΖ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 127|>","<|""VertexLabel"" -> ""6.10"", ""Text"" -> ""To cut a given uncut straight line similarly to a given cut straight line."", ""TextWordCount"" -> 14, ""GreekText"" -> ""τὴν δοθεῖσαν εὐθεῖαν ἄτμητον τῇ δοθείσῃ τετμημένῃ ὁμοίως τεμεῖν."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 1, ""Theorem"" -> 34}, {""Book"" -> 6, ""Theorem"" -> 2}}, ""Proof"" -> ""Let AB be the given uncut straight line, and AC the straight line cut at the points D, E; and let them be so placed as to contain any angle; let CB be joined, and through D, E let DF, EG be drawn parallel to BC, and through D let DHK be drawn parallel to AB. [I. 31] Therefore each of the figures FH, HB is a parallelogram; therefore DH is equal to FG and HK to GB. [I. 34] Now, since the straight line HE has been drawn parallel to KC, one of the sides of the triangle DKC, therefore, proportionally, as CE is to ED, so is KH to HD. [VI. 2] But KH is equal to BG, and HD to GF; therefore, as CE is to ED, so is BG to GF. Again, since FD has been drawn parallel to GE, one of the sides of the triangle AGE, therefore, proportionally, as ED is to DA, so is GF to FA. [VI. 2] But it was also proved that, as CE is to ED, so is BG to GF; therefore, as CE is to ED, so is BG to GF, and, as ED is to DA, so is GF to FA. Therefore the given uncut straight line AB has been cut similarly to the given cut straight line AC."", ""ProofWordCount"" -> 222, ""GreekProof"" -> ""ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἄτμητος ἡ ΑΒ, ἡ δὲ τετμημένη ἡ ΑΓ κατὰ τὰ Δ, Ε σημεῖα, καὶ κείσθωσαν ὥστε γωνίαν τυχοῦσαν περιέχειν, καὶ ἐπεζεύχθω ἡ ΓΒ, καὶ διὰ τῶν Δ, Ε τῇ ΒΓ παράλληλοι ἤχθωσαν αἱ ΔΖ, ΕΗ, διὰ δὲ τοῦ Δ τῇ ΑΒ παράλληλος ἤχθω ἡ ΔΘΚ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΖΘ, ΘΒ: ἴση ἄρα ἡ μὲν ΔΘ τῇ ΖΗ, ἡ δὲ ΘΚ τῇ ΗΒ. καὶ ἐπεὶ τριγώνου τοῦ ΔΚΓ παρὰ μίαν τῶν πλευρῶν τὴν ΚΓ εὐθεῖα ἦκται ἡ ΘΕ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΚΘ πρὸς τὴν ΘΔ. ἴση δὲ ἡ μὲν ΚΘ τῇ ΒΗ, ἡ δὲ ΘΔ τῇ ΗΖ. ἔστιν ἄρα ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ. πάλιν, ἐπεὶ τριγώνου τοῦ ΑΗΕ παρὰ μίαν τῶν πλευρῶν τὴν ΗΕ ἦκται ἡ ΖΔ, ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΕΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ. ἐδείχθη δὲ καὶ ὡς ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ: ἔστιν ἄρα ὡς μὲν ἡ ΓΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΒΗ πρὸς τὴν ΗΖ, ὡς δὲ ἡ ΕΔ πρὸς τὴν ΔΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ. ἡ ἄρα δοθεῖσα εὐθεῖα ἄτμητος ἡ ΑΒ τῇ δοθείσῃ εὐθείᾳ τετμημένῃ τῇ ΑΓ ὁμοίως τέτμηται: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 215|>","<|""VertexLabel"" -> ""6.11"", ""Text"" -> ""To two given straight lines to find a third proportional."", ""TextWordCount"" -> 10, ""GreekText"" -> ""δύο δοθεισῶν εὐθειῶν τρίτην ἀνάλογον προσευρεῖν."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 3}, {""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 6, ""Theorem"" -> 2}}, ""Proof"" -> ""Let BA, AC be the two given straight lines, and let them be placed so as to contain any angle; thus it is required to find a third proportional to BA, AC. For let them be produced to the points D, E, and let BD be made equal to AC; [I. 3] let BC be joined, and through D let DE be drawn parallel to it. [I. 31] Since, then, BC has been drawn parallel to DE, one of the sides of the triangle ADE, proportionally, as AB is to BD, so is AC to CE. [VI. 2] But BD is equal to AC; therefore, as AB is to AC, so is AC to CE. Therefore to two given straight lines AB, AC a third proportional to them, CE, has been found."", ""ProofWordCount"" -> 132, ""GreekProof"" -> ""ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΒΑ, ΑΓ καὶ κείσθωσαν γωνίαν περιέχουσαι τυχοῦσαν. δεῖ δὴ τῶν ΒΑ, ΑΓ τρίτην ἀνάλογον προσευρεῖν. ἐκβεβλήσθωσαν γὰρ ἐπὶ τὰ Δ, Ε σημεῖα, καὶ κείσθω τῇ ΑΓ ἴση ἡ ΒΔ, καὶ ἐπεζεύχθω ἡ ΒΓ, καὶ διὰ τοῦ Δ παράλληλος αὐτῇ ἤχθω ἡ ΔΕ. ἐπεὶ οὖν τριγώνου τοῦ ΑΔΕ παρὰ μίαν τῶν πλευρῶν τὴν ΔΕ ἦκται ἡ ΒΓ, ἀνάλογόν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΔ, οὕτως ἡ ΑΓ πρὸς τὴν ΓΕ. ἴση δὲ ἡ ΒΔ τῇ ΑΓ. ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΑΓ, οὕτως ἡ ΑΓ πρὸς τὴν ΓΕ. δύο ἄρα δοθεισῶν εὐθειῶν τῶν ΑΒ, ΑΓ τρίτη ἀνάλογον αὐταῖς προσεύρηται ἡ ΓΕ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 112|>","<|""VertexLabel"" -> ""6.12"", ""Text"" -> ""To three given straight lines to find a fourth proportional."", ""TextWordCount"" -> 10, ""GreekText"" -> ""τριῶν δοθεισῶν εὐθειῶν τετάρτην ἀνάλογον προσευρεῖν."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 6, ""Theorem"" -> 2}}, ""Proof"" -> ""Let A, B, C be the three given straight lines; thus it is required to find a fourth proportional to A, B, C. Let two straight lines DE, DF be set out containing any angle EDF; let DG be made equal to A, GE equal to B, and further DH equal to C; let GH be joined, and let EF be drawn through E parallel to it. [I. 31] Since, then, GH has been drawn parallel to EF, one of the sides of the triangle DEF, therefore, as DG is to GE, so is DH to HF. [VI. 2] But DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF. Therefore to the three given straight lines A, B, C a fourth proportional HF has been found."", ""ProofWordCount"" -> 140, ""GreekProof"" -> ""ἔστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ: δεῖ δὴ τῶν Α, Β, Γ τετάρτην ἀνάλογον προσευρεῖν. Ἐκκείσθωσαν δύο εὐθεῖαι αἱ ΔΕ, ΔΖ γωνίαν περιέχουσαι τυχοῦσαν τὴν ὑπὸ ΕΔΖ: καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΗ, τῇ δὲ Β ἴση ἡ ΗΕ, καὶ ἔτι τῇ Γ ἴση ἡ ΔΘ: καὶ ἐπιζευχθείσης τῆς ΗΘ παράλληλος αὐτῇ ἤχθω διὰ τοῦ Ε ἡ ΕΖ. ἐπεὶ οὖν τριγώνου τοῦ ΔΕΖ παρὰ μίαν τὴν ΕΖ ἦκται ἡ ΗΘ, ἔστιν ἄρα ὡς ἡ ΔΗ πρὸς τὴν ΗΕ, οὕτως ἡ ΔΘ πρὸς τὴν ΘΖ. ἴση δὲ ἡ μὲν ΔΗ τῇ Α, ἡ δὲ ΗΕ τῇ Β, ἡ δὲ ΔΘ τῇ Γ: ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν ΘΖ. τριῶν ἄρα δοθεισῶν εὐθειῶν τῶν Α, Β, Γ τετάρτη ἀνάλογον προσεύρηται ἡ ΘΖ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 136|>","<|""VertexLabel"" -> ""6.13"", ""Text"" -> ""To two given straight lines to find a mean proportional."", ""TextWordCount"" -> 10, ""GreekText"" -> ""δύο δοθεισῶν εὐθειῶν μέσην ἀνάλογον προσευρεῖν."", ""GreekTextWordCount"" -> 6, ""References"" -> {{""Book"" -> 3, ""Theorem"" -> 31}, {""Book"" -> 6, ""Theorem"" -> 8}}, ""Proof"" -> ""Let AB, BC be the two given straight lines; thus it is required to find a mean proportional to AB, BC. Let them be placed in a straight line, and let the semicircle ADC be described on AC; let BD be drawn from the point B at right angles to the straight line AC, and let AD, DC be joined. Since the angle ADC is an angle in a semicircle, it is right. [III. 31] And, since, in the right-angled triangle ADC, DB has been drawn from the right angle perpendicular to the base, therefore DB is a mean proportional between the segments of the base, AB, BC. [VI. 8] Therefore to the two given straight lines AB, BC a mean proportional DB has been found."", ""ProofWordCount"" -> 126, ""GreekProof"" -> ""ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΒ, ΒΓ: δεῖ δὴ τῶν ΑΒ, ΒΓ μέσην ἀνάλογον προσευρεῖν. κείσθωσαν ἐπ᾽ εὐθείας, καὶ γεγράφθω ἐπὶ τῆς ΑΓ ἡμικύκλιον τὸ ΑΔΓ, καὶ ἤχθω ἀπὸ τοῦ Β σημείου τῇ ΑΓ εὐθείᾳ πρὸς ὀρθὰς ἡ ΒΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ. ἐπεὶ ἐν ἡμικυκλίῳ γωνία ἐστὶν ἡ ὑπὸ ΑΔΓ, ὀρθή ἐστιν. καὶ ἐπεὶ ἐν ὀρθογωνίῳ τριγώνῳ τῷ ΑΔΓ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἦκται ἡ ΔΒ, ἡ ΔΒ ἄρα τῶν τῆς βάσεως τμημάτων τῶν ΑΒ, ΒΓ μέση ἀνάλογόν ἐστιν. δύο ἄρα δοθεισῶν εὐθειῶν τῶν ΑΒ, ΒΓ μέση ἀνάλογον προσεύρηται ἡ ΔΒ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 102|>","<|""VertexLabel"" -> ""6.14"", ""Text"" -> ""In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal."", ""TextWordCount"" -> 30, ""GreekText"" -> ""τῶν ἴσων τε καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: καὶ ὧν ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα."", ""GreekTextWordCount"" -> 29, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 14}, {""Book"" -> 5, ""Theorem"" -> 7}, {""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 1}}, ""Proof"" -> ""Let AB, BC be equal and equiangular parallelograms having the angles at B equal, and let DB, BE be placed in a straight line; therefore FB, BG are also in a straight line. [I. 14] I say that, in AB, BC, the sides about the equal angles are reciprocally proportional, that is to say, that, as DB is to BE, so is GB to BF. For let the parallelogram FE be completed. Since, then, the parallelogram AB is equal to the parallelogram BC, and FE is another area, therefore, as AB is to FE, so is BC to FE. [V. 7] But, as AB is to FE, so is DB to BE, [VI. 1] and, as BC is to FE, so is GB to BF. [id.]therefore also, as DB is to BE, so is GB to BF. [V. 11] Therefore in the parallelograms AB, BC the sides about the equal angles are reciprocally proportional. Next, let GB be to BF as DB to BE; I say that the parallelogram AB is equal to the parallelogram BC. For since, as DB is to BE, so is GB to BF, while, as DB is to BE, so is the parallelogram AB to the parallelogram FE, [VI. 1] and, as GB is to BF, so is the parallelogram BC to the parallelogram FE, [VI. 1] therefore also, as AB is to FE, so is BC to FE; [V. 11] therefore the parallelogram AB is equal to the parallelogram BC. [V. 9]"", ""ProofWordCount"" -> 249, ""GreekProof"" -> ""ἔστω ἴσα τε καὶ ἰσογώνια παραλληλόγραμμα τὰ ΑΒ, ΒΓ ἴσας ἔχοντα τὰς πρὸς τῷ Β γωνίας, καὶ κείσθωσαν ἐπ᾽ εὐθείας αἱ ΔΒ, ΒΕ: ἐπ᾽ εὐθείας ἄρα εἰσὶ καὶ αἱ ΖΒ, ΒΗ. λέγω, ὅτι τῶν ΑΒ, ΒΓ ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, τουτέστιν, ὅτι ἐστὶν ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ. συμπεπληρώσθω γὰρ τὸ ΖΕ παραλληλόγραμμον. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ, ἄλλο δέ τι τὸ ΖΕ, ἔστιν ἄρα ὡς τὸ ΑΒ πρὸς τὸ ΖΕ, οὕτως τὸ ΒΓ πρὸς τὸ ΖΕ. ἀλλ᾽ ὡς μὲν τὸ ΑΒ πρὸς τὸ ΖΕ, οὕτως ἡ ΔΒ πρὸς τὴν ΒΕ, ὡς δὲ τὸ ΒΓ πρὸς τὸ ΖΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ: καὶ ὡς ἄρα ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ. τῶν ἄρα ΑΒ, ΒΓ παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ἀλλὰ δὴ ἔστω ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ. ἐπεὶ γάρ ἐστιν ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ, ἀλλ᾽ ὡς μὲν ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως τὸ ΑΒ παραλληλόγραμμον πρὸς τὸ ΖΕ παραλληλόγραμμον, ὡς δὲ ἡ ΗΒ πρὸς τὴν ΒΖ, οὕτως τὸ ΒΓ παραλληλόγραμμον πρὸς τὸ ΖΕ παραλληλόγραμμον, καὶ ὡς ἄρα τὸ ΑΒ πρὸς τὸ ΖΕ, οὕτως τὸ ΒΓ πρὸς τὸ ΖΕ: ἴσον ἄρα ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ. τῶν ἄρα ἴσων τε καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: καὶ ὧν ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 278|>","<|""VertexLabel"" -> ""6.15"", ""Text"" -> ""In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal."", ""TextWordCount"" -> 45, ""GreekText"" -> ""τῶν ἴσων καὶ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: καὶ ὧν μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα."", ""GreekTextWordCount"" -> 36, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 14}, {""Book"" -> 5, ""Theorem"" -> 7}, {""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 1}}, ""Proof"" -> ""Let ABC, ADE be equal triangles having one angle equal to one angle, namely the angle BAC to the angle DAE; I say that in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional, that is to say, that, as CA is to AD, so is EA to AB. For let them be placed so that CA is in a straight line with AD; therefore EA is also in a straight line with AB. [I. 14] Let BD be joined. Since then the triangle ABC is equal to the triangle ADE, and BAD is another area, therefore, as the triangle CAB is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 7] But, as CAB is to BAD, so is CA to AD, [VI. 1] and, as EAD is to BAD, so is EA to AB. [id.] Therefore also, as CA is to AD, so is EA to AB. [V. 11] Therefore in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional. Next, let the sides of the triangles ABC, ADE be reciprocally proportional, that is to say, let EA be to AB as CA to AD; I say that the triangle ABC is equal to the triangle ADE. For, if BD be again joined, since, as CA is to AD, so is EA to AB, while, as CA is to AD, so is the triangle ABC to the triangle BAD, and, as EA is to AB, so is the triangle EAD to the triangle BAD, [VI. 1] therefore, as the triangle ABC is to the triangle BAD, so is the triangle EAD to the triangle BAD. [V. 11] Therefore each of the triangles ABC, EAD has the same ratio to BAD. Therefore the triangle ABC is equal to the triangle EAD. [V. 9]"", ""ProofWordCount"" -> 307, ""GreekProof"" -> ""ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΑΔΕ μίαν μιᾷ ἴσην ἔχοντα γωνίαν τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΔΑΕ: λέγω, ὅτι τῶν ΑΒΓ, ΑΔΕ τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, τουτέστιν, ὅτι ἐστὶν ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ. κείσθω γὰρ ὥστε ἐπ᾽ εὐθείας εἶναι τὴν ΓΑ τῇ ΑΔ: ἐπ᾽ εὐθείας ἄρα ἐστὶ καὶ ἡ ΕΑ τῇ ΑΒ. καὶ ἐπεζεύχθω ἡ ΒΔ. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΔΕ τριγώνῳ, ἄλλο δέ τι τὸ ΒΑΔ, ἔστιν ἄρα ὡς τὸ ΓΑΒ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον. ἀλλ᾽ ὡς μὲν τὸ ΓΑΒ πρὸς τὸ ΒΑΔ, οὕτως ἡ ΓΑ πρὸς τὴν ΑΔ, ὡς δὲ τὸ ΕΑΔ πρὸς τὸ ΒΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ. καὶ ὡς ἄρα ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ. τῶν ΑΒΓ, ΑΔΕ ἄρα τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ἀλλὰ δὴ ἀντιπεπονθέτωσαν αἱ πλευραὶ τῶν ΑΒΓ, ΑΔΕ τριγώνων, καὶ ἔστω ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΔΕ τριγώνῳ. Ἐπιζευχθείσης γὰρ πάλιν τῆς ΒΔ, ἐπεί ἐστιν ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ, ἀλλ᾽ ὡς μὲν ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, ὡς δὲ ἡ ΕΑ πρὸς τὴν ΑΒ, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον. ἑκάτερον ἄρα τῶν ΑΒΓ, ΕΑΔ πρὸς τὸ ΒΑΔ τὸν αὐτὸν ἔχει λόγον. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΕΑΔ τριγώνῳ. τῶν ἄρα ἴσων καὶ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: καὶ ὧν μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἐκεῖνα ἴσα ἐστίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 321|>","<|""VertexLabel"" -> ""6.16"", ""Text"" -> ""If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines will be proportional."", ""TextWordCount"" -> 45, ""GreekText"" -> ""ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ: κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ, αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται."", ""GreekTextWordCount"" -> 39, ""References"" -> {{""Book"" -> 6, ""Theorem"" -> 14}}, ""Proof"" -> ""Let the four straight lines AB, CD, E, F be proportional, so that, as AB is to CD, so is E to F; I say that the rectangle contained by AB, F is equal to the rectangle contained by CD, E. Let AG, CH be drawn from the points A, C at right angles to the straight lines AB, CD, and let AG be made equal to F, and CH equal to E. Let the parallelograms BG, DH be completed. Then since, as AB is to CD, so is E to F, while E is equal to CH, and F to AG, therefore, as AB is to CD, so is CH to AG. Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional. But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [VI. 14] therefore the parallelogram BG is equal to the parallelogram DH. And BG is the rectangle AB, F, for AG is equal to F; and DH is the rectangle CD, E, for E is equal to CH; therefore the rectangle contained by AB, F is equal to the rectangle contained by CD, E. Next, let the rectangle contained by AB, F be equal to the rectangle contained by CD, E; I say that the four straight lines will be proportional, so that, as AB is to CD, so is E to F. For, with the same construction, since the rectangle AB, F is equal to the rectangle CD, E, and the rectangle AB, F is BG, for AG is equal to F, and the rectangle CD, E is DH, for CH is equal to E, therefore BG is equal to DH. And they are equiangular But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [VI. 14] Therefore, as AB is to CD, so is CH to AG. But CH is equal to E, and AG to F; therefore, as AB is to CD, so is E to F."", ""ProofWordCount"" -> 340, ""GreekProof"" -> ""Ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, Ε, Ζ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ: λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ. Ἤχθωσαν [γὰρ] ἀπὸ τῶν Α, Γ σημείων ταῖς ΑΒ, ΓΔ εὐθείαις πρὸς ὀρθὰς αἱ ΑΗ, ΓΘ, καὶ κείσθω τῇ μὲν Ζ ἴση ἡ ΑΓ, τῇ δὲ Ε ἴση ἡ ΓΘ. καὶ συμπεπληρώσθω τὰ ΒΗ, ΔΘ παραλληλόγραμμα.Καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ, ἴση δὲ ἡ μὲν Ε τῇ ΓΘ, ἡ δὲ Ζ τῇ ΑΗ, ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΘ πρὸς τὴν ΑΗ. τῶν ΒΗ, ΔΘ ἄρα παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ὧν δὲ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα: ἴσον ἄρα ἐστὶ τὸ ΒΗ παραλληλόγραμμον τῷ ΔΘ παραλληλογράμμῳ. καί ἐστι τὸ μὲν ΒΗ τὸ ὑπὸ τῶν ΑΒ, Ζ: ἴση γὰρ ἡ ΑΗ τῇ Ζ: τὸ δὲ ΔΘ τὸ ὑπὸ τῶν ΓΔ, Ε: ἴση γὰρ ἡ Ε τῇ ΓΘ: τὸ ἄρα ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ. Ἀλλὰ δὴ τὸ ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἔστω τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ: λέγω, ὅτι αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ. Τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ τὸ ὑπὸ τῶν ΑΒ, Ζ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε, καί ἐστι τὸ μὲν ὑπὸ τῶν ΑΒ, Ζ τὸ ΒΗ: ἴση γάρ ἐστιν ἡ ΑΗ τῇ Ζ: τὸ δὲ ὑπὸ τῶν ΓΔ, Ε τὸ ΔΘ: ἴση γὰρ ἡ ΓΘ τῇ Ε: τὸ ἄρα ΒΗ ἴσον ἐστὶ τῷ ΔΘ. καί ἐστιν ἰσογώνια. τῶν δὲ ἴσων καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΘ πρὸς τὴν ΑΗ. ἴση δὲ ἡ μὲν ΓΘ τῇ Ε, ἡ δὲ ΑΗ τῇ Ζ: ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ. Ἐὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ: κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ, αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 391|>","<|""VertexLabel"" -> ""6.17"", ""Text"" -> ""If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes be equal to the square on the mean, the three straight lines will be proportional. "", ""TextWordCount"" -> 43, ""GreekText"" -> ""ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης τετραγώνῳ: κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι ἀνάλογον ἔσονται."", ""GreekTextWordCount"" -> 37, ""References"" -> {{""Book"" -> 6, ""Theorem"" -> 16}}, ""Proof"" -> ""Let the three straight lines A, B, C be proportional, so that, as A is to B, so is B to C; I say that the rectangle contained by A, C is equal to the square on B. Let D be made equal to B. Then, since, as A is to B, so is B to C, and B is equal to D, therefore, as A is to B, so is D to C. But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means.[VI. 16] Therefore the rectangle A, C is equal to the rectangle B, D. But the rectangle B, D is the square on B, for B is equal to D; therefore the rectangle contained by A, C is equal to the square on B. Next, let the rectangle A, C be equal to the square on B; I say that, as A is to B, so is B to C. For, with the same construction, since the rectangle A, C is equal to the square on B, while the square on B is the rectangle B, D, for B is equal to D, therefore the rectangle A, C is equal to the rectangle B, D. But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [VI. 16] Therefore, as A is to B, so is D to C. But B is equal to D; therefore, as A is to B, so is B to C."", ""ProofWordCount"" -> 263, ""GreekProof"" -> ""ἔστωσαν τρεῖς εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ: λέγω, ὅτι τὸ ὑπὸ τῶν Α, Γ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς β τετραγώνῳ. κείσθω τῇ Β ἴση ἡ Δ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ, ἴση δὲ ἡ Β τῇ Δ, ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, ἡ Δ πρὸς τὴν Γ. ἐὰν δὲ τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ. τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ὑπὸ τῶν Β, Δ. ἀλλὰ τὸ ὑπὸ τῶν Β, Δ τὸ ἀπὸ τῆς Β ἐστιν: ἴση γὰρ ἡ Β τῇ Δ: τὸ ἄρα ὑπὸ τῶν Α, Γ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς Β τετραγώνῳ. ἀλλὰ δὴ τὸ ὑπὸ τῶν Α, Γ ἴσον ἔστω τῷ ἀπὸ τῆς Β: λέγω, ὅτι ἐστὶν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ τὸ ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ἀπὸ τῆς Β, ἀλλὰ τὸ ἀπὸ τῆς Β τὸ ὑπὸ τῶν Β, Δ ἐστιν: ἴση γὰρ ἡ Β τῇ Δ: τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ὑπὸ τῶν Β, Δ. ἐὰν δὲ τὸ ὑπὸ τῶν ἄκρων ἴσον ᾖ τῷ ὑπὸ τῶν μέσων, αἱ τέσσαρες εὐθεῖαι ἀνάλογόν εἰσιν. ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Δ πρὸς τὴν Γ. ἴση δὲ ἡ Β τῇ Δ: ὡς ἄρα ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ. ἐὰν ἄρα τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης τετραγώνῳ: κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι ἀνάλογον ἔσονται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 306|>","<|""VertexLabel"" -> ""6.18"", ""Text"" -> ""On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure."", ""TextWordCount"" -> 19, ""GreekText"" -> ""ἀπὸ τῆς δοθείσης εὐθείας τῷ δοθέντι εὐθυγράμμῳ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγράψαι."", ""GreekTextWordCount"" -> 14, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 23}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 6, ""Definition"" -> 1}, {""Book"" -> 6, ""Theorem"" -> 4}}, ""Proof"" -> ""Let AB be the given straight line and CE the given rectilineal figure; thus it is required to describe on the straight line AB a rectilineal figure similar and similarly situated to the rectilineal figure CE. Let DF be joined, and on the straight line AB, and at the points A, B on it, let the angle GAB be constructed equal to the angle at C, and the angle ABG equal to the angle CDF. [I. 23] Therefore the remaining angle CFD is equal to the angle AGB; [I. 32] therefore the triangle FCD is equiangular with the triangle GAB. Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB. Again, on the straight line BG, and at the points B, G on it, let the angle BGH be constructed equal to the angle DFE, and the angle GBH equal to the angle FDE. [I. 23] Therefore the remaining angle at E is equal to the remaining angle at H; [I. 32] therefore the triangle FDE is equiangular with the triangle GBH; therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [VI. 4] But it was also proved that, as FD is to GB, so is FC to GA, and CD to AB; therefore also, as FC is to AG, so is CD to AB, and FE to GH, and further ED to HB. And, since the angle CFD is equal to the angle AGB, and the angle DFE to the angle BGH, therefore the whole angle CFE is equal to the whole angle AGH. For the same reason the angle CDE is also equal to the angle ABH. And the angle at C is also equal to the angle at A, and the angle at E to the angle at H. Therefore AH is equiangular with CE; and they have the sides about their equal angles proportional; therefore the rectilineal figure AH is similar to the rectilineal figure CE. [VI. Def. 1] Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE."", ""ProofWordCount"" -> 360, ""GreekProof"" -> ""ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν εὐθύγραμμον τὸ ΓΕ: δεῖ δὴ ἀπὸ τῆς ΑΒ εὐθείας τῷ ΓΕ εὐθυγράμμῳ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγράψαι. ἐπεζεύχθω ἡ ΔΖ, καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Α, Β τῇ μὲν πρὸς τῷ Γ γωνίᾳ ἴση ἡ ὑπὸ ΗΑΒ, τῇ δὲ ὑπὸ ΓΔΖ ἴση ἡ ὑπὸ ΑΒΗ. λοιπὴ ἄρα ἡ ὑπὸ ΓΖΔ τῇ ὑπὸ ΑΗΒ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΖΓΔ τρίγωνον τῷ ΗΑΒ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΓ πρὸς τὴν ΗΑ, καὶ ἡ ΓΔ πρὸς τὴν ΑΒ. πάλιν συνεστάτω πρὸς τῇ ΒΗ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Β, Η τῇ μὲν ὑπὸ ΔΖΕ γωνίᾳ ἴση ἡ ὑπὸ ΒΗΘ, τῇ δὲ ὑπὸ ΖΔΕ ἴση ἡ ὑπὸ ΗΒΘ. λοιπὴ ἄρα ἡ πρὸς τῷ Ε λοιπῇ τῇ πρὸς τῷ Θ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΖΔΕ τρίγωνον τῷ ΗΘΒ τριγώνῳ: ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΕ πρὸς τὴν ΗΘ καὶ ἡ ΕΔ πρὸς τὴν ΘΒ. ἐδείχθη δὲ καὶ ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΓ πρὸς τὴν ΗΑ καὶ ἡ ΓΔ πρὸς τὴν ΑΒ: καὶ ὡς ἄρα ἡ ΖΓ πρὸς τὴν ΑΗ, οὕτως ἥ τε ΓΔ πρὸς τὴν ΑΒ καὶ ἡ ΖΕ πρὸς τὴν ΗΘ καὶ ἔτι ἡ ΕΔ πρὸς τὴν ΘΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΓΖΔ γωνία τῇ ὑπὸ ΑΗΒ, ἡ δὲ ὑπὸ ΔΖΕ τῇ ὑπὸ ΒΗΘ, ὅλη ἄρα ἡ ὑπὸ ΓΖΕ ὅλῃ τῇ ὑπὸ ΑΗΘ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΒΘ ἐστιν ἴση. ἔστι δὲ καὶ ἡ μὲν πρὸς τῷ Γ τῇ πρὸς τῷ Α ἴση, ἡ δὲ πρὸς τῷ Ε τῇ πρὸς τῷ Θ. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΘ τῷ ΓΕ: καὶ τὰς περὶ τὰς ἴσας γωνίας αὐτῶν πλευρὰς ἀνάλογον ἔχει: ὅμοιον ἄρα ἐστὶ τὸ ΑΘ εὐθύγραμμον τῷ ΓΕ εὐθυγράμμῳ. ἀπὸ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ ΓΕ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγέγραπται τὸ ΑΘ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 342|>","<|""VertexLabel"" -> ""6.19"", ""Text"" -> ""Similar triangles are to one another in the duplicate ratio of the corresponding sides."", ""TextWordCount"" -> 14, ""GreekText"" -> ""τὰ ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Book"" -> 5, ""Definition"" -> 9}, {""Book"" -> 5, ""Definition"" -> 11}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 5, ""Theorem"" -> 16}, {""Book"" -> 6, ""Theorem"" -> 1}, {""Book"" -> 6, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 15}}, ""Proof"" -> ""Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, sois DE to EF, so that BC corresponds to EF; [V. Def. 11] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF. For let a third proportional BG be taken to BC, EF, so that, as BC is to EF, so is EF to BG; [VI. 11]and let AG be joined. Since then, as AB is to BC, so is DE to EF, therefore, alternately, as AB is to DE, so is BC to EF. [V. 16] But, as BC is to EF, so is EF to BG; therefore also, as AB is to DE, so is EF to BG. [V. 11] Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional. But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [VI. 15]therefore the triangle ABG is equal to the triangle DEF. Now since, as BC is to EF, so is EF to BG, and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [V. Def. 9] therefore BC has to BG a ratio duplicate of that which CB has to EF. But, as CB is to BG, so is the triangle ABC to the triangle ABG; [VI. 1] therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF. But the triangle ABG is equal to the triangle DEF; therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF."", ""ProofWordCount"" -> 313, ""GreekProof"" -> ""ἔστω ὅμοια τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἴσην ἔχοντα τὴν πρὸς τῷ Β γωνίαν τῇ πρὸς τῷ Ε, ὡς δὲ τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, ὥστε ὁμόλογον εἶναι τὴν ΒΓ τῇ ΕΖ: λέγω, ὅτι τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. εἰλήφθω γὰρ τῶν ΒΓ, ΕΖ τρίτη ἀνάλογον ἡ ΒΗ, ὥστε εἶναι ὡς τὴν ΒΓ πρὸς τὴν ΕΖ, οὕτως τὴν ΕΖ πρὸς τὴν ΒΗ: καὶ ἐπεζεύχθω ἡ ΑΗ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΕ πρὸς τὴν ΕΖ, ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΔΕ, οὕτως ἡ ΒΓ πρὸς τὴν ΕΖ. ἀλλ᾽ ὡς ἡ ΒΓ πρὸς ΕΖ, οὕτως ἐστὶν ἡ ΕΖ πρὸς ΒΗ. καὶ ὡς ἄρα ἡ ΑΒ πρὸς ΔΕ, οὕτως ἡ ΕΖ πρὸς ΒΗ: τῶν ΑΒΗ, ΔΕΖ ἄρα τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ὧν δὲ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα. ἴσον ἄρα ἐστὶ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΕΖ, οὕτως ἡ ΕΖ πρὸς τὴν ΒΗ, ἐὰν δὲ τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἡ πρώτη πρὸς τὴν τρίτην διπλασίονα λόγον ἔχει ἤπερ πρὸς τὴν δευτέραν, ἡ ΒΓ ἄρα πρὸς τὴν ΒΗ διπλασίονα λόγον ἔχει ἤπερ ἡ ΓΒ πρὸς τὴν ΕΖ. ὡς δὲ ἡ ΓΒ πρὸς τὴν ΒΗ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΒΗ τρίγωνον: καὶ τὸ ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΑΒΗ διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἴσον δὲ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ: καὶ τὸ ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. τὰ ἄρα ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον ἐπείπερ ἐδείχθη, ὡς ἡ ΓΒ πρὸς ΒΗ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΒΗ τρίγωνον, τουτέστι τὸ ΔΕΖ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 355|>","<|""VertexLabel"" -> ""6.20"", ""Text"" -> ""Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side."", ""TextWordCount"" -> 42, ""GreekText"" -> ""τὰ ὅμοια πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν."", ""GreekTextWordCount"" -> 34, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 5, ""Theorem"" -> 12}, {""Book"" -> 5, ""Theorem"" -> 22}, {""Book"" -> 6, ""Definition"" -> 1}, {""Book"" -> 6, ""Definition"" -> 1}, {""Book"" -> 6, ""Theorem"" -> 1}, {""Book"" -> 6, ""Theorem"" -> 4}, {""Book"" -> 6, ""Theorem"" -> 6}, {""Book"" -> 6, ""Theorem"" -> 19}}, ""Proof"" -> ""Let ABCDE, FGHKL be similar polygons, and let AB correspond to FG; I say that the polygons ABCDE, FGHKL are divided into similar triangles, and into triangles equal in multitude and inthe same ratio as the wholes, and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which AB has to FG. Let BE, EC, GL, LH be joined. Now, since the polygon ABCDE is similar to the polygonFGHKL, the angle BAE is equal to the angle GFL; and, as BA is to AE, so is GF to FL. [VI. Def. 1] Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal anglesproportional, therefore the triangle ABE is equiangular with the triangle FGL; [VI. 6] so that it is also similar; [VI. 4 and Def. 1]therefore the angle ABE is equal to the angle FGL. But the whole angle ABC is also equal to the whole angle FGH because of the similarity of the polygons; therefore the remaining angle EBC is equal to the angle LGH. And, since, because of the similarity of the triangles ABE,FGL, as EB is to BA, so is LG to GF, and moreover also, because of the similarity of the polygons, as AB is to BC, so is FG to GH, therefore, ex aequali, as EB is to BC, so is LG to GH; [V. 22]that is, the sides about the equal angles EBC, LGH are proportional; therefore the triangle EBC is equiangular with the triangle LGH, [VI. 6] so that the triangle EBC is also similar to the triangle LGH. [VI. 4 and Def. 1] For the same reason the triangle ECD is also similar to the triangle LHK. Therefore the similar polygons ABCDE, FGHKL have been divided into similar triangles, and into triangles equal inmultitude. I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and ABE, EBC, ECD are antecedents, while FGL, LGH, LHK are their consequents, and that the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG. For let AC, FH be joined. Then since, because of the similarity of the polygons,the angle ABC is equal to the angle FGH, and, as AB is to BC, so is FG to GH, the triangle ABC is equiangular with the triangle FGH; [VI. 6]therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF. And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, therefore the remaining angle AMB is also equal to the remaining angle FNG; [I. 32] therefore the triangle ABM is equiangular with the triangleFGN. Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH. Therefore, proportionally, as AM is to MB, so is FN to NG,and, as BM is to MC, so is GN to NH; so that, in addition, ex aequali, as AM is to MC, so is FN to NH. But, as AM is to MC, so is the triangle ABM to MBC, and AME to EMC; for they are to one another as theirbases. [VI. 1] Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [V. 12] therefore, as the triangle AMB is to BMC, so is ABE to CBE. But, as AMB is to BMC, so is AM to MC; therefore also, as AM is to MC, so is the triangle ABE to the triangle EBC. For the same reason also, as FN is to NH, so is the triangle FGL to the triangleGLH. And, as AM is to MC, so is FN to NH; therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH; and, alternately, as the triangle ABE is to the triangle FGL,so is the triangle BEC to the triangle GLH. Similarly we can prove, if BD, GK be joined, that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK. And since, as the triangle ABE is to the triangle FGL,so is EBC to LGH, and further ECD to LHK, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [V. 12] therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL. But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG; for similar triangles are in the duplicate ratio of the corresponding sides. [VI. 19] Therefore the polygon ABCDE also has to the polygonFGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG."", ""ProofWordCount"" -> 855, ""GreekProof"" -> ""ἔστω ὅμοια πολύγωνα τὰ ΑΒΓΔΕ, ΖΗΘΚΛ, ὁμόλογος δὲ ἔστω ἡ ΑΒ τῇ ΖΗ: λέγω, ὅτι τὰ ΑΒΓΔΕ, ΖΗΘΚΛ πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΖΗ. ἐπεζεύχθωσαν αἱ ΒΕ, ΕΓ, ΗΛ, ΛΘ. καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓΔΕ πολύγωνον τῷ ΖΗΘΚΛ πολυγώνῳ, ἴση ἐστὶν ἡ ὑπὸ ΒΑΕ γωνία τῇ ὑπὸ ΗΖΛ. καί ἐστιν ὡς ἡ ΒΑ πρὸς ΑΕ, οὕτως ἡ ΗΖ πρὸς ΖΛ. ἐπεὶ οὖν δύο τρίγωνά ἐστι τὰ ΑΒΕ, ΖΗΛ μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΗΛ τριγώνῳ: ὥστε καὶ ὅμοιον: ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΒΕ γωνία τῇ ὑπὸ ΖΗΛ. ἔστι δὲ καὶ ὅλη ἡ ὑπὸ ΑΒΓ ὅλῃ τῇ ὑπὸ ΖΗΘ ἴση διὰ τὴν ὁμοιότητα τῶν πολυγώνων: λοιπὴ ἄρα ἡ ὑπὸ ΕΒΓ γωνία τῇ ὑπὸ ΛΗΘ ἐστιν ἴση. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν ΑΒΕ, ΖΗΛ τριγώνων ἐστὶν ὡς ἡ ΕΒ πρὸς ΒΑ, οὕτως ἡ ΛΗ πρὸς ΗΖ, ἀλλὰ μὴν καὶ διὰ τὴν ὁμοιότητα τῶν πολυγώνων ἐστὶν ὡς ἡ ΑΒ πρὸς ΒΓ, οὕτως ἡ ΖΗ πρὸς ΗΘ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ ΕΒ πρὸς ΒΓ, οὕτως ἡ ΛΗ πρὸς ΗΘ, καὶ περὶ τὰς ἴσας γωνίας τὰς ὑπὸ ΕΒΓ, ΛΗΘ αἱ πλευραὶ ἀνάλογόν εἰσιν: ἰσογώνιον ἄρα ἐστὶ τὸ ΕΒΓ τρίγωνον τῷ ΛΗΘ τριγώνῳ: ὥστε καὶ ὅμοιόν ἐστι τὸ ΕΒΓ τρίγωνον τῷ ΛΗΘ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΕΓΔ τρίγωνον ὅμοιόν ἐστι τῷ ΛΘΚ τριγώνῳ. τὰ ἄρα ὅμοια πολύγωνα τὰ ΑΒΓΔΕ, ΖΗΘΚΛ εἴς τε ὅμοια τρίγωνα διῄρηται καὶ εἰς ἴσα τὸ πλῆθος. λέγω, ὅτι καὶ ὁμόλογα τοῖς ὅλοις, τουτέστιν ὥστε ἀνάλογον εἶναι τὰ τρίγωνα, καὶ ἡγούμενα μὲν εἶναι τὰ ΑΒΕ, ΕΒΓ, ΕΓΔ, ἑπόμενα δὲ αὐτῶν τὰ ΖΗΛ, ΛΗΘ, ΛΘΚ, καὶ ὅτι τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἡ ΑΒ πρὸς τὴν ΖΗ. ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΖΘ. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν πολυγώνων ἴση ἐστὶν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΖΗΘ, καί ἐστιν ὡς ἡ ΑΒ πρὸς ΒΓ, οὕτως ἡ ΖΗ πρὸς ΗΘ, ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΖΗΘ τριγώνῳ: ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΗΖΘ, ἡ δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΗΘΖ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΑΜ γωνία τῇ ὑπὸ ΗΖΝ, ἔστι δὲ καὶ ἡ ὑπὸ ΑΒΜ τῇ ὑπὸ ΖΗΝ ἴση, καὶ λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΖΝΗ ἴση ἐστίν: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΜ τρίγωνον τῷ ΖΗΝ τριγώνῳ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τὸ ΒΜΓ τρίγωνον ἰσογώνιόν ἐστι τῷ ΗΝΘ τριγώνῳ. ἀνάλογον ἄρα ἐστίν, ὡς μὲν ἡ ΑΜ πρὸς ΜΒ, οὕτως ἡ ΖΝ πρὸς ΝΗ, ὡς δὲ ἡ ΒΜ πρὸς ΜΓ, οὕτως ἡ ΗΝ πρὸς ΝΘ: ὥστε καὶ δι᾽ ἴσου, ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ. ἀλλ᾽ ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως τὸ ΑΒΜ τρίγωνον πρὸς τὸ ΜΒΓ, καὶ τὸ ΑΜΕ πρὸς τὸ ΕΜΓ: πρὸς ἄλληλα γάρ εἰσιν ὡς αἱ βάσεις. καὶ ὡς ἄρα ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ὡς ἄρα τὸ ΑΜΒ τρίγωνον πρὸς τὸ ΒΜΓ, οὕτως τὸ ΑΒΕ πρὸς τὸ ΓΒΕ. ἀλλ᾽ ὡς τὸ ΑΜΒ πρὸς τὸ ΒΜΓ, οὕτως ἡ ΑΜ πρὸς ΜΓ: καὶ ὡς ἄρα ἡ ΑΜ πρὸς ΜΓ, οὕτως τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΕΒΓ τρίγωνον. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ ΖΝ πρὸς ΝΘ, οὕτως τὸ ΖΗΛ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον. καί ἐστιν ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ: καὶ ὡς ἄρα τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΒΕΓ τρίγωνον, οὕτως τὸ ΖΗΛ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον, καὶ ἐναλλὰξ, ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΒΕΓ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον. ὁμοίως δὴ δείξομεν ἐπιζευχθεισῶν τῶν ΒΔ, ΗΚ, ὅτι καὶ ὡς τὸ ΒΕΓ τρίγωνον πρὸς τὸ ΛΗΘ τρίγωνον, οὕτως τὸ ΕΓΔ τρίγωνον πρὸς τὸ ΛΘΚ τρίγωνον. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΕΒΓ πρὸς τὸ ΛΗΘ, καὶ ἔτι τὸ ΕΓΔ πρὸς τὸ ΛΘΚ, καὶ ὡς ἄρα ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα: ἔστιν ἄρα ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. ἀλλὰ τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον πλευράν: τὰ γὰρ ὅμοια τρίγωνα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν. καὶ τὸ ΑΒΓΔΕ ἄρα πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον πλευράν. τὰ ἄρα ὅμοια πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν: ὅπερ ἔδει δεῖξαι. Πόρισμα ὡσαύτως δὲ καὶ ἐπὶ τῶν ὁμοίων τετραπλεύρων δειχθήσεται, ὅτι ἐν διπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. ἐδείχθη δὲ καὶ ἐπὶ τῶν τριγώνων: ὥστε καὶ καθόλου τὰ ὅμοια εὐθύγραμμα σχήματα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. ὅπερ ἔδει δεῖξαι. πόρισμα β # καὶ ἐὰν τῶν ΑΒ, ΖΗ τρίτην ἀνάλογον λάβωμεν τὴν Ξ, ἡ ΒΑ πρὸς τὴν Ξ διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΖΗ. ἔχει δὲ καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον ἢ τὸ τετράπλευρον πρὸς τὸ τετράπλευρον διπλασίονα λόγον ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἡ ΑΒ πρὸς τὴν ΖΗ: ἐδείχθη δὲ τοῦτο καὶ ἐπὶ τῶν τριγώνων: ὥστε καὶ καθόλου φανερόν, ὅτι, ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔσται ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον."", ""GreekProofWordCount"" -> 950|>","<|""VertexLabel"" -> ""6.21"", ""Text"" -> ""Figures which are similar to the same rectilineal figure are also similar to one another."", ""TextWordCount"" -> 15, ""GreekText"" -> ""τὰ τῷ αὐτῷ εὐθυγράμμῳ ὅμοια καὶ ἀλλήλοις ἐστὶν ὅμοια."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 6, ""Definition"" -> 1}}, ""Proof"" -> ""For let each of the rectilineal figures A, B be similar to C; I say that A is also similar to B. For, since A is similar to C, it is equiangular with it and has the sides about the equal angles proportional. [VI. Def. 1] Again, since B is similar to C, it is equiangular with it and has the sides about the equal angles proportional. Therefore each of the figures A, B is equiangular with C and with C has the sides about the equal angles proportional; therefore A is similar to B."", ""ProofWordCount"" -> 95, ""GreekProof"" -> ""ἔστω γὰρ ἑκάτερον τῶν Α, Β εὐθυγράμμων τῷ Γ ὅμοιον: λέγω, ὅτι καὶ τὸ Α τῷ Β ἐστιν ὅμοιον. ἐπεὶ γὰρ ὅμοιόν ἐστι τὸ Α τῷ Γ, ἰσογώνιόν τέ ἐστιν αὐτῷ καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. πάλιν, ἐπεὶ ὅμοιόν ἐστι τὸ Β τῷ Γ, ἰσογώνιόν τέ ἐστιν αὐτῷ καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. ἑκάτερον ἄρα τῶν Α, Β τῷ Γ ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει ὥστε καὶ τὸ Α τῷ Β ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. ὅμοιον ἄρα ἐστὶ τὸ Α τῷ Β: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 108|>","<|""VertexLabel"" -> ""6.22"", ""Text"" -> ""If four straight lines be proportional, the rectilineal figures similar and similarly described upon them will also be proportional; and, if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves also be proportional."", ""TextWordCount"" -> 40, ""GreekText"" -> ""ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπ᾽ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ἔσται: κἂν τὰ ἀπ᾽ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ᾖ, καὶ αὐταὶ αἱ εὐθεῖαι ἀνάλογον ἔσονται."", ""GreekTextWordCount"" -> 37, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 5, ""Theorem"" -> 22}, {""Book"" -> 6, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 12}, {""Book"" -> 6, ""Theorem"" -> 18}, {""Book"" -> 6, ""Theorem"" -> 19}}, ""Proof"" -> ""Let the four straight lines AB, CD, EF, GH be proportional, so that, as AB is to CD, so is EF to GH, and let there be described on AB, CD the similar and similarly situated rectilineal figures KAB, LCD, and on EF, GH the similar and similarly situated rectilineal figures MF, NH; I say that, as KAB is to LCD, so is MF to NH. For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11] Then since, as AB is to CD, so is EF to GH, and, as CD is to O, so is GH to P, therefore, ex aequali, as AB is to O, so is EF to P. [V. 22] But, as AB is to O, so is KAB to LCD, [VI. 19] and, as EF is to P, so is MF to NH; therefore also, as KAB is to LCD, so is MF to NH. [V. 11] Next, let MF be to NH as KAB is to LCD; I say also that, as AB is to CD, so is EF to GH. For, if EF is not to GH as AB to CD, let EF be to QR as AB to CD, [VI. 12]and on QR let the rectilineal figure SR be described similar and similarly situated to either of the two MF, NH. [VI. 18] Since then, as AB is to CD, so is EF to QR, and there have been described on AB, CD the similar and similarly situated figures KAB, LCD, and on EF, QR the similar and similarly situated figures MF, SR, therefore, as KAB is to LCD, so is MF to SR. But also, by hypothesis, as KAB is to LCD, so is MF to NH; therefore also, as MF is to SR, so is MF to NH. [V. 11] Therefore MF has the same ratio to each of the figures NH, SR; therefore NH is equal to SR. [V. 9] But it is also similar and similarly situated to it; therefore GH is equal to QR. And, since, as AB is to CD, so is EF to QR, while QR is equal to GH, therefore, as AB is to CD, so is EF to GH."", ""ProofWordCount"" -> 379, ""GreekProof"" -> ""ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, ΕΖ, ΗΘ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, καὶ ἀναγεγράφθωσαν ἀπὸ μὲν τῶν ΑΒ, ΓΔ ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΚΑΒ, ΛΓΔ, ἀπὸ δὲ τῶν ΕΖ, ΗΘ ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΜΖ, ΝΘ: λέγω, ὅτι ἐστὶν ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. εἰλήφθω γὰρ τῶν μὲν ΑΒ, ΓΔ τρίτη ἀνάλογον ἡ Ξ, τῶν δὲ ΕΖ, ΗΘ τρίτη ἀνάλογον ἡ Ο. καὶ ἐπεί ἐστιν ὡς μὲν ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, ὡς δὲ ἡ ΓΔ πρὸς τὴν Ξ, οὕτως ἡ ΗΘ πρὸς τὴν Ο, δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν Ξ, οὕτως ἡ ΕΖ πρὸς τὴν Ο. ἀλλ᾽ ὡς μὲν ἡ ΑΒ πρὸς τὴν Ξ, οὕτως καὶ τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, ὡς δὲ ἡ ΕΖ πρὸς τὴν Ο, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ: καὶ ὡς ἄρα τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. ἀλλὰ δὴ ἔστω ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ: λέγω, ὅτι ἐστὶ καὶ ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. εἰ γὰρ μή ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, ἔστω ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΠΡ ὁποτέρῳ τῶν ΜΖ, ΝΘ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον τὸ ΣΡ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, καὶ ἀναγέγραπται ἀπὸ μὲν τῶν ΑΒ, ΓΔ ὅμοιά τε καὶ ὁμοίως κείμενα τὰ ΚΑΒ, ΛΓΔ, ἀπὸ δὲ τῶν ΕΖ, ΠΡ ὅμοιά τε καὶ ὁμοίως κείμενα τὰ ΜΖ, ΣΡ, ἔστιν ἄρα ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΣΡ. ὑπόκειται δὲ καὶ ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ: καὶ ὡς ἄρα τὸ ΜΖ πρὸς τὸ ΣΡ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. τὸ ΜΖ ἄρα πρὸς ἑκάτερον τῶν ΝΘ, ΣΡ τὸν αὐτὸν ἔχει λόγον: ἴσον ἄρα ἐστὶ τὸ ΝΘ τῷ ΣΡ. ἔστι δὲ αὐτῷ καὶ ὅμοιον καὶ ὁμοίως κείμενον: ἴση ἄρα ἡ ΗΘ τῇ ΠΡ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, ἴση δὲ ἡ ΠΡ τῇ ΗΘ, ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. ἐὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπ᾽ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ἔσται: κἂν τὰ ἀπ᾽ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ᾖ, καὶ αὐταὶ αἱ εὐθεῖαι ἀνάλογον ἔσονται: ὅπερ ἔδει δεῖξαι. λῆμμα ὅτι δέ, ἐὰν εὐθύγραμμα ἴσα ᾖ καὶ ὅμοια, αἱ ὁμόλογοι αὐτῶν πλευραὶ ἴσαι ἀλλήλαις εἰσίν, δείξομεν οὕτως. ἔστω ἴσα καὶ ὅμοια εὐθύγραμμα τὰ ΝΘ, ΣΡ, καὶ ἔστω ὡς ἡ ΘΗ πρὸς τὴν ΗΝ, οὕτως ἡ ΡΠ πρὸς τὴν ΠΣ: λέγω, ὅτι ἴση ἐστὶν ἡ ΡΠ τῇ ΘΗ. εἰ γὰρ ἄνισοί εἰσιν, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ΡΠ τῆς ΘΗ. καὶ ἐπεί ἐστιν ὡς ἡ ΡΠ πρὸς ΠΣ, οὕτως ἡ ΘΗ πρὸς τὴν ΗΝ, καὶ ἐναλλάξ, ὡς ἡ ΡΠ πρὸς τὴν ΘΗ, οὕτως ἡ ΠΣ πρὸς τὴν ΗΝ, μείζων δὲ ἡ ΠΡ τῆς ΘΗ, μείζων ἄρα καὶ ἡ ΠΣ τῆς ΗΝ: ὥστε καὶ τὸ ΡΣ μεῖζόν ἐστι τοῦ ΘΝ. ἀλλὰ καὶ ἴσον: ὅπερ ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ΠΡ τῇ ΗΘ: ἴση ἄρα: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 574|>","<|""VertexLabel"" -> ""6.23"", ""Text"" -> ""Equiangular parallelograms have to one another the ratio compounded of the ratios of their sides."", ""TextWordCount"" -> 15, ""GreekText"" -> ""τὰ ἰσογώνια παραλληλόγραμμα πρὸς ἄλληλα λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 1}, {""Book"" -> 6, ""Theorem"" -> 12}}, ""Proof"" -> ""Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG;I say that the parallelogram AC has to the parallelogram CF the ratio compounded of the ratios of the sides. For let them be placed so that BC is in a straight line with CG; therefore DC is also in a straight line with CE. Let the parallelogram DG be completed; let a straight line K be set out, and let it be contrived that, as BC is to CG, so is K to L, and, as DC is to CE, so is L to M. [VI. 12] Then the ratios of K to L and of L to M are the sameas the ratios of the sides, namely of BC to CG and of DC to CE. But the ratio of K to M is compounded of the ratio of K to L and of that of L to M; so that K has also to M the ratio compounded of the ratiosof the sides. Now since, as BC is to CG, so is the parallelogram AC to the parallelogram CH, [VI. 1] while, as BC is to CG, so is K to L, therefore also, as K is to L, so is AC to CH. [V. 11] Again, since, as DC is to CE, so is the parallelogram CH to CF, [VI. 1] while, as DC is to CE, so is L to M, therefore also, as L is to M, so is the parallelogram CH to the parallelogram CF. [V. 11] Since then it was proved that, as K is to L, so is the parallelogram AC to the parallelogram CH, and, as L is to M, so is the parallelogram CH to the parallelogram CF, therefore, ex aequali, as K is to M, so is AC to the parallelogram CF. But K has to M the ratio compounded of the ratios of the sides; therefore AC also has to CF the ratio compounded of the ratios of the sides."", ""ProofWordCount"" -> 338, ""GreekProof"" -> ""ἔστω ἰσογώνια παραλληλόγραμμα τὰ ΑΓ, ΓΖ ἴσην ἔχοντα τὴν ὑπὸ ΒΓΔ γωνίαν τῇ ὑπὸ ΕΓΗ: λέγω, ὅτι τὸ ΑΓ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν. κείσθω γὰρ ὥστε ἐπ᾽ εὐθείας εἶναι τὴν ΒΓ τῇ ΓΗ: ἐπ᾽ εὐθείας ἄρα ἐστὶ καὶ ἡ ΔΓ τῇ ΓΕ. καὶ συμπεπληρώσθω τὸ ΔΗ παραλληλόγραμμον, καὶ ἐκκείσθω τις εὐθεῖα ἡ Κ, καὶ γεγονέτω ὡς μὲν ἡ ΒΓ πρὸς τὴν ΓΗ, οὕτως ἡ Κ πρὸς τὴν Λ, ὡς δὲ ἡ ΔΓ πρὸς τὴν ΓΕ, οὕτως ἡ Λ πρὸς τὴν Μ. οἱ ἄρα λόγοι τῆς τε Κ πρὸς τὴν Λ καὶ τῆς Λ πρὸς τὴν Μ οἱ αὐτοί εἰσι τοῖς λόγοις τῶν πλευρῶν, τῆς τε ΒΓ πρὸς τὴν ΓΗ καὶ τῆς ΔΓ πρὸς τὴν ΓΕ. ἀλλ᾽ ὁ τῆς Κ πρὸς Μ λόγος σύγκειται ἔκ τε τοῦ τῆς Κ πρὸς Λ λόγου καὶ τοῦ τῆς Λ πρὸς Μ: ὥστε καὶ ἡ Κ πρὸς τὴν Μ λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΓΗ, οὕτως τὸ ΑΓ παραλληλόγραμμον πρὸς τὸ ΓΘ, ἀλλ᾽ ὡς ἡ ΒΓ πρὸς τὴν ΓΗ, οὕτως ἡ Κ πρὸς τὴν λ, καὶ ὡς ἄρα ἡ Κ πρὸς τὴν Λ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΘ. πάλιν, ἐπεί ἐστιν ὡς ἡ ΔΓ πρὸς τὴν ΓΕ, οὕτως τὸ ΓΘ παραλληλόγραμμον πρὸς τὸ ΓΖ, ἀλλ᾽ ὡς ἡ ΔΓ πρὸς τὴν ΓΕ, οὕτως ἡ Λ πρὸς τὴν Μ, καὶ ὡς ἄρα ἡ Λ πρὸς τὴν Μ, οὕτως τὸ ΓΘ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον. ἐπεὶ οὖν ἐδείχθη, ὡς μὲν ἡ Κ πρὸς τὴν Λ, οὕτως τὸ ΑΓ παραλληλόγραμμον πρὸς τὸ ΓΘ παραλληλόγραμμον, ὡς δὲ ἡ Λ πρὸς τὴν Μ, οὕτως τὸ ΓΘ παραλληλόγραμμον πρὸς τὸ ΓΖ παραλληλόγραμμον, δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ Κ πρὸς τὴν Μ, οὕτως τὸ ΑΓ πρὸς τὸ ΓΖ παραλληλόγραμμον. ἡ δὲ Κ πρὸς τὴν Μ λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν: καὶ τὸ ΑΓ ἄρα πρὸς τὸ ΓΖ λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν. τὰ ἄρα ἰσογώνια παραλληλόγραμμα πρὸς ἄλληλα λόγον ἔχει τὸν συγκείμενον ἐκ τῶν πλευρῶν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 346|>","<|""VertexLabel"" -> ""6.24"", ""Text"" -> ""In any parallelogram the parallelograms about the diameter are similar both to the whole and to one another."", ""TextWordCount"" -> 18, ""GreekText"" -> ""παντὸς παραλληλογράμμου τὰ περὶ τὴν διάμετρον παραλληλόγραμμα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις."", ""GreekTextWordCount"" -> 14, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 16}, {""Book"" -> 5, ""Theorem"" -> 18}, {""Book"" -> 5, ""Theorem"" -> 22}, {""Book"" -> 6, ""Definition"" -> 1}, {""Book"" -> 6, ""Theorem"" -> 2}, {""Book"" -> 6, ""Theorem"" -> 21}}, ""Proof"" -> ""Let ABCD be a parallelogram, and AC its diameter, and let EG, HK be parallelograms about AC; I say that each of the parallelograms EG, HK is similar both to the whole ABCD and to the other. For, since EF has been drawn parallel to BC, one of the sides of the triangle ABC, proportionally, as BE is to EA, so is CF to FA. [VI. 2] Again, since FG has been drawn parallel to CD, one of the sides of the triangle ACD, proportionally, as CF is to FA, so is DG to GA. [VI. 2] But it was proved that, as CF is to FA, so also is BE to EA; therefore also, as BE is to EA, so is DG to GA, and therefore, componendo, as BA is to AE, so is DA to AG, [V. 18]and, alternately, as BA is to AD, so is EA to AG. [V. 16] Therefore in the parallelograms ABCD, EG, the sides about the common angle BAD are proportional. And, since GF is parallel to DC, the angle AFG is equal to the angle DCA; and the angle DAC is common to the two triangles ADC, AGF; therefore the triangle ADC is equiangular with the triangle AGF. For the same reason the triangle ACB is also equiangular with the triangle AFE, and the whole parallelogram ABCD is equiangular with the parallelogram EG. Therefore, proportionally, as AD is to DC, so is AG to GF, as DC is to CA, so is GF to FA, as AC is to CB, so is AF to FE, and further, as CB is to BA, so is FE to EA. And, since it was proved that, as DC is to CA, so is GF to FA, and, as AC is to CB, so is AF to FE, therefore, ex aequali, as DC is to CB, so is GF to FE. [V. 22] Therefore in the parallelograms ABCD, EG the sides about the equal angles are proportional; therefore the parallelogram ABCD is similar to the parallelogram EG. [VI. Def. 1] For the same reason the parallelogram ABCD is also similar to the parallelogram KH; therefore each of the parallelograms EG, HK is similar to ABCD. But figures similar to the same rectilineal figure are also similar to one another; [VI. 21] therefore the parallelogram EG is also similar to the parallelogram HK."", ""ProofWordCount"" -> 397, ""GreekProof"" -> ""ἔστω παραλληλόγραμμον τὸ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἡ ΑΓ, περὶ δὲ τὴν ΑΓ παραλληλόγραμμα ἔστω τὰ ΕΗ, ΘΚ: λέγω, ὅτι ἑκάτερον τῶν ΕΗ, ΘΚ παραλληλογράμμων ὅμοιόν ἐστι ὅλῳ τῷ ΑΒΓΔ καὶ ἀλλήλοις. ἐπεὶ γὰρ τριγώνου τοῦ ΑΒΓ παρὰ μίαν τῶν πλευρῶν τὴν ΒΓ ἦκται ἡ ΕΖ, ἀνάλογόν ἐστιν ὡς ἡ ΒΕ πρὸς τὴν ΕΑ, οὕτως, ἡ ΓΖ πρὸς τὴν ΖΑ. πάλιν, ἐπεὶ τριγώνου τοῦ ΑΓΔ παρὰ μίαν τὴν ΓΔ ἦκται ἡ ΖΗ, ἀνάλογόν ἐστιν ὡς ἡ ΓΖ πρὸς τὴν ΖΑ, οὕτως ἡ ΔΗ πρὸς τὴν ΗΑ. ἀλλ᾽ ὡς ἡ ΓΖ πρὸς τὴν ΖΑ, οὕτως ἐδείχθη καὶ ἡ ΒΕ πρὸς τὴν ΕΑ: καὶ ὡς ἄρα ἡ ΒΕ πρὸς τὴν ΕΑ, οὕτως ἡ ΔΗ πρὸς τὴν ΗΑ, καὶ συνθέντι ἄρα ὡς ἡ ΒΑ πρὸς ΑΕ, οὕτως ἡ ΔΑ πρὸς ΑΗ, καὶ ἐναλλὰξ ὡς ἡ ΒΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΗ. τῶν ἄρα ΑΒΓΔ, ΕΗ παραλληλογράμμων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὴν κοινὴν γωνίαν τὴν ὑπὸ ΒΑΔ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΗΖ τῇ ΔΓ, ἴση ἐστὶν ἡ μὲν ὑπὸ ΑΖΗ γωνία τῇ ὑπὸ ΔΓΑ: καὶ κοινὴ τῶν δύο τριγώνων τῶν ΑΔΓ, ΑΗΖ ἡ ὑπὸ ΔΑΓ γωνία: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΔΓ τρίγωνον τῷ ΑΗΖ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΓΒ τρίγωνον ἰσογώνιόν ἐστι τῷ ΑΖΕ τριγώνῳ, καὶ ὅλον τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΗ παραλληλογράμμῳ ἰσογώνιόν ἐστιν. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΑΔ πρὸς τὴν ΔΓ, οὕτως ἡ ΑΗ πρὸς τὴν ΗΖ, ὡς δὲ ἡ ΔΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ, ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΑΖ πρὸς τὴν ΖΕ, καὶ ἔτι ὡς ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΖΕ πρὸς τὴν ΕΑ. καὶ ἐπεὶ ἐδείχθη ὡς μὲν ἡ ΔΓ πρὸς τὴν ΓΑ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΑ, ὡς δὲ ἡ ΑΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΑΖ πρὸς τὴν ΖΕ, δι᾽ ἴσου ἄρα ἐστὶν ὡς ἡ ΔΓ πρὸς τὴν ΓΒ, οὕτως ἡ ΗΖ πρὸς τὴν ΖΕ. τῶν ἄρα ΑΒΓΔ, ΕΗ παραλληλογράμμων ἀνάλογόν εἰσιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: ὅμοιον ἄρα ἐστὶ τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΗ παραλληλογράμμῳ. διὰ τὰ αὐτὰ δὴ τὸ ΑΒΓΔ παραλληλόγραμμον καὶ τῷ ΚΘ παραλληλογράμμῳ ὅμοιόν ἐστιν: ἑκάτερον ἄρα τῶν ΕΗ, ΘΚ παραλληλογράμμων τῷ ΑΒΓΔ παραλληλογράμμῳ ὅμοιόν ἐστιν. τὰ δὲ τῷ αὐτῷ εὐθυγράμμῳ ὅμοια καὶ ἀλλήλοις ἐστὶν ὅμοια: καὶ τὸ ΕΗ ἄρα παραλληλόγραμμον τῷ ΘΚ παραλληλογράμμῳ ὅμοιόν ἐστιν. παντὸς ἄρα παραλληλογράμμου τὰ περὶ τὴν διάμετρον παραλληλόγραμμα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 408|>","<|""VertexLabel"" -> ""6.25"", ""Text"" -> ""To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure."", ""TextWordCount"" -> 20, ""GreekText"" -> ""τῷ δοθέντι εὐθυγράμμῳ ὅμοιον καὶ ἄλλῳ τῷ δοθέντι ἴσον τὸ αὐτὸ συστήσασθαι."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 44}, {""Book"" -> 1, ""Theorem"" -> 45}, {""Book"" -> 5, ""Theorem"" -> 16}, {""Book"" -> 6, ""Theorem"" -> 1}, {""Book"" -> 6, ""Theorem"" -> 13}, {""Book"" -> 6, ""Theorem"" -> 18}, {""Book"" -> 6, ""Theorem"" -> 19}}, ""Proof"" -> ""Let ABC be the given rectilineal figure to which the figure to be constructed must be similar, and D that to which it must be equal; thus it is required to construct one and the same figure similar to ABC and equal to D. Let there be applied to BC the parallelogram BE equal to the triangle ABC [I. 44], and to CE the parallelogram CM equal to D in the angle FCE which is equal to the angle CBL. [I. 45] Therefore BC is in a straight line with CF, and LE with EM. Now let GH be taken a mean proportional to BC, CF [VI. 13], and on GH let KGH be described similar and similarly situated to ABC. [VI. 18] Then, since, as BC is to GH, so is GH to CF, and, if three straight lines be proportional, as the first is to the third, so is the figure on the first to the similar and similarly situated figure described on the second, [VI. 19] therefore, as BC is to CF, so is the triangle ABC to the triangle KGH. But, as BC is to CF, so also is the parallelogram BE to the parallelogram EF. [VI. 1] Therefore also, as the triangle ABC is to the triangle KGH, so is the parallelogram BE to the parallelogram EF; therefore, alternately, as the triangle ABC is to the parallelogram BE, so is the triangle KGH to the parallelogram EF. [V. 16] But the triangle ABC is equal to the parallelogram BE; therefore the triangle KGH is also equal to the parallelogram EF. But the parallelogram EF is equal to D; therefore KGH is also equal to D. And KGH is also similar to ABC. Therefore one and the same figure KGH has been constructed similar to the given rectilineal figure ABC and equal to the other given figure D."", ""ProofWordCount"" -> 312, ""GreekProof"" -> ""ἔστω τὸ μὲν δοθὲν εὐθύγραμμον, ᾧ δεῖ ὅμοιον συστήσασθαι, τὸ ΑΒΓ, ᾧ δὲ δεῖ ἴσον, τὸ Δ: δεῖ δὴ τῷ μὲν ΑΒΓ ὅμοιον, τῷ δὲ Δ ἴσον τὸ αὐτὸ συστήσασθαι. παραβεβλήσθω γὰρ παρὰ μὲν τὴν ΒΓ τῷ ΑΒΓ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΒΕ, παρὰ δὲ τὴν ΓΕ τῷ Δ ἴσον παραλληλόγραμμον τὸ ΓΜ ἐν γωνίᾳ τῇ ὑπὸ ΖΓΕ, ἥ ἐστιν ἴση τῇ ὑπὸ ΓΒΛ. ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ μὲν ΒΓ τῇ ΓΖ, ἡ δὲ ΛΕ τῇ ΕΜ. καὶ εἰλήφθω τῶν ΒΓ, ΓΖ μέση ἀνάλογον ἡ ΗΘ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΗΘ τῷ ΑΒΓ ὅμοιόν τε καὶ ὁμοίως κείμενον τὸ ΚΗΘ. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΗΘ, οὕτως ἡ ΗΘ πρὸς τὴν ΓΖ, ἐὰν δὲ τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον, ἔστιν ἄρα ὡς ἡ ΒΓ πρὸς τὴν ΓΖ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΚΗΘ τρίγωνον. ἀλλὰ καὶ ὡς ἡ ΒΓ πρὸς τὴν ΓΖ, οὕτως τὸ ΒΕ παραλληλόγραμμον πρὸς τὸ ΕΖ παραλληλόγραμμον. καὶ ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΚΗΘ τρίγωνον, οὕτως τὸ ΒΕ παραλληλόγραμμον πρὸς τὸ ΕΖ παραλληλόγραμμον: ἐναλλὰξ ἄρα ὡς τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΒΕ παραλληλόγραμμον, οὕτως τὸ ΚΗΘ τρίγωνον πρὸς τὸ ΕΖ παραλληλόγραμμον. ἴσον δὲ τὸ ΑΒΓ τρίγωνον τῷ ΒΕ παραλληλογράμμῳ: ἴσον ἄρα καὶ τὸ ΚΗΘ τρίγωνον τῷ ΕΖ παραλληλογράμμῳ. ἀλλὰ τὸ ΕΖ παραλληλόγραμμον τῷ Δ ἐστιν ἴσον: καὶ τὸ ΚΗΘ ἄρα τῷ Δ ἐστιν ἴσον. ἔστι δὲ τὸ ΚΗΘ καὶ τῷ ΑΒΓ ὅμοιον. τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ ΑΒΓ ὅμοιον καὶ ἄλλῳ τῷ δοθέντι τῷ Δ ἴσον τὸ αὐτὸ συνέσταται τὸ ΚΗΘ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 277|>","<|""VertexLabel"" -> ""6.26"", ""Text"" -> ""If from a parallelogram there be taken away a parallelogram similar and similarly situated to the whole and having a common angle with it, it is about the same diameter with the whole."", ""TextWordCount"" -> 33, ""GreekText"" -> ""ἐὰν ἀπὸ παραλληλογράμμου παραλληλόγραμμον ἀφαιρεθῇ ὅμοιόν τε τῷ ὅλῳ καὶ ὁμοίως κείμενον κοινὴν γωνίαν ἔχον αὐτῷ, περὶ τὴν αὐτὴν διάμετρόν ἐστι τῷ ὅλῳ."", ""GreekTextWordCount"" -> 23, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 31}, {""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 5, ""Theorem"" -> 11}, {""Book"" -> 6, ""Theorem"" -> 24}}, ""Proof"" -> ""For suppose it is not, but, if possible, let AHC be the diameter < of ABCD >, let GF be produced and carried through to H, and let HK be drawn through H parallel to either of the straight lines AD, BC. [I. 31] Since, then, ABCD is about the same diameter with KG, therefore, as DA is to AB, so is GA to AK. [VI. 24] But also, because of the similarity of ABCD, EG, as DA is to AB, so is GA to AE; therefore also, as GA is to AK, so is GA to AE. [V. 11] Therefore GA has the same ratio to each of the straight lines AK, AE. Therefore AE is equal to AK [V. 9], the less to the greater: which is impossible. Therefore ABCD cannot but be about the same diameter with AF; therefore the parallelogram ABCD is about the same diameter with the parallelogram AF."", ""ProofWordCount"" -> 152, ""GreekProof"" -> ""ἀπὸ γὰρ παραλληλογράμμου τοῦ ΑΒΓΔ παραλληλόγραμμον ἀφῃρήσθω τὸ ΑΖ ὅμοιον τῷ ΑΒΓΔ καὶ ὁμοίως κείμενον κοινὴν γωνίαν ἔχον αὐτῷ τὴν ὑπὸ ΔΑΒ: λέγω, ὅτι περὶ τὴν αὐτὴν διάμετρόν ἐστι τὸ ΑΒΓΔ τῷ ΑΖ. μὴ γάρ, ἀλλ᾽ εἰ δυνατόν, ἔστω αὐτῶν διάμετρος ἡ ΑΘΓ, καὶ ἐκβληθεῖσα ἡ ΗΖ διήχθω ἐπὶ τὸ Θ, καὶ ἤχθω διὰ τοῦ Θ ὁποτέρᾳ τῶν ΑΔ, ΒΓ παράλληλος ἡ ΘΚ. ἐπεὶ οὖν περὶ τὴν αὐτὴν διάμετρόν ἐστι τὸ ΑΒΓΔ τῷ ΚΗ, ἔστιν ἄρα ὡς ἡ ΔΑ πρὸς τὴν ΑΒ, οὕτως ἡ ΗΑ πρὸς τὴν ΑΚ. ἔστι δὲ καὶ διὰ τὴν ὁμοιότητα τῶν ΑΒΓΔ, ΕΗ καὶ ὡς ἡ ΔΑ πρὸς τὴν ΑΒ, οὕτως ἡ ΗΑ πρὸς τὴν ΑΕ: καὶ ὡς ἄρα ἡ ΗΑ πρὸς τὴν ΑΚ, οὕτως ἡ ΗΑ πρὸς τὴν ΑΕ. ἡ ΗΑ ἄρα πρὸς ἑκατέραν τῶν ΑΚ, ΑΕ τὸν αὐτὸν ἔχει λόγον. ἴση ἄρα ἐστὶν ἡ ΑΕ τῇ ΑΚ ἡ ἐλάττων τῇ μείζονι: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οὔκ ἐστι περὶ τὴν αὐτὴν διάμετρον τὸ ΑΒΓΔ τῷ ΑΖ: περὶ τὴν αὐτὴν ἄρα ἐστὶ διάμετρον τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΑΖ παραλληλογράμμῳ. ἐὰν ἄρα ἀπὸ παραλληλογράμμου παραλληλόγραμμον ἀφαιρεθῇ ὅμοιόν τε τῷ ὅλῳ καὶ ὁμοίως κείμενον κοινὴν γωνίαν ἔχον αὐτῷ, περὶ τὴν αὐτὴν διάμετρόν ἐστι τῷ ὅλῳ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 202|>","<|""VertexLabel"" -> ""6.27"", ""Text"" -> ""Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect."", ""TextWordCount"" -> 49, ""GreekText"" -> ""πάντων τῶν παρὰ τὴν αὐτὴν εὐθεῖαν παραβαλλομένων παραλληλογράμμων καὶ ἐλλειπόντων εἴδεσι παραλληλογράμμοις ὁμοίοις τε καὶ ὁμοίως κειμένοις τῷ ἀπὸ τῆς ἡμισείας ἀναγραφομένῳ μέγιστόν ἐστι τὸ ἀπὸ τῆς ἡμισείας παραβαλλόμενον παραλληλόγραμμον ὅμοιον ὂν τῷ ἐλλείμματι."", ""GreekTextWordCount"" -> 34, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 36}, {""Book"" -> 1, ""Theorem"" -> 43}, {""Book"" -> 6, ""Theorem"" -> 26}}, ""Proof"" -> ""Let AB be a straight line and let it be bisected at C; let there be applied to the straight line AB the parallelogram AD deficient by the parallelogrammic figure DB described on the half of AB, that is, CB; I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB, AD is greatest. For let there be applied to the straight line AB the parallelogram AF deficient by the parallelogrammic figure FB similar and similarly situated to DB; I say that AD is greater than AF. For, since the parallelogram DB is similar to the parallelogram FB, they are about the same diameter. [VI. 26] Let their diameter DB be drawn, and let the figure be described. Then, since CF is equal to FE, [I. 43] and FB is common, therefore the whole CH is equal to the whole KE. But CH is equal to CG, since AC is also equal to CB. [I. 36] Therefore GC is also equal to EK. Let CF be added to each; therefore the whole AF is equal to the gnomon LMN; so that the parallelogram DB, that is, AD, is greater than the parallelogram AF."", ""ProofWordCount"" -> 203, ""GreekProof"" -> ""ἔστω εὐθεῖα ἡ ΑΒ καὶ τετμήσθω δίχα κατὰ τὸ Γ, καὶ παραβεβλήσθω παρὰ τὴν ΑΒ εὐθεῖαν τὸ ΑΔ παραλληλόγραμμον ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΔΒ ἀναγραφέντι ἀπὸ τῆς ἡμισείας τῆς ΑΒ, τουτέστι τῆς ΓΒ: λέγω, ὅτι πάντων τῶν παρὰ τὴν ΑΒ παραβαλλομένων παραλληλογράμμων καὶ ἐλλειπόντων εἴδεσι παραλληλογράμμοις ὁμοίοις τε καὶ ὁμοίως κειμένοις τῷ ΔΒ μέγιστόν ἐστι τὸ ΑΔ. παραβεβλήσθω γὰρ παρὰ τὴν ΑΒ εὐθεῖαν τὸ ΑΖ παραλληλόγραμμον ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΖΒ ὁμοίῳ τε καὶ ὁμοίως κειμένῳ τῷ ΔΒ: λέγω, ὅτι μεῖζόν ἐστι τὸ ΑΔ τοῦ ΑΖ. ἐπεὶ γὰρ ὅμοιόν ἐστι τὸ ΔΒ παραλληλόγραμμον τῷ ΖΒ παραλληλογράμμῳ, περὶ τὴν αὐτήν εἰσι διάμετρον. ἤχθω αὐτῶν διάμετρος ἡ ΔΒ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΓΖ τῷ ΖΕ, κοινὸν δὲ τὸ ΖΒ, ὅλον ἄρα τὸ ΓΘ ὅλῳ τῷ ΚΕ ἐστιν ἴσον. ἀλλὰ τὸ ΓΘ τῷ ΓΗ ἐστιν ἴσον, ἐπεὶ καὶ ἡ ΑΓ τῇ ΓΒ. καὶ τὸ ΗΓ ἄρα τῷ ΕΚ ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΓΖ: ὅλον ἄρα τὸ ΑΖ τῷ ΛΜΝ γνώμονί ἐστιν ἴσον: ὥστε τὸ ΔΒ παραλληλόγραμμον, τουτέστι τὸ ΑΔ, τοῦ ΑΖ παραλληλογράμμου μεῖζόν ἐστιν. πάντων ἄρα τῶν παρὰ τὴν αὐτὴν εὐθεῖαν παραβαλλομένων παραλληλογράμμων καὶ ἐλλειπόντων εἴδεσι παραλληλογράμμοις ὁμοίοις τε καὶ ὁμοίως κειμένοις τῷ ἀπὸ τῆς ἡμισείας ἀναγραφομένῳ μέγιστόν ἐστι τὸ ἀπὸ τῆς ἡμισείας παραβληθέν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 210|>","<|""VertexLabel"" -> ""6.28"", ""Text"" -> ""To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one: thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect."", ""TextWordCount"" -> 51, ""GreekText"" -> ""παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ ἴσον παραλληλόγραμμον παραβαλεῖν ἐλλεῖπον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ δοθέντι: δεῖ δὲ τὸ διδόμενον εὐθύγραμμον ᾧ δεῖ ἴσον παραβαλεῖν μὴ μεῖζον εἶναι τοῦ ἀπὸ τῆς ἡμισείας ἀναγραφομένου ὁμοίου τῷ ἐλλείμματι τοῦ τε ἀπὸ τῆς ἡμισείας καὶ ᾧ δεῖ ὅμοιον ἐλλείπειν."", ""GreekTextWordCount"" -> 46, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 36}, {""Book"" -> 6, ""Theorem"" -> 18}, {""Book"" -> 6, ""Theorem"" -> 21}, {""Book"" -> 6, ""Theorem"" -> 25}, {""Book"" -> 6, ""Theorem"" -> 26}}, ""Proof"" -> ""Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D. Let AB be bisected at the point E, and on EB let EBFG be described similar and similarly situated to D; [VI. 18] let the parallelogram AG be completed. If then AG is equal to C, that which was enjoined will have been done; for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D. But, if not, let HE be greater than C. Now HE is equal to GB; therefore GB is also greater than C. Let KLMN be constructed at once equal to the excess by which GB is greater than C and similar and similarly situated to D. [VI. 25] But D is similar to GB; therefore KM is also similar to GB. [VI. 21] Let, then, KL correspond to GE, and LM to GF. Now, since GB is equal to C, KM, therefore GB is greater than KM; therefore also GE is greater than KL, and GF than LM. Let GO be made equal to KL, and GP equal to LM; and let the parallelogram OGPQ be completed; therefore it is equal and similar to KM. Therefore GQ is also similar to GB; [VI. 21] therefore GQ is about the same diameter with GB. [VI. 26] Let GQB be their diameter, and let the figure be described. Then, since BG is equal to C, KM, and in them GQ is equal to KM, therefore the remainder, the gnomon UWV, is equal to the remainder C. And, since PR is equal to OS, let QB be added to each; therefore the whole PB is equal to the whole OB. But OB is equal to TE, since the side AE is also equal to the side EB; [I. 36] therefore TE is also equal to PB. Let OS be added to each; therefore the whole TS is equal to the whole, the gnomon VWU. But the gnomon VWU was proved equal to C; therefore TS is also equal to C. Therefore to the given straight line AB there has been applied the parallelogram ST equal to the given rectilineal figure C and deficient by a parallelogrammic figure QB which is similar to D."", ""ProofWordCount"" -> 465, ""GreekProof"" -> ""ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν εὐθύγραμμον, ᾧ δεῖ ἴσον παρὰ τὴν ΑΒ παραβαλεῖν, τὸ Γ μὴ μεῖζον ὂν τοῦ ἀπὸ τῆς ἡμισείας τῆς ΑΒ ἀναγραφομένου ὁμοίου τῷ ἐλλείμματι, ᾧ δὲ δεῖ ὅμοιον ἐλλείπειν, τὸ Δ: δεῖ δὴ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον παραβαλεῖν ἐλλεῖπον εἴδει παραλληλογράμμῳ ὁμοίῳ ὄντι τῷ Δ. τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ε σημεῖον, καὶ ἀναγεγράφθω ἀπὸ τῆς ΕΒ τῷ Δ ὅμοιον καὶ ὁμοίως κείμενον τὸ ΕΒΖΗ, καὶ συμπεπληρώσθω τὸ ΑΗ παραλληλόγραμμον. εἰ μὲν οὖν ἴσον ἐστὶ τὸ ΑΗ τῷ Γ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν: παραβέβληται γὰρ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον τὸ ΑΗ ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΗΒ ὁμοίῳ ὄντι τῷ Δ. εἰ δὲ οὔ, μεῖζον ἔστω τὸ ΘΕ τοῦ Γ. ἴσον δὲ τὸ ΘΕ τῷ ΗΒ: μεῖζον ἄρα καὶ τὸ ΗΒ τοῦ Γ. ᾧ δὴ μεῖζόν ἐστι τὸ ΗΒ τοῦ Γ, ταύτῃ τῇ ὑπεροχῇ ἴσον, τῷ δὲ Δ ὅμοιον καὶ ὁμοίως κείμενον τὸ αὐτὸ συνεστάτω τὸ ΚΛΜΝ. ἀλλὰ τὸ Δ τῷ ΗΒ ἐστιν ὅμοιον: καὶ τὸ ΚΜ ἄρα τῷ ΗΒ ἐστιν ὅμοιον. ἔστω οὖν ὁμόλογος ἡ μὲν ΚΛ τῇ ΗΕ, ἡ δὲ ΛΜ τῇ ΗΖ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΗΒ τοῖς Γ, ΚΜ, μεῖζον ἄρα ἐστὶ τὸ ΗΒ τοῦ ΚΜ: μείζων ἄρα ἐστὶ καὶ ἡ μὲν ΗΕ τῆς ΚΛ, ἡ δὲ ΗΖ τῆς ΛΜ. κείσθω τῇ μὲν ΚΛ ἴση ἡ ΗΞ, τῇ δὲ ΛΜ ἴση ἡ ΗΟ, καὶ συμπεπληρώσθω τὸ ΞΗΟΠ παραλληλόγραμμον: ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΗΠ τῷ ΚΜ ἀλλὰ τὸ ΚΜ τῷ ΗΒ ὅμοιόν ἐστιν. καὶ τὸ ΗΠ ἄρα τῷ ΗΒ ὅμοιόν ἐστιν: περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὸ ΗΠ τῷ ΗΒ. ἔστω αὐτῶν διάμετρος ἡ ΗΠΒ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΒΗ τοῖς Γ, ΚΜ, ὧν τὸ ΗΠ τῷ ΚΜ ἐστιν ἴσον, λοιπὸς ἄρα ὁ ΥΧΦ γνώμων λοιπῷ τῷ Γ ἴσος ἐστίν. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΟΡ τῷ ΞΣ, κοινὸν προσκείσθω τὸ ΠΒ: ὅλον ἄρα τὸ ΟΒ ὅλῳ τῷ ΞΒ ἴσον ἐστίν. ἀλλὰ τὸ ΞΒ τῷ ΤΕ ἐστιν ἴσον, ἐπεὶ καὶ πλευρὰ ἡ ΑΕ πλευρᾷ τῇ ΕΒ ἐστιν ἴση: καὶ τὸ ΤΕ ἄρα τῷ ΟΒ ἐστιν ἴσον. κοινὸν προσκείσθω τὸ ΞΣ: ὅλον ἄρα τὸ ΤΣ ὅλῳ τῷ ΦΧΥ γνώμονί ἐστιν ἴσον. ἀλλ᾽ ὁ ΦΧΥ γνώμων τῷ Γ ἐδείχθη ἴσος: καὶ τὸ ΤΣ ἄρα τῷ Γ ἐστιν ἴσον. παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται τὸ ΣΤ ἐλλεῖπον εἴδει παραλληλογράμμῳ τῷ ΠΒ ὁμοίῳ ὄντι τῷ Δ ἐπειδήπερ τὸ ΠΒ τῷ ΗΠ ὅμοιόν ἐστιν: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 433|>","<|""VertexLabel"" -> ""6.29"", ""Text"" -> ""To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one."", ""TextWordCount"" -> 26, ""GreekText"" -> ""παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι εὐθυγράμμῳ ἴσον παραλληλόγραμμον παραβαλεῖν ὑπερβάλλον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ δοθέντι."", ""GreekTextWordCount"" -> 16, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 36}, {""Book"" -> 1, ""Theorem"" -> 43}, {""Book"" -> 6, ""Theorem"" -> 21}, {""Book"" -> 6, ""Theorem"" -> 24}, {""Book"" -> 6, ""Theorem"" -> 25}, {""Book"" -> 6, ""Theorem"" -> 26}}, ""Proof"" -> ""Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, and D that to which the excess is required to be similar; thus it is required to apply to the straight line AB a parallelogram equal to the rectilineal figure C and exceeding by a parallelogrammic figure similar to D. Let AB be bisected at E; let there be described on EB the parallelogram BF similar and similarly situated to D; and let GH be constructed at once equal to the sum of BF, C and similar and similarly situated to D. [VI. 25] Let KH correspond to FL and KG to FE. Now, since GH is greater than FB, therefore KH is also greater than FL, and KG than FE. Let FL, FE be produced, let FLM be equal to KH, and FEN to KG, and let MN be completed; therefore MN is both equal and similar to GH. But GH is similar to EL; therefore MN is also similar to EL; [VI. 21]therefore EL is about the same diameter with MN. [VI. 26] Let their diameter FO be drawn, and let the figure be described. Since GH is equal to EL, C, while GH is equal to MN, therefore MN is also equal to EL, C. Let EL be subtracted from each; therefore the remainder, the gnomon XWV, is equal to C. Now, since AE is equal to EB, AN is also equal to NB [I. 36], that is, to LP [I. 43]. Let EO be added to each; therefore the whole AO is equal to the gnomon VWX. But the gnomon VWX is equal to C; therefore AO is also equal to C. Therefore to the given straight line AB there has been applied the parallelogram AO equal to the given rectilineal figure C and exceeding by a parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [VI. 24]."", ""ProofWordCount"" -> 335, ""GreekProof"" -> ""ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν εὐθύγραμμον, ᾧ δεῖ ἴσον παρὰ τὴν ΑΒ παραβαλεῖν, τὸ Γ, ᾧ δὲ δεῖ ὅμοιον ὑπερβάλλειν, τὸ Δ: δεῖ δὴ παρὰ τὴν ΑΒ εὐθεῖαν τῷ Γ εὐθυγράμμῳ ἴσον παραλληλόγραμμον παραβαλεῖν ὑπερβάλλον εἴδει παραλληλογράμμῳ ὁμοίῳ τῷ Δ. τετμήσθω ἡ ΑΒ δίχα κατὰ τὸ Ε, καὶ ἀναγεγράφθω ἀπὸ τῆς ΕΒ τῷ Δ ὅμοιον καὶ ὁμοίως κείμενον παραλληλόγραμμον τὸ ΒΖ, καὶ συναμφοτέροις μὲν τοῖς ΒΖ, Γ ἴσον, τῷ δὲ Δ ὅμοιον καὶ ὁμοίως κείμενον τὸ αὐτὸ συνεστάτω τὸ ΗΘ. ὁμόλογος δὲ ἔστω ἡ μὲν ΚΘ τῇ ΖΛ, ἡ δὲ ΚΗ τῇ ΖΕ. καὶ ἐπεὶ μεῖζόν ἐστι τὸ ΗΘ τοῦ ΖΒ, μείζων ἄρα ἐστὶ καὶ ἡ μὲν ΚΘ τῆς ΖΛ, ἡ δὲ ΚΗ τῆς ΖΕ. ἐκβεβλήσθωσαν αἱ ΖΛ, ΖΕ, καὶ τῇ μὲν ΚΘ ἴση ἔστω ἡ ΖΛΜ, τῇ δὲ ΚΗ ἴση ἡ ΖΕΝ, καὶ συμπεπληρώσθω τὸ ΜΝ: τὸ ΜΝ ἄρα τῷ ΗΘ ἴσον τέ ἐστι καὶ ὅμοιον. ἀλλὰ τὸ ΗΘ τῷ ΕΛ ἐστιν ὅμοιον: καὶ τὸ ΜΝ ἄρα τῷ ΕΛ ὅμοιόν ἐστιν: περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὸ ΕΛ τῷ ΜΝ. ἤχθω αὐτῶν διάμετρος ἡ ΖΞ, καὶ καταγεγράφθω τὸ σχῆμα. ἐπεὶ ἴσον ἐστὶ τὸ ΗΘ τοῖς ΕΛ, Γ, ἀλλὰ τὸ ΗΘ τῷ ΜΝ ἴσον ἐστίν, καὶ τὸ ΜΝ ἄρα τοῖς ΕΛ, Γ ἴσον ἐστίν. κοινὸν ἀφῃρήσθω τὸ ΕΛ: λοιπὸς ἄρα ὁ ΨΧΦ γνώμων τῷ Γ ἐστιν ἴσος. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΕ τῇ ΕΒ, ἴσον ἐστὶ καὶ τὸ ΑΝ τῷ ΝΒ, τουτέστι τῷ ΛΟ. κοινὸν προσκείσθω τὸ ΕΞ: ὅλον ἄρα τὸ ΑΞ ἴσον ἐστὶ τῷ ΦΧΨ γνώμονι. ἀλλὰ ὁ ΦΧΨ γνώμων τῷ Γ ἴσος ἐστίν: καὶ τὸ ΑΞ ἄρα τῷ Γ ἴσον ἐστίν. παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται τὸ ΑΞ ὑπερβάλλον εἴδει παραλληλογράμμῳ τῷ ΠΟ ὁμοίῳ ὄντι τῷ Δ, ἐπεὶ καὶ τῷ ΕΛ ἐστιν ὅμοιον τὸ ΟΠ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 307|>","<|""VertexLabel"" -> ""6.30"", ""Text"" -> ""To cut a given finite straight line in extreme and mean ratio."", ""TextWordCount"" -> 12, ""GreekText"" -> ""τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην ἄκρον καὶ μέσον λόγον τεμεῖν."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 6, ""Theorem"" -> 14}, {""Book"" -> 6, ""Theorem"" -> 29}}, ""Proof"" -> ""Let AB be the given finite straight line; thus it is required to cut AB in extreme and mean ratio. On AB let the square BC be described; and let there be applied to AC the parallelogram CD equal to BC and exceeding by the figure AD similar to BC. [VI. 29] Now BC is a square; therefore AD is also a square. And, since BC is equal to CD, let CE be subtracted from each; therefore the remainder BF is equal to the remainder AD. But it is also equiangular with it; therefore in BF, AD the sides about the equal angles are reciprocally proportional; [VI. 14] therefore, as FE is to ED, so is AE to EB. But FE is equal to AB, and ED to AE. Therefore, as BA is to AE, so is AE to EB. And AB is greater than AE; therefore AE is also greater than EB. Therefore the straight line AB has been cut in extreme and mean ratio at E, and the greater segment of it is AE."", ""ProofWordCount"" -> 176, ""GreekProof"" -> ""ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ: δεῖ δὴ τὴν ΑΒ εὐθεῖαν ἄκρον καὶ μέσον λόγον τεμεῖν. Ἀναγεγράφθω ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΒΓ, καὶ παραβεβλήσθω παρὰ τὴν ΑΓ τῇ ΒΓ ἴσον παραλληλόγραμμον τὸ ΓΔ ὑπερβάλλον εἴδει τῷ ΑΔ ὁμοίῳ τῷ ΒΓ. τετράγωνον δέ ἐστι τὸ ΒΓ: τετράγωνον ἄρα ἐστὶ καὶ τὸ ΑΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΒΓ τῷ ΓΔ, κοινὸν ἀφῃρήσθω τὸ ΓΕ: λοιπὸν ἄρα τὸ ΒΖ λοιπῷ τῷ ΑΔ ἐστιν ἴσον. ἔστι δὲ αὐτῷ καὶ ἰσογώνιον: τῶν ΒΖ, ΑΔ ἄρα ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας: ἔστιν ἄρα ὡς ἡ ΖΕ πρὸς τὴν ΕΔ, οὕτως ἡ ΑΕ πρὸς τὴν ΕΒ. ἴση δὲ ἡ μὲν ΖΕ τῇ ΑΒ, ἡ δὲ ΕΔ τῇ ΑΕ. ἔστιν ἄρα ὡς ἡ ΒΑ πρὸς τὴν ΑΕ, οὕτως ἡ ΑΕ πρὸς τὴν ΕΒ. μείζων δὲ ἡ ΑΒ τῆς ΑΕ: μείζων ἄρα καὶ ἡ ΑΕ τῆς ΕΒ. ἡ ἄρα ΑΒ εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Ε, καὶ τὸ μεῖζον αὐτῆς τμῆμά ἐστι τὸ ΑΕ: ὅπερ ἔδει ποιῆσαι."", ""GreekProofWordCount"" -> 167|>","<|""VertexLabel"" -> ""6.31"", ""Text"" -> ""In right - angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle."", ""TextWordCount"" -> 29, ""GreekText"" -> ""ἐν τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις."", ""GreekTextWordCount"" -> 30, ""References"" -> {{""Book"" -> 6, ""Definition"" -> 1}, {""Book"" -> 6, ""Theorem"" -> 8}, {""Book"" -> 6, ""Theorem"" -> 19}}, ""Proof"" -> ""Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC. Let AD be drawn perpendicular. Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another. [VI. 8] And, since ABC is similar to ABD, therefore, as CB is to BA, so is AB to BD. [VI. Def. 1] And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second. [VI. 19] Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA. For the same reason also, as BC is to CD, so is the figure on BC to that on CA; so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC. But BC is equal to BD, DC; therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC."", ""ProofWordCount"" -> 223, ""GreekProof"" -> ""ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν: λέγω, ὅτι τὸ ἀπὸ τῆς ΒΓ εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν ΒΑ, ΑΓ εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις. ἤχθω κάθετος ἡ ΑΔ. ἐπεὶ οὖν ἐν ὀρθογωνίῳ τριγώνῳ τῷ ΑΒΓ ἀπὸ τῆς πρὸς τῷ Α ὀρθῆς γωνίας ἐπὶ τὴν ΒΓ βάσιν κάθετος ἦκται ἡ ΑΔ, τὰ ΑΒΔ, ΑΔΓ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ τῷ ΑΒΓ καὶ ἀλλήλοις. καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓ τῷ ΑΒΔ, ἔστιν ἄρα ὡς ἡ ΓΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΑΒ πρὸς τὴν ΒΔ. καὶ ἐπεὶ τρεῖς εὐθεῖαι ἀνάλογόν εἰσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον. ὡς ἄρα ἡ ΓΒ πρὸς τὴν ΒΔ, οὕτως τὸ ἀπὸ τῆς ΓΒ εἶδος πρὸς τὸ ἀπὸ τῆς ΒΑ τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ ΒΓ πρὸς τὴν ΓΔ, οὕτως τὸ ἀπὸ τῆς ΒΓ εἶδος πρὸς τὸ ἀπὸ τῆς ΓΑ. ὥστε καὶ ὡς ἡ ΒΓ πρὸς τὰς ΒΔ, ΔΓ, οὕτως τὸ ἀπὸ τῆς ΒΓ εἶδος πρὸς τὰ ἀπὸ τῶν ΒΑ, ΑΓ τὰ ὅμοια καὶ ὁμοίως ἀναγραφόμενα. ἴση δὲ ἡ ΒΓ ταῖς ΒΔ, ΔΓ: ἴσον ἄρα καὶ τὸ ἀπὸ τῆς ΒΓ εἶδος τοῖς ἀπὸ τῶν ΒΑ, ΑΓ εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις. ἐν ἄρα τοῖς ὀρθογωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνίαν ὑποτεινούσης πλευρᾶς εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τῶν τὴν ὀρθὴν γωνίαν περιεχουσῶν πλευρῶν εἴδεσι τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 257|>","<|""VertexLabel"" -> ""6.32"", ""Text"" -> ""If two triangles having two sides proportional to two sides be placed together at one angle so that their corresponding sides are also parallel, the remaining sides of the triangles will be in a straight line."", ""TextWordCount"" -> 36, ""GreekText"" -> ""ἐὰν δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν τὰς δύο πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα ὥστε τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ τῶν τριγώνων πλευραὶ ἐπ᾽ εὐθείας ἔσονται."", ""GreekTextWordCount"" -> 32, ""References"" -> {{""Book"" -> 1, ""Theorem"" -> 14}, {""Book"" -> 1, ""Theorem"" -> 29}, {""Book"" -> 1, ""Theorem"" -> 32}, {""Book"" -> 6, ""Theorem"" -> 6}}, ""Proof"" -> ""Let ABC, DCE be two triangles having the two sides BA, AC proportional to the two sides DC, DE, so that, as AB is to AC, so is DC to DE, and AB parallel to DC, and AC to DE; I say that BC is in a straight line with CE. For, since AB is parallel to DC, and the straight line AC has fallen upon them, the alternate angles BAC, ACD are equal to one another. [I. 29] For the same reason the angle CDE is also equal to the angle ACD; so that the angle BAC is equal to the angle CDE. And, since ABC, DCE are two triangles having one angle, the angle at A, equal to one angle, the angle at D, and the sides about the equal angles proportional, so that, as BA is to AC, so is CD to DE, therefore the triangle ABC is equiangular with the triangle DCE; [VI. 6]therefore the angle ABC is equal to the angle DCE. But the angle ACD was also proved equal to the angle BAC; therefore the whole angle ACE is equal to the two angles ABC, BAC. Let the angle ACB be added to each; therefore the angles ACE, ACB are equal to the angles BAC, ACB, CBA. But the angles BAC, ABC, ACB are equal to two right angles; [I. 32] therefore the angles ACE, ACB are also equal to two right angles. Therefore with a straight line AC, and at the point C on it, the two straight lines BC, CE not lying on the same side make the adjacent angles ACE, ACB equal to two right angles; therefore BC is in a straight line with CE. [I. 14]"", ""ProofWordCount"" -> 287, ""GreekProof"" -> ""ἔστω δύο τρίγωνα τὰ ΑΒΓ, ΔΓΕ τὰς δύο πλευρὰς τὰς ΒΑ, ΑΓ ταῖς δυσὶ πλευραῖς ταῖς ΔΓ, ΔΕ ἀνάλογον ἔχοντα, ὡς μὲν τὴν ΑΒ πρὸς τὴν ΑΓ, οὕτως τὴν ΔΓ πρὸς τὴν ΔΕ, παράλληλον δὲ τὴν μὲν ΑΒ τῇ ΔΓ, τὴν δὲ ΑΓ τῇ ΔΕ: λέγω, ὅτι ἐπ᾽ εὐθείας ἐστὶν ἡ ΒΓ τῇ ΓΕ. ἐπεὶ γὰρ παράλληλός ἐστιν ἡ ΑΒ τῇ ΔΓ, καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΑΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ, ΑΓΔ ἴσαι ἀλλήλαις εἰσίν. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΓΔ ἴση ἐστίν. ὥστε καὶ ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση. καὶ ἐπεὶ δύο τρίγωνά ἐστι τὰ ΑΒΓ, ΔΓΕ μίαν γωνίαν τὴν πρὸς τῷ Α μιᾷ γωνίᾳ τῇ πρὸς τῷ Δ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ὡς τὴν ΒΑ πρὸς τὴν ΑΓ, οὕτως τὴν ΓΔ πρὸς τὴν ΔΕ, ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΓΕ τριγώνῳ: ἴση ἄρα ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΔΓΕ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΔ τῇ ὑπὸ ΒΑΓ ἴση: ὅλη ἄρα ἡ ὑπὸ ΑΓΕ δυσὶ ταῖς ὑπὸ ΑΒΓ, ΒΑΓ ἴση ἐστίν. κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ: αἱ ἄρα ὑπὸ ΑΓΕ, ΑΓΒ ταῖς ὑπὸ ΒΑΓ, ΑΓΒ, ΓΒΑ ἴσαι εἰσίν. ἀλλ᾽ αἱ ὑπὸ ΒΑΓ, ΑΒΓ, ΑΓΒ δυσὶν ὀρθαῖς ἴσαι εἰσίν: καὶ αἱ ὑπὸ ΑΓΕ, ΑΓΒ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν. πρὸς δή τινι εὐθείᾳ τῇ ΑΓ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Γ δύο εὐθεῖαι αἱ ΒΓ, ΓΕ μὴ ἐπὶ τὰ αὐτὰ μέρη κείμεναι τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΑΓΕ, ΑΓΒ δυσὶν ὀρθαῖς ἴσας ποιοῦσιν: ἐπ᾽ εὐθείας ἄρα ἐστὶν ἡ ΒΓ τῇ ΓΕ. ἐὰν ἄρα δύο τρίγωνα συντεθῇ κατὰ μίαν γωνίαν τὰς δύο πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα ὥστε τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι, αἱ λοιπαὶ τῶν τριγώνων πλευραὶ ἐπ᾽ εὐθείας ἔσονται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 303|>","<|""VertexLabel"" -> ""6.33"", ""Text"" -> ""In equal circles angles have the same ratio as the circumferences on which they stand, whether they stand at the centres or at the circumferences."", ""TextWordCount"" -> 25, ""GreekText"" -> ""ἐν τοῖς ἴσοις κύκλοις αἱ γωνίαι τὸν αὐτὸν ἔχουσι λόγον ταῖς περιφερείαις, ἐφ᾽ ὧν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι."", ""GreekTextWordCount"" -> 28, ""References"" -> {{""Book"" -> 3, ""Theorem"" -> 27}, {""Book"" -> 5, ""Definition"" -> 5}}, ""Proof"" -> ""Let ABC, DEF be equal circles, and let the angles BGC, EHF be angles at their centres G, H, and the angles BAC, EDF angles at the circumferences; I say that, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. For let any number of consecutive circumferences CK, KL be made equal to the circumference BC, and any number of consecutive circumferences FM, MN equal to the circumference EF; and let GK, GL, HM, HN be joined. Then, since the circumferences BC, CK, KL are equal to one another, the angles BGC, CGK, KGL are also equal to one another; [III. 27] therefore, whatever multiple the circumference BL is of BC, that multiple also is the angle BGL of the angle BGC. For the same reason also, whatever multiple the circumference NE is of EF, that multiple also is the angle NHE of the angle EHF. If then the circumference BL is equal to the circumference EN, the angle BGL is also equal to the angle EHN; [III. 27] if the circumference BL is greater than the circumference EN, the angle BGL is also greater than the angle EHN; and, if less, less. There being then four magnitudes, two circumferences BC, EF, and two angles BGC, EHF, there have been taken, of the circumference BC and the angle BGC equimultiples, namely the circumference BL and the angle BGL, and of the circumference EF and the angle EHF equimultiples, namely the circumference EN and the angle EHN. And it has been proved that, if the circumference BL is in excess of the circumference EN, the angle BGL is also in excess of the angle EHN; if equal, equal; and if less, less. Therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF. [V. Def. 5] But, as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF; for they are doubles respectively. Therefore also, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF."", ""ProofWordCount"" -> 370, ""GreekProof"" -> ""ἔστωσαν ἴσοι κύκλοι οἱ ΑΒΓ, ΔΕΖ, καὶ πρὸς μὲν τοῖς κέντροις αὐτῶν τοῖς Η, Θ γωνίαι ἔστωσαν αἱ ὑπὸ ΒΗΓ, ΕΘΖ, πρὸς δὲ ταῖς περιφερείαις αἱ ὑπὸ ΒΑΓ, ΕΔΖ: λέγω, ὅτι ἐστὶν ὡς ἡ ΒΓ περιφέρεια πρὸς τὴν ΕΖ περιφέρειαν, οὕτως ἥ τε ὑπὸ ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ καὶ ἡ ὑπὸ ΒΑΓ πρὸς τὴν ὑπὸ ΕΔΖ. κείσθωσαν γὰρ τῇ μὲν ΒΓ περιφερείᾳ ἴσαι κατὰ τὸ ἑξῆς ὁσαιδηποτοῦν αἱ ΓΚ, ΚΛ, τῇ δὲ ΕΖ περιφερείᾳ ἴσαι ὁσαιδηποτοῦν αἱ ΖΜ, ΜΝ, καὶ ἐπεζεύχθωσαν αἱ ΗΚ, ΗΛ, ΘΜ, ΘΝ. ἐπεὶ οὖν ἴσαι εἰσὶν αἱ ΒΓ, ΓΚ, ΚΛ περιφέρειαι ἀλλήλαις, ἴσαι εἰσὶ καὶ αἱ ὑπὸ ΒΗΓ, ΓΗΚ, ΚΗΛ γωνίαι ἀλλήλαις: ὁσαπλασίων ἄρα ἐστὶν ἡ ΒΛ περιφέρεια τῆς ΒΓ, τοσαυταπλασίων ἐστὶ καὶ ἡ ὑπὸ ΒΗΛ γωνία τῆς ὑπὸ ΒΗΓ. διὰ τὰ αὐτὰ δὴ καὶ ὁσαπλασίων ἐστὶν ἡ ΝΕ περιφέρεια τῆς ΕΖ, τοσαυταπλασίων ἐστὶ καὶ ἡ ὑπὸ ΝΘΕ γωνία τῆς ὑπὸ ΕΘΖ. εἰ ἄρα ἴση ἐστὶν ἡ ΒΛ περιφέρεια τῇ ΕΝ περιφερείᾳ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΒΗΛ τῇ ὑπὸ ΕΘΝ, καὶ εἰ μείζων ἐστὶν ἡ ΒΛ περιφέρεια τῆς ΕΝ περιφερείας, μείζων ἐστὶ καὶ ἡ ὑπὸ ΒΗΛ γωνία τῆς ὑπὸ ΕΘΝ, καὶ εἰ ἐλάσσων, ἐλάσσων. τεσσάρων δὴ ὄντων μεγεθῶν, δύο μὲν περιφερειῶν τῶν ΒΓ, ΕΖ, δύο δὲ γωνιῶν τῶν ὑπὸ ΒΗΓ, ΕΘΖ, εἴληπται τῆς μὲν ΒΓ περιφερείας καὶ τῆς ὑπὸ ΒΗΓ γωνίας ἰσάκις πολλαπλασίων ἥ τε ΒΛ περιφέρεια καὶ ἡ ὑπὸ ΒΗΛ γωνία, τῆς δὲ ΕΖ περιφερείας καὶ τῆς ὑπὸ ΕΘΖ γωνίας ἥ τε ΕΝ περιφέρεια καὶ ἡ ὑπὸ ΕΘΝ γωνία. καὶ δέδεικται, ὅτι εἰ ὑπερέχει ἡ ΒΛ περιφέρεια τῆς ΕΝ περιφερείας, ὑπερέχει καὶ ἡ ὑπὸ ΒΗΛ γωνία τῆς ὑπὸ ΕΘΝ γωνίας, καὶ εἰ ἴση, ἴση, καὶ εἰ ἐλάσσων, ἐλάσσων. ἔστιν ἄρα, ὡς ἡ ΒΓ περιφέρεια πρὸς τὴν ΕΖ, οὕτως ἡ ὑπὸ ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ. ἀλλ᾽ ὡς ἡ ὑπὸ ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ, οὕτως ἡ ὑπὸ ΒΑΓ πρὸς τὴν ὑπὸ ΕΔΖ: διπλασία γὰρ ἑκατέρα ἑκατέρας. καὶ ὡς ἄρα ἡ ΒΓ περιφέρεια πρὸς τὴν ΕΖ περιφέρειαν, οὕτως ἥ τε ὑπὸ ΒΗΓ γωνία πρὸς τὴν ὑπὸ ΕΘΖ καὶ ἡ ὑπὸ ΒΑΓ πρὸς τὴν ὑπὸ ΕΔΖ. ἐν ἄρα τοῖς ἴσοις κύκλοις αἱ γωνίαι τὸν αὐτὸν ἔχουσι λόγον ταῖς περιφερείαις, ἐφ᾽ ὧν βεβήκασιν, ἐάν τε πρὸς τοῖς κέντροις ἐάν τε πρὸς ταῖς περιφερείαις ὦσι βεβηκυῖαι: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 378|>","<|""VertexLabel"" -> ""7.1"", ""Text"" -> ""Two unequal numbers being set out, and the less being continually subtracted in turn from the greater, if the number which is left never measures the one before it until an unit is left, the original numbers will be prime to one another."", ""TextWordCount"" -> 43, ""GreekText"" -> ""δύο ἀριθμῶν ἀνίσων ἐκκειμένων, ἀνθυφαιρουμένου δὲ ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος, ἐὰν ὁ λειπόμενος μηδέποτε καταμετρῇ τὸν πρὸ ἑαυτοῦ, ἕως οὗ λειφθῇ μονάς, οἱ ἐξ ἀρχῆς ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ἔσονται."", ""GreekTextWordCount"" -> 32, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 12}}, ""Proof"" -> ""For, if AB, CD are not prime to one another, some number will measure them. Let a number measure them, and let it be E; let CD, measuring BF, leave FA less than itself, let AF, measuring DG, leave GC less than itself, and let GC, measuring FH, leave an unit HA. Since, then, E measures CD, and CD measures BF, therefore E also measures BF. But it also measures the whole BA; therefore it will also measure the remainder AF. But AF measures DG; therefore E also measures DG. But it also measures the whole DC therefore it will also measure the remainder CG. But CG measures FH; therefore E also measures FH. But it also measures the whole FA; therefore it will also measure the remainder, the unit AH, though it is a number: which is impossible. Therefore no number will measure the numbers AB, CD; therefore AB, CD are prime to one another. [VII. Def. 12]"", ""ProofWordCount"" -> 159, ""GreekProof"" -> ""δύο γὰρ ἀνίσων ἀριθμῶν τῶν ΑΒ, ΓΔ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος ὁ λειπόμενος μηδέποτε καταμετρείτω τὸν πρὸ ἑαυτοῦ, ἕως οὗ λειφθῇ μονάς: λέγω, ὅτι οἱ ΑΒ, ΓΔ πρῶτοι πρὸς ἀλλήλους εἰσίν, τουτέστιν ὅτι τοὺς ΑΒ, ΓΔ μονὰς μόνη μετρεῖ. εἰ γὰρ μή εἰσιν οἱ ΑΒ, ΓΔ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις αὐτοὺς ἀριθμός. μετρείτω, καὶ ἔστω ὁ Ε: καὶ ὁ μὲν ΓΔ τὸν ΒΖ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΖΑ, ὁ δὲ ΑΖ τὸν ΔΗ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΗΓ, ὁ δὲ ΗΓ τὸν ΖΘ μετρῶν λειπέτω μονάδα τὴν ΘΑ. ἐπεὶ οὖν ὁ Ε τὸν ΓΔ μετρεῖ, ὁ δὲ ΓΔ τὸν ΒΖ μετρεῖ καὶ ὁ Ε ἄρα τὸν ΒΖ μετρεῖ: μετρεῖ δὲ καὶ ὅλον τὸν ΒΑ: καὶ λοιπὸν ἄρα τὸν ΑΖ μετρήσει. ὁ δὲ ΑΖ τὸν ΔΗ μετρεῖ: καὶ ὁ Ε ἄρα τὸν ΔΗ μετρεῖ: μετρεῖ δὲ καὶ ὅλον τὸν ΔΓ: καὶ λοιπὸν ἄρα τὸν ΓΗ μετρήσει. ὁ δὲ ΓΗ τὸν ΖΘ μετρεῖ: καὶ ὁ Ε ἄρα τὸν ΖΘ μετρεῖ: μετρεῖ δὲ καὶ ὅλον τὸν ΖΑ: καὶ λοιπὴν ἄρα τὴν ΑΘ μονάδα μετρήσει ἀριθμὸς ὤν: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς ΑΒ, ΓΔ ἀριθμοὺς μετρήσει τις ἀριθμός: οἱ ΑΒ, ΓΔ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 202|>","<|""VertexLabel"" -> ""7.2"", ""Text"" -> ""Given two numbers not prime to one another, to find their greatest common measure."", ""TextWordCount"" -> 14, ""GreekText"" -> ""δύο ἀριθμῶν δοθέντων μὴ πρώτων πρὸς ἀλλήλους τὸ μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν."", ""GreekTextWordCount"" -> 13, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 1}}, ""Proof"" -> ""Let AB, CD be the two given numbers not prime to one another. Thus it is required to find the greatest common measure of AB, CD. If now CD measures AB —and it also measures itself— CD is a common measure of CD, AB. And it is manifest that it is also the greatest; for no greater number than CD will measure CD. But, if CD does not measure AB, then, the less of the numbers AB, CD being continually subtracted from the greater, some number will be left which will measure the one before it. For an unit will not be left; otherwise AB, CD will be prime to one another [VII. 1], which is contrary to the hypothesis. Therefore some number will be left which will measure the one before it. Now let CD, measuring BE, leave EA less than itself, let EA, measuring DF, leave FC less than itself, and let CF measure AE. Since then, CF measures AE, and AE measures DF, therefore CF will also measure DF. But it also measures itself; therefore it will also measure the whole CD. But CD measures BE; therefore CF also measures BE. But it also measures EA; therefore it will also measure the whole BA. But it also measures CD; therefore CF measures AB, CD. Therefore CF is a common measure of AB, CD. I say next that it is also the greatest. For, if CF is not the greatest common measure of AB, CD, some number which is greater than CF will measure the numbers AB, CD. Let such a number measure them, and let it be G. Now, since G measures CD, while CD measures BE, G also measures BE. But it also measures the whole BA; therefore it will also measure the remainder AE. But AE measures DF; therefore G will also measure DF. But it also measures the whole DC; therefore it will also measure the remainder CF, that is, the greater will measure the less: which is impossible. Therefore no number which is greater than CF will measure the numbers AB, CD; therefore CF is the greatest common measure of AB, CD."", ""ProofWordCount"" -> 359, ""GreekProof"" -> ""ἔστωσαν οἱ δοθέντες δύο ἀριθμοὶ μὴ πρῶτοι πρὸς ἀλλήλους οἱ ΑΒ, ΓΔ. δεῖ δὴ τῶν ΑΒ, ΓΔ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν. εἰ μὲν οὖν ὁ ΓΔ τὸν ΑΒ μετρεῖ, μετρεῖ δὲ καὶ ἑαυτόν, ὁ ΓΔ ἄρα τῶν ΓΔ, ΑΒ κοινὸν μέτρον ἐστίν. καὶ φανερόν, ὅτι καὶ μέγιστον: οὐδεὶς γὰρ μείζων τοῦ ΓΔ τὸν ΓΔ μετρήσει. εἰ δὲ οὐ μετρεῖ ὁ ΓΔ τὸν ΑΒ, τῶν ΑΒ, ΓΔ ἀνθυφαιρουμένου ἀεὶ τοῦ ἐλάσσονος ἀπὸ τοῦ μείζονος λειφθήσεταί τις ἀριθμός, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ. μονὰς μὲν γὰρ οὐ λειφθήσεται: εἰ δὲ μή, ἔσονται οἱ ΑΒ, ΓΔ πρῶτοι πρὸς ἀλλήλους: ὅπερ οὐχ ὑπόκειται. λειφθήσεταί τις ἄρα ἀριθμός, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ. καὶ ὁ μὲν ΓΔ τὸν ΒΕ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΕΑ, ὁ δὲ ΕΑ τὸν ΔΖ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΖΓ, ὁ δὲ ΓΖ τὸν ΑΕ μετρείτω. ἐπεὶ οὖν ὁ ΓΖ τὸν ΑΕ μετρεῖ, ὁ δὲ ΑΕ τὸν ΔΖ μετρεῖ, καὶ ὁ ΓΖ ἄρα τὸν ΔΖ μετρήσει: μετρεῖ δὲ καὶ ἑαυτόν: καὶ ὅλον ἄρα τὸν ΓΔ μετρήσει. ὁ δὲ ΓΔ τὸν ΒΕ μετρεῖ: καὶ ὁ ΓΖ ἄρα τὸν ΒΕ μετρεῖ: μετρεῖ δὲ καὶ τὸν ΕΑ: καὶ ὅλον ἄρα τὸν ΒΑ μετρήσει: μετρεῖ δὲ καὶ τὸν ΓΔ: ὁ ΓΖ ἄρα τοὺς ΑΒ, ΓΔ μετρεῖ. ὁ ΓΖ ἄρα τῶν ΑΒ, ΓΔ κοινὸν μέτρον ἐστίν. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή ἐστιν ὁ ΓΖ τῶν ΑΒ, ΓΔ μέγιστον κοινὸν μέτρον, μετρήσει τις τοὺς ΑΒ, ΓΔ ἀριθμοὺς ἀριθμὸς μείζων ὢν τοῦ ΓΖ. μετρείτω, καὶ ἔστω ὁ Η. καὶ ἐπεὶ ὁ Η τὸν ΓΔ μετρεῖ, ὁ δὲ ΓΔ τὸν ΒΕ μετρεῖ, καὶ ὁ Η ἄρα τὸν ΒΕ μετρεῖ: μετρεῖ δὲ καὶ ὅλον τὸν ΒΑ: καὶ λοιπὸν ἄρα τὸν ΑΕ μετρήσει. ὁ δὲ ΑΕ τὸν ΔΖ μετρεῖ: καὶ ὁ Η ἄρα τὸν ΔΖ μετρήσει: μετρεῖ δὲ καὶ ὅλον τὸν ΔΓ: καὶ λοιπὸν ἄρα τὸν ΓΖ μετρήσει ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα τοὺς ΑΒ, ΓΔ ἀριθμοὺς ἀριθμός τις μετρήσει μείζων ὢν τοῦ ΓΖ: ὁ ΓΖ ἄρα τῶν ΑΒ, ΓΔ μέγιστόν ἐστι κοινὸν μέτρον: ὅπερ ἔδει δεῖξαι. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἀριθμὸς δύο ἀριθμοὺς μετρῇ, καὶ τὸ μέγιστον αὐτῶν κοινὸν μέτρον μετρήσει: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 357|>","<|""VertexLabel"" -> ""7.3"", ""Text"" -> ""Given three numbers not prime to one another, to find their greatest common measure."", ""TextWordCount"" -> 14, ""GreekText"" -> ""τριῶν ἀριθμῶν δοθέντων μὴ πρώτων πρὸς ἀλλήλους τὸ μέγιστον αὐτῶν κοινὸν μέτρον εὑρεῖν."", ""GreekTextWordCount"" -> 13, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 2}}, ""Proof"" -> ""Let A, B, C be the three given numbers not prime to one another; thus it is required to find the greatest common measure of A, B, C. For let the greatest common measure, D, of the two numbers A, B be taken; [VII. 2] then D either measures, or does not measure, C. First, let it measure it. But it measures A, B also; therefore D measures A, B, C; therefore D is a common measure of A, B, C. I say that it is also the greatest. For, if D is not the greatest common measure of A, B, C, some number which is greater than D will measure the numbers A, B, C. Let such a number measure them, and let it be E. Since then E measures A, B, C, it will also measure A, B; therefore it will also measure the greatest common measure of A, B. [VII. 2] But the greatest common measure of A, B is D; therefore E measures D, the greater the less: which is impossible. Therefore no number which is greater than D will measure the numbers A, B, C; therefore D is the greatest common measure of A, B, C. Next, let D not measure C; I say first that C, D are not prime to one another. For, since A, B, C are not prime to one another, some number will measure them. Now that which measures A, B, C will also measure A, B, and will measure D, the greatest common measure of A, B. [VII. 2] But it measures C also; therefore some number will measure the numbers D, C; therefore D, C are not prime to one another. Let then their greatest common measure E be taken. [VII. 2] Then, since E measures D, and D measures A, B, therefore E also measures A, B. But it measures C also; therefore E measures A, B, C; therefore E is a common measure of A, B, C. I say next that it is also the greatest. For, if E is not the greatest common measure of A, B, C, some number which is greater than E will measure the numbers A, B, C. Let such a number measure them, and let it be F. Now, since F measures A, B, C, it also measures A, B; therefore it will also measure the greatest common measure of A, B. [VII. 2] But the greatest common measure of A, B is D; therefore F measures D. And it measures C also; therefore F measures D, C; therefore it will also measure the greatest common measure of D, C. [VII. 2] But the greatest common measure of D, C is E; therefore F measures E, the greater the less: which is impossible. Therefore no number which is greater than E will measure the numbers A, B, C; therefore E is the greatest common measure of A, B, C."", ""ProofWordCount"" -> 489, ""GreekProof"" -> ""ἔστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ μὴ πρῶτοι πρὸς ἀλλήλους οἱ α, Β, Γ: δεῖ δὴ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον εὑρεῖν. εἰλήφθω γὰρ δύο τῶν Α, Β τὸ μέγιστον κοινὸν μέτρον ὁ Δ: ὁ δὴ Δ τὸν Γ ἤτοι μετρεῖ ἢ οὐ μετρεῖ. μετρείτω πρότερον: μετρεῖ δὲ καὶ τοὺς Α, Β: ὁ Δ ἄρα τοὺς Α, Β, Γ μετρεῖ: ὁ Δ ἄρα τῶν Α, Β, Γ κοινὸν μέτρον ἐστίν. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή ἐστιν ὁ Δ τῶν Α, Β, Γ μέγιστον κοινὸν μέτρον, μετρήσει τις τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμὸς μείζων ὢν τοῦ Δ. μετρείτω, καὶ ἔστω ὁ Ε. ἐπεὶ οὖν ὁ Ε τοὺς Α, Β, Γ μετρεῖ, καὶ τοὺς Α, Β ἄρα μετρήσει: καὶ τὸ τῶν Α, Β ἄρα μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Α, Β μέγιστον κοινὸν μέτρον ἐστὶν ὁ Δ: ὁ Ε ἄρα τὸν Δ μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμός τις μετρήσει μείζων ὢν τοῦ Δ: ὁ Δ ἄρα τῶν Α, Β, Γ μέγιστόν ἐστι κοινὸν μέτρον. μὴ μετρείτω δὴ ὁ Δ τὸν Γ: λέγω πρῶτον, ὅτι οἱ Γ, Δ οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους. ἐπεὶ γὰρ οἱ Α, Β, Γ οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους, μετρήσει τις αὐτοὺς ἀριθμός. ὁ δὴ τοὺς Α, Β, Γ μετρῶν καὶ τοὺς Α, Β μετρήσει, καὶ τὸ τῶν Α, Β μέγιστον κοινὸν μέτρον τὸν Δ μετρήσει: μετρεῖ δὲ καὶ τὸν Γ: τοὺς Δ, Γ ἄρα ἀριθμοὺς ἀριθμός τις μετρήσει: οἱ Δ, Γ ἄρα οὔκ εἰσι πρῶτοι πρὸς ἀλλήλους. εἰλήφθω οὖν αὐτῶν τὸ μέγιστον κοινὸν μέτρον ὁ Ε. καὶ ἐπεὶ ὁ Ε τὸν Δ μετρεῖ, ὁ δὲ Δ τοὺς Α, Β μετρεῖ, καὶ ὁ Ε ἄρα τοὺς Α, Β μετρεῖ: μετρεῖ δὲ καὶ τὸν Γ: ὁ Ε ἄρα τοὺς Α, Β, Γ μετρεῖ: ὁ Ε ἄρα τῶν Α, Β, Γ κοινόν ἐστι μέτρον. λέγω δή, ὅτι καὶ μέγιστον. εἰ γὰρ μή ἐστιν ὁ Ε τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον, μετρήσει τις τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμὸς μείζων ὢν τοῦ Ε. μετρείτω, καὶ ἔστω ὁ Ζ. καὶ ἐπεὶ ὁ Ζ τοὺς Α, Β, Γ μετρεῖ, καὶ τοὺς Α, Β μετρεῖ: καὶ τὸ τῶν Α, Β ἄρα μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Α, Β μέγιστον κοινὸν μέτρον ἐστὶν ὁ Δ: ὁ Ζ ἄρα τὸν Δ μετρεῖ: μετρεῖ δὲ καὶ τὸν Γ: ὁ Ζ ἄρα τοὺς Δ, Γ μετρεῖ: καὶ τὸ τῶν Δ, Γ ἄρα μέγιστον κοινὸν μέτρον μετρήσει. τὸ δὲ τῶν Δ, Γ μέγιστον κοινὸν μέτρον ἐστὶν ὁ Ε: ὁ Ζ ἄρα τὸν Ε μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Α, Β, Γ ἀριθμοὺς ἀριθμός τις μετρήσει μείζων ὢν τοῦ Ε: ὁ Ε ἄρα τῶν Α, Β, Γ μέγιστόν ἐστι κοινὸν μέτρον: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 464|>","<|""VertexLabel"" -> ""7.4"", ""Text"" -> ""Any number is either a part or parts of any number, the less of the greater."", ""TextWordCount"" -> 16, ""GreekText"" -> ""ἅπας ἀριθμὸς παντὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ μείζονος ἤτοι μέρος ἐστὶν ἢ μέρη."", ""GreekTextWordCount"" -> 13, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 2}}, ""Proof"" -> ""Let A, BC be two numbers, and let BC be the less; I say that BC is either a part, or parts, of A. For A, BC are either prime to one another or not. First, let A, BC be prime to one another. Then, if BC be divided into the units in it, each unit of those in BC will be some part of A; so that BC is parts of A. Next let A, BC not be prime to one another; then BC either measures, or does not measure, A. If now BC measures A, BC is a part of A. But, if not, let the greatest common measure D of A, BC be taken; [VII. 2] and let BC be divided into the numbers equal to D, namely BE, EF, FC. Now, since D measures A, D is a part of A. But D is equal to each of the numbers BE, EF, FC; therefore each of the numbers BE, EF, FC is also a part of A; so that BC is parts of A."", ""ProofWordCount"" -> 178, ""GreekProof"" -> ""ἔστωσαν δύο ἀριθμοὶ οἱ Α, ΒΓ, καὶ ἔστω ἐλάσσων ὁ ΒΓ: λέγω, ὅτι ὁ ΒΓ τοῦ Α ἤτοι μέρος ἐστὶν ἢ μέρη. οἱ Α, ΒΓ γὰρ ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν ἢ οὔ. ἔστωσαν πρότερον οἱ Α, ΒΓ πρῶτοι πρὸς ἀλλήλους. διαιρεθέντος δὴ τοῦ ΒΓ εἰς τὰς ἐν αὐτῷ μονάδας ἔσται ἑκάστη μονὰς τῶν ἐν τῷ ΒΓ μέρος τι τοῦ Α: ὥστε μέρη ἐστὶν ὁ ΒΓ τοῦ Α. μὴ ἔστωσαν δὴ οἱ Α, ΒΓ πρῶτοι πρὸς ἀλλήλους: ὁ δὴ ΒΓ τὸν Α ἤτοι μετρεῖ ἢ οὐ μετρεῖ. εἰ μὲν οὖν ὁ ΒΓ τὸν Α μετρεῖ, μέρος ἐστὶν ὁ ΒΓ τοῦ Α. εἰ δὲ οὔ, εἰλήφθω τῶν Α, ΒΓ μέγιστον κοινὸν μέτρον ὁ Δ, καὶ διῃρήσθω ὁ ΒΓ εἰς τοὺς τῷ Δ ἴσους τοὺς ΒΕ, ΕΖ, ΖΓ. καὶ ἐπεὶ ὁ Δ τὸν Α μετρεῖ, μέρος ἐστὶν ὁ Δ τοῦ Α: ἴσος δὲ ὁ Δ ἑκάστῳ τῶν ΒΕ, ΕΖ, ΖΓ: καὶ ἕκαστος ἄρα τῶν ΒΕ, ΕΖ, ΖΓ τοῦ Α μέρος ἐστίν: ὥστε μέρη ἐστὶν ὁ ΒΓ τοῦ Α. ἅπας ἄρα ἀριθμὸς παντὸς ἀριθμοῦ ὁ ἐλάσσων τοῦ μείζονος ἤτοι μέρος ἐστὶν ἢ μέρη: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 183|>","<|""VertexLabel"" -> ""7.5"", ""Text"" -> ""If a number be a part of a number, and another be the same part of another, the sum will also be the same part of the sum that the one is of the one."", ""TextWordCount"" -> 35, ""GreekText"" -> ""ἐὰν ἀριθμός ἀριθμοῦ μέρος ᾖ, καὶ ἕτερος ἑτέρου τὸ αὐτὸ μέρος ᾖ, καὶ συναμφότερος συναμφοτέρου τὸ αὐτὸ μέρος ἔσται, ὅπερ ὁ εἷς τοῦ ἑνός."", ""GreekTextWordCount"" -> 24, ""References"" -> {}, ""Proof"" -> ""For let the number A be a part of BC, and another, D, the same part of another EF that A is of BC; I say that the sum of A, D is also the same part of the sum of BC, EF that A is of BC. For since, whatever part A is of BC, D is also the same part of EF, therefore, as many numbers as there are in BC equal to A, so many numbers are there also in EF equal to D. Let BC be divided into the numbers equal to A, namely BG, GC, and EF into the numbers equal to D, namely EH, HF; then the multitude of BG, GC will be equal to the multitude of EH, HF. And, since BG is equal to A, and EH to D, therefore BG, EH are also equal to A, D. For the same reason GC, HF are also equal to A, D. Therefore, as many numbers as there are in BC equal to A, so many are there also in BC, EF equal to A, D. Therefore, whatever multiple BC is of A, the same multiple also is the sum of BC, EF of the sum of A, D. Therefore, whatever part A is of BC, the same part also is the sum of A, D of the sum of BC, EF."", ""ProofWordCount"" -> 228, ""GreekProof"" -> ""ἀριθμὸς γὰρ ὁ Α ἀριθμοῦ τοῦ ΒΓ μέρος ἔστω, καὶ ἕτερος ὁ Δ ἑτέρου τοῦ ΕΖ τὸ αὐτὸ μέρος, ὅπερ ὁ Α τοῦ ΒΓ: λέγω, ὅτι καὶ συναμφότερος ὁ Α, Δ συναμφοτέρου τοῦ ΒΓ, ΕΖ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὁ Α τοῦ ΒΓ. ἐπεὶ γάρ, ὃ μέρος ἐστὶν ὁ Α τοῦ ΒΓ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Δ τοῦ ΕΖ, ὅσοι ἄρα εἰσὶν ἐν τῷ ΒΓ ἀριθμοὶ ἴσοι τῷ Α, τοσοῦτοί εἰσι καὶ ἐν τῷ ΕΖ ἀριθμοὶ ἴσοι τῷ Δ. διῃρήσθω ὁ μὲν ΒΓ εἰς τοὺς τῷ Α ἴσους τοὺς ΒΗ, ΗΓ, ὁ δὲ ΕΖ εἰς τοὺς τῷ Δ ἴσους τοὺς ΕΘ, ΘΖ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΓ τῷ πλήθει τῶν ΕΘ, ΘΖ. καὶ ἐπεὶ ἴσος ἐστὶν ὁ μὲν ΒΗ τῷ Α, ὁ δὲ ΕΘ τῷ Δ, καὶ οἱ ΒΗ, ΕΘ ἄρα τοῖς Α, Δ ἴσοι. διὰ τὰ αὐτὰ δὴ καὶ οἱ ΗΓ, ΘΖ τοῖς Α, Δ. ὅσοι ἄρα εἰσὶν ἐν τῷ ΒΓ ἀριθμοὶ ἴσοι τῷ Α, τοσοῦτοί εἰσι καὶ ἐν τοῖς ΒΓ, ΕΖ ἴσοι τοῖς α, Δ. ὁσαπλασίων ἄρα ἐστὶν ὁ ΒΓ τοῦ Α, τοσαυταπλασίων ἐστὶ καὶ συναμφότερος ὁ ΒΓ, ΕΖ συναμφοτέρου τοῦ Α, Δ. ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ ΒΓ, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ Α, Δ συναμφοτέρου τοῦ ΒΓ, ΕΖ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 215|>","<|""VertexLabel"" -> ""7.6"", ""Text"" -> ""If a number be parts of a number, and another be the same parts of another, the sum will also be the same parts of the sum that the one is of the one."", ""TextWordCount"" -> 34, ""GreekText"" -> ""ἐὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, καὶ ἕτερος ἑτέρου τὰ αὐτὰ μέρη ᾖ, καὶ συναμφότερος συναμφοτέρου τὰ αὐτὰ μέρη ἔσται, ὅπερ ὁ εἷς τοῦ ἑνός."", ""GreekTextWordCount"" -> 24, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 5}}, ""Proof"" -> ""For let the number AB be parts of the number C, and another, DE, the same parts of another, F, that AB is of C; I say that the sum of AB, DE is also the same parts of the sum of C, F that AB is of C. For since, whatever parts AB is of C, DE is also the same parts of F, therefore, as many parts of C as there are in AB, so many parts of F are there also in DE. Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE; thus the multitude of AG, GB will be equal to the multitude of DH, HE. And since, whatever part AG is of C, the same part is DH of F also, therefore, whatever part AG is of C, the same part also is the sum of AG, DH of the sum of C, F. [VII. 5] For the same reason, whatever part GB is of C, the same part also is the sum of GB, HE of the sum of C, F. Therefore, whatever parts AB is of C, the same parts also is the sum of AB, DE of the sum of C, F."", ""ProofWordCount"" -> 213, ""GreekProof"" -> ""ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ Γ μέρη ἔστω, καὶ ἕτερος ὁ ΔΕ ἑτέρου τοῦ Ζ τὰ αὐτὰ μέρη, ἅπερ ὁ ΑΒ τοῦ Γ: λέγω, ὅτι καὶ συναμφότερος ὁ ΑΒ, ΔΕ συναμφοτέρου τοῦ Γ, Ζ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὁ ΑΒ τοῦ Γ. ἐπεὶ γάρ, ἃ μέρη ἐστὶν ὁ ΑΒ τοῦ Γ, τὰ αὐτὰ μέρη καὶ ὁ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μέρη τοῦ Γ, τοσαῦτά ἐστι καὶ ἐν τῷ ΔΕ μέρη τοῦ Ζ. διῃρήσθω ὁ μὲν ΑΒ εἰς τὰ τοῦ Γ μέρη τὰ ΑΗ, ΗΒ, ὁ δὲ ΔΕ εἰς τὰ τοῦ Ζ μέρη τὰ ΔΘ, ΘΕ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει τῶν ΔΘ, ΘΕ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΔΘ τοῦ Ζ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΗ τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ ΑΗ, ΔΘ συναμφοτέρου τοῦ Γ, Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὃ μέρος ἐστὶν ὁ ΗΒ τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ ΗΒ, ΘΕ συναμφοτέρου τοῦ Γ, Ζ. ἃ ἄρα μέρη ἐστὶν ὁ ΑΒ τοῦ Γ, τὰ αὐτὰ μέρη ἐστὶ καὶ συναμφότερος ὁ ΑΒ, ΔΕ συναμφοτέρου τοῦ Γ, Ζ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 203|>","<|""VertexLabel"" -> ""7.7"", ""Text"" -> ""If a number be that part of a number, which a number subtracted is of a number subtracted, the remainder will also be the same part of the remainder that the whole is of the whole."", ""TextWordCount"" -> 36, ""GreekText"" -> ""ἐὰν ἀριθμὸς ἀριθμοῦ μέρος ᾖ, ὅπερ ἀφαιρεθεὶς ἀφαιρεθέντος, καὶ ὁ λοιπὸς τοῦ λοιποῦ τὸ αὐτὸ μέρος ἔσται, ὅπερ ὁ ὅλος τοῦ ὅλου."", ""GreekTextWordCount"" -> 22, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 5}}, ""Proof"" -> ""For let the number AB be that part of the number CD which AE subtracted is of CF subtracted; I say that the remainder EB is also the same part of the remainder FD that the whole AB is of the whole CD. For, whatever part AE is of CF, the same part also let EB be of CG. Now since, whatever part AE is of CF, the same part also is EB of CG, therefore, whatever part AE is of CF, the same part also is AB of GF. [VII. 5] But, whatever part AE is of CF, the same part also, by hypothesis, is AB of CD; therefore, whatever part AB is of GF, the same part is it of CD also; therefore GF is equal to CD. Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD. Now since, whatever part AE is of CF, the same part also is EB of GC, while GC is equal to FD, therefore, whatever part AE is of CF, the same part also is EB of FD. But, whatever part AE is of CF, the same part also is AB of CD; therefore also the remainder EB is the same part of the remainder FD that the whole AB is of the whole CD."", ""ProofWordCount"" -> 220, ""GreekProof"" -> ""ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ ΓΔ μέρος ἔστω, ὅπερ ἀφαιρεθεὶς ὁ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ: λέγω, ὅτι καὶ λοιπὸς ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΑΒ ὅλου τοῦ ΓΔ. ὃ γὰρ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἔστω καὶ ὁ ΕΒ τοῦ ΓΗ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΒ τοῦ ΓΗ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΗΖ. ὃ δὲ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ὑπόκειται καὶ ὁ ΑΒ τοῦ ΓΔ: ὃ ἄρα μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΗΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ τοῦ ΓΔ: ἴσος ἄρα ἐστὶν ὁ ΗΖ τῷ ΓΔ. κοινὸς ἀφῃρήσθω ὁ ΓΖ: λοιπὸς ἄρα ὁ ΗΓ λοιπῷ τῷ ΖΔ ἐστιν ἴσος. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΒ τοῦ ΗΓ, ἴσος δὲ ὁ ΗΓ τῷ ΖΔ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΒ τοῦ ΖΔ. ἀλλὰ ὃ μέρος ἐστὶν ὁ ΑΕ τοῦ ΓΖ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΓΔ: καὶ λοιπὸς ἄρα ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΑΒ ὅλου τοῦ ΓΔ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 221|>","<|""VertexLabel"" -> ""7.8"", ""Text"" -> ""If a number be the same parts of a number that a number subtracted is of a number subtracted, the remainder will also be the same parts of the remainder that the whole is of the whole."", ""TextWordCount"" -> 37, ""GreekText"" -> ""ἐὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, ἅπερ ἀφαιρεθεὶς ἀφαιρεθέντος, καὶ ὁ λοιπὸς τοῦ λοιποῦ τὰ αὐτὰ μέρη ἔσται, ἅπερ ὁ ὅλος τοῦ ὅλου."", ""GreekTextWordCount"" -> 22, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 7}}, ""Proof"" -> ""For let the number AB be the same parts of the number CD that AE subtracted is of CF subtracted; I say that the remainder EB is also the same parts of the remainder FD that the whole AB is of the whole CD. For let GH be made equal to AB. Therefore, whatever parts GH is of CD, the same parts also is AE of CF. Let GH be divided into the parts of CD, namely GK, KH, and AE into the parts of CF, namely AL, LE; thus the multitude of GK, KH will be equal to the multitude of AL, LE. Now since, whatever part GK is of CD, the same part also is AL of CF, while. CD is greater than CF, therefore GK is also greater than AL. Let GM be made equal to AL. Therefore, whatever part GK is of CD, the same part also is GM of CF; therefore also the remainder MK is the same part of the remainder FD that the whole GK is of the whole CD. [VII. 7] Again, since, whatever part KH is of CD, the same part also is EL of CF, while CD is greater than CF, therefore HK is also greater than EL. Let KN be made equal to EL. Therefore, whatever part KH is of CD, the same part also is KN of CF; therefore also the remainder NH is the same part of the remainder FD that the whole KH is of the whole CD. [VII. 7] But the remainder MK was also proved to be the same part of the remainder FD that the whole GK is of the whole CD; therefore also the sum of MK, NH is the same parts of DF that the whole HG is of the whole CD. But the sum of MK, NH is equal to EB, and HG is equal to BA; therefore the remainder EB is the same parts of the remainder FD that the whole AB is of the whole CD."", ""ProofWordCount"" -> 338, ""GreekProof"" -> ""ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ ΓΔ μέρη ἔστω, ἅπερ ἀφαιρεθεὶς ὁ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ: λέγω, ὅτι καὶ λοιπὸς ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὅλος ὁ ΑΒ ὅλου τοῦ ΓΔ. κείσθω γὰρ τῷ ΑΒ ἴσος ὁ ΗΘ. ἃ ἄρα μέρη ἐστὶν ὁ ΗΘ τοῦ ΓΔ, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ ΑΕ τοῦ ΓΖ. διῃρήσθω ὁ μὲν ΗΘ εἰς τὰ τοῦ ΓΔ μέρη τὰ ΗΚ, ΚΘ, ὁ δὲ ΑΕ εἰς τὰ τοῦ ΓΖ μέρη τὰ ΑΛ, ΛΕ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΗΚ, ΚΘ τῷ πλήθει τῶν ΑΛ, ΛΕ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΗΚ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΛ τοῦ ΓΖ, μείζων δὲ ὁ ΓΔ τοῦ ΓΖ, μείζων ἄρα καὶ ὁ ΗΚ τοῦ ΑΛ. κείσθω τῷ ΑΛ ἴσος ὁ ΗΜ. ὃ ἄρα μέρος ἐστὶν ὁ ΗΚ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΗΜ τοῦ ΓΖ: καὶ λοιπὸς ἄρα ὁ ΜΚ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΗΚ ὅλου τοῦ ΓΔ. πάλιν ἐπεί, ὃ μέρος ἐστὶν ὁ ΚΘ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΕΛ τοῦ ΓΖ, μείζων δὲ ὁ ΓΔ τοῦ ΓΖ, μείζων ἄρα καὶ ὁ ΘΚ τοῦ ΕΛ. κείσθω τῷ ΕΛ ἴσος ὁ ΚΝ. ὃ ἄρα μέρος ἐστὶν ὁ ΚΘ τοῦ ΓΔ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΚΝ τοῦ ΓΖ: καὶ λοιπὸς ἄρα ὁ ΝΘ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὅλος ὁ ΚΘ ὅλου τοῦ ΓΔ. ἐδείχθη δὲ καὶ λοιπὸς ὁ ΜΚ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ὤν, ὅπερ ὅλος ὁ ΗΚ ὅλου τοῦ ΓΔ: καὶ συναμφότερος ἄρα ὁ ΜΚ, ΝΘ τοῦ ΔΖ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὅλος ὁ ΘΗ ὅλου τοῦ ΓΔ. ἴσος δὲ συναμφότερος μὲν ὁ ΜΚ, ΝΘ τῷ ΕΒ, ὁ δὲ ΘΗ τῷ ΒΑ: καὶ λοιπὸς ἄρα ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὅλος ὁ ΑΒ ὅλου τοῦ ΓΔ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 317|>","<|""VertexLabel"" -> ""7.9"", ""Text"" -> ""If a number be a part of a number, and another be the same part of another, alternately also, whatever part or parts the first is of the third, the same part, or the same parts, will the second also be of the fourth."", ""TextWordCount"" -> 44, ""GreekText"" -> ""ἐὰν ἀριθμὸς ἀριθμοῦ μέρος ᾖ, καὶ ἕτερος ἑτέρου τὸ αὐτὸ μέρος ᾖ, καὶ ἐναλλάξ, ὃ μέρος ἐστὶν ἢ μέρη ὁ πρῶτος τοῦ τρίτου, τὸ αὐτὸ μέρος ἔσται ἢ τὰ αὐτὰ μέρη καὶ ὁ δεύτερος τοῦ τετάρτου."", ""GreekTextWordCount"" -> 36, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 5}, {""Book"" -> 7, ""Theorem"" -> 6}}, ""Proof"" -> ""For let the number A be a part of the number BC, and another, D, the same part of another, EF, that A is of BC; I say that, alternately also, whatever part or parts A is of D, the same part or parts is BC of EF also. For since, whatever part A is of BC, the same part also is D of EF, therefore, as many numbers as there are in BC equal to A, so many also are there in EF equal to D. Let BC be divided into the numbers equal to A, namely BG, GC, and EF into those equal to D, namely EH, HF; thus the multitude of BG, GC will be equal to the multitude of EH, HF. Now, since the numbers BG, GC are equal to one another, and the numbers EH, HF are also equal to one another, while the multitude of BG, GC is equal to the multitude of EH, HF, therefore, whatever part or parts BG is of EH, the same part or the same parts is GC of HF also; so that, in addition, whatever part or parts BG is of EH, the same part also, or the same parts, is the sum BC of the sum EF. [VII. 5, 6] But BG is equal to A, and EH to D; therefore, whatever part or parts A is of D, the same part or the same parts is BC of EF also."", ""ProofWordCount"" -> 244, ""GreekProof"" -> ""ἀριθμὸς γὰρ ὁ Α ἀριθμοῦ τοῦ ΒΓ μέρος ἔστω, καὶ ἕτερος ὁ Δ ἑτέρου τοῦ ΕΖ τὸ αὐτὸ μέρος, ὅπερ ὁ Α τοῦ ΒΓ: λέγω, ὅτι καὶ ἐναλλάξ, ὃ μέρος ἐστὶν ὁ Α τοῦ Δ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΒΓ τοῦ ΕΖ ἢ μέρη. ἐπεὶ γὰρ ὃ μέρος ἐστὶν ὁ Α τοῦ ΒΓ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Δ τοῦ ΕΖ, ὅσοι ἄρα εἰσὶν ἐν τῷ ΒΓ ἀριθμοὶ ἴσοι τῷ Α, τοσοῦτοί εἰσι καὶ ἐν τῷ ΕΖ ἴσοι τῷ Δ. διῃρήσθω ὁ μὲν ΒΓ εἰς τοὺς τῷ Α ἴσους τοὺς ΒΗ, ΗΓ, ὁ δὲ ΕΖ εἰς τοὺς τῷ Δ ἴσους τοὺς ΕΘ, ΘΖ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΓ τῷ πλήθει τῶν ΕΘ, ΘΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΒΗ, ΗΓ ἀριθμοὶ ἀλλήλοις, εἰσὶ δὲ καὶ οἱ ΕΘ, ΘΖ ἀριθμοὶ ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΓ τῷ πλήθει τῶν ΕΘ, ΘΖ, ὃ ἄρα μέρος ἐστὶν ὁ ΒΗ τοῦ ΕΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΗΓ τοῦ ΘΖ ἢ τὰ αὐτὰ μέρη: ὥστε καὶ ὃ μέρος ἐστὶν ὁ ΒΗ τοῦ ΕΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ συναμφότερος ὁ ΒΓ συναμφοτέρου τοῦ ΕΖ ἢ τὰ αὐτὰ μέρη. ἴσος δὲ ὁ μὲν ΒΗ τῷ Α, ὁ δὲ ΕΘ τῷ Δ: ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ Δ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΒΓ τοῦ ΕΖ ἢ τὰ αὐτὰ μέρη: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 239|>","<|""VertexLabel"" -> ""7.10"", ""Text"" -> ""If a number be parts of a number, and another be the same parts of another, alternately also, whatever parts or part the first is of the third, the same parts or the same part will the second also be of the fourth."", ""TextWordCount"" -> 43, ""GreekText"" -> ""ἐὰν ἀριθμὸς ἀριθμοῦ μέρη ᾖ, καὶ ἕτερος ἑτέρου τὰ αὐτὰ μέρη ᾖ, καὶ ἐναλλάξ, ἃ μέρη ἐστὶν ὁ πρῶτος τοῦ τρίτου ἢ μέρος, τὰ αὐτὰ μέρη ἔσται καὶ ὁ δεύτερος τοῦ τετάρτου ἢ τὸ αὐτὸ μέρος."", ""GreekTextWordCount"" -> 36, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 5}, {""Book"" -> 7, ""Theorem"" -> 6}, {""Book"" -> 7, ""Theorem"" -> 9}}, ""Proof"" -> ""For let the number AB be parts of the number C, and another, DE, the same parts of another, F; I say that, alternately also, whatever parts or part AB is of DE, the same parts or the same part is C of F also. For since, whatever parts AB is of C, the same parts also is DE of F, therefore, as many parts of C as there are in AB, so many parts also of F are there in DE. Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE; thus the multitude of AG, GB will be equal to the multitude of DH, HE. Now since, whatever part AG is of C, the same part also is DH of F, alternately also, whatever part or parts AG is of DH, the same part or the same parts is C of F also. [VII. 9] For the same reason also, whatever part or parts GB is of HE, the same part or the same parts is C of F also; so that, in addition, whatever parts or part AB is of DE, the same parts also, or the same part, is C of F. [VII. 5, 6]"", ""ProofWordCount"" -> 211, ""GreekProof"" -> ""ἀριθμὸς γὰρ ὁ ΑΒ ἀριθμοῦ τοῦ Γ μέρη ἔστω, καὶ ἕτερος ὁ ΔΕ ἑτέρου τοῦ Ζ τὰ αὐτὰ μέρη: λέγω, ὅτι καὶ ἐναλλάξ, ἃ μέρη ἐστὶν ὁ ΑΒ τοῦ ΔΕ ἢ μέρος, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ Γ τοῦ Ζ ἢ τὸ αὐτὸ μέρος. ἐπεὶ γάρ, ἃ μέρη ἐστὶν ὁ ΑΒ τοῦ Γ, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μέρη τοῦ Γ, τοσαῦτα καὶ ἐν τῷ ΔΕ μέρη τοῦ Ζ. διῃρήσθω ὁ μὲν ΑΒ εἰς τὰ τοῦ Γ μέρη τὰ ΑΗ, ΗΒ, ὁ δὲ ΔΕ εἰς τὰ τοῦ Ζ μέρη τὰ ΔΘ, ΘΕ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει τῶν ΔΘ, ΘΕ. καὶ ἐπεί, ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ Γ, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΔΘ τοῦ Ζ, καὶ ἐναλλάξ, ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Ζ ἢ τὰ αὐτὰ μέρη. διὰ τὰ αὐτὰ δὴ καί, ὃ μέρος ἐστὶν ὁ ΗΒ τοῦ ΘΕ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Ζ ἢ τὰ αὐτὰ μέρη: ὥστε καί ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΗΒ τοῦ ΘΕ ἢ τὰ αὐτὰ μέρη: καὶ ὃ ἄρα μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΒ τοῦ ΔΕ ἢ τὰ αὐτὰ μέρη: ἀλλ᾽ ὃ μέρος ἐστὶν ὁ ΑΗ τοῦ ΔΘ ἢ μέρη, τὸ αὐτὸ μέρος ἐδείχθη καὶ ὁ Γ τοῦ Ζ ἢ τὰ αὐτὰ μέρη, καὶ ἃ ἄρα μέρη ἐστὶν ὁ ΑΒ τοῦ ΔΕ ἢ μέρος, τὰ αὐτὰ μέρη ἐστὶ καὶ ὁ Γ τοῦ Ζ ἢ τὸ αὐτὸ μέρος: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 284|>","<|""VertexLabel"" -> ""7.11"", ""Text"" -> ""If, as whole is to whole, so is a number subtracted to a number subtracted, the remainder will also be to the remainder as whole to whole."", ""TextWordCount"" -> 27, ""GreekText"" -> ""ἐὰν ᾖ ὡς ὅλος πρὸς ὅλον, οὕτως ἀφαιρεθεὶς πρὸς ἀφαιρεθέντα, καὶ ὁ λοιπὸς πρὸς τὸν λοιπὸν ἔσται, ὡς ὅλος πρὸς ὅλον."", ""GreekTextWordCount"" -> 21, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 20}, {""Book"" -> 7, ""Theorem"" -> 7}, {""Book"" -> 7, ""Theorem"" -> 8}}, ""Proof"" -> ""As the whole AB is to the whole CD, so let AE subtracted be to CF subtracted; I say that the remainder EB is also to the remainder FD as the whole AB to the whole CD. Since, as AB is to CD, so is AE to CF, whatever part or parts AB is of CD, the same part or the same parts is AE of CF also; [VII. Def. 20] Therefore also the remainder EB is the same part or parts of FD that AB is of CD. [VII. 7, 8] Therefore, as EB is to FD, so is AB to CD. [VII. Def. 20]"", ""ProofWordCount"" -> 106, ""GreekProof"" -> ""ἔστω ὡς ὅλος ὁ ΑΒ πρὸς ὅλον τὸν ΓΔ, οὕτως ἀφαιρεθεὶς ὁ ΑΕ πρὸς ἀφαιρεθέντα τὸν ΓΖ: λέγω, ὅτι καὶ λοιπὸς ὁ ΕΒ πρὸς λοιπὸν τὸν ΖΔ ἐστιν, ὡς ὅλος ὁ ΑΒ πρὸς ὅλον τὸν ΓΔ. ἐπεί ἐστιν ὡς ὁ ΑΒ πρὸς τὸν ΓΔ, οὕτως ὁ ΑΕ πρὸς τὸν ΓΖ, ὃ ἄρα μέρος ἐστὶν ὁ ΑΒ τοῦ ΓΔ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ ΑΕ τοῦ ΓΖ ἢ τὰ αὐτὰ μέρη. καὶ λοιπὸς ἄρα ὁ ΕΒ λοιποῦ τοῦ ΖΔ τὸ αὐτὸ μέρος ἐστὶν ἢ μέρη, ἅπερ ὁ ΑΒ τοῦ ΓΔ. ἔστιν ἄρα ὡς ὁ ΕΒ πρὸς τὸν ΖΔ, οὕτως ὁ ΑΒ πρὸς τὸν ΓΔ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 109|>","<|""VertexLabel"" -> ""7.12"", ""Text"" -> ""If there be as many numbers as we please in proportion, then, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents."", ""TextWordCount"" -> 32, ""GreekText"" -> ""ἐὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ ἀνάλογον, ἔσται ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους."", ""GreekTextWordCount"" -> 22, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 20}, {""Book"" -> 7, ""Theorem"" -> 5}, {""Book"" -> 7, ""Theorem"" -> 6}}, ""Proof"" -> ""Let A, B, C, D be as many numbers as we please in proportion, so that, as A is to B, so is C to D; I say that, as A is to B, so are A, C to B, D. For since, as A is to B, so is C to D, whatever part or parts A is of B, the same part or parts is C of D also. [VII. Def. 20] Therefore also the sum of A, C is the same part or the same parts of the sum of B, D that A is of B. [VII. 5, 6] Therefore, as A is to B, so are A, C to B, D. [VII. Def. 20]"", ""ProofWordCount"" -> 119, ""GreekProof"" -> ""ἔστωσαν ὁποσοιοῦν ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ: λέγω, ὅτι ἐστὶν ὡς ὁ Α πρὸς τὸν Β, οὕτως οἱ Α, Γ πρὸς τοὺς Β, Δ. ἐπεὶ γάρ ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ, ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ Β ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Δ ἢ μέρη. καὶ συναμφότερος ἄρα ὁ Α, Γ συναμφοτέρου τοῦ Β, Δ τὸ αὐτὸ μέρος ἐστὶν ἢ τὰ αὐτὰ μέρη, ἅπερ ὁ Α τοῦ Β. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως οἱ Α, Γ πρὸς τοὺς Β, Δ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 116|>","<|""VertexLabel"" -> ""7.13"", ""Text"" -> ""If four numbers be proportional, they will also be proportional alternately."", ""TextWordCount"" -> 11, ""GreekText"" -> ""ἐὰν τέσσαρες ἀριθμοὶ ἀνάλογον ὦσιν, καὶ ἐναλλὰξ ἀνάλογον ἔσονται."", ""GreekTextWordCount"" -> 9, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 20}, {""Book"" -> 7, ""Theorem"" -> 10}}, ""Proof"" -> ""Let the four numbers A, B, C, D be proportional, so that, as A is to B, so is C to D; I say that they will also be proportional alternately, so that, as A is to C, so will B be to D. For since, as A is to B, so is C to D, therefore, whatever part or parts A is of B, the same part or the same parts is C of D also. [VII. Def. 20] Therefore, alternately, whatever part or parts A is of C, the same part or the same parts is B of D also. [VII. 10] Therefore, as A is to C, so is B to D. [VII. Def. 20]"", ""ProofWordCount"" -> 118, ""GreekProof"" -> ""ἔστωσαν τέσσαρες ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ: λέγω, ὅτι καὶ ἐναλλὰξ ἀνάλογον ἔσονται, ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Β πρὸς τὸν Δ. ἐπεὶ γάρ ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ, ὃ ἄρα μέρος ἐστὶν ὁ Α τοῦ Β ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Γ τοῦ Δ ἢ τὰ αὐτὰ μέρη. ἐναλλὰξ ἄρα, ὃ μέρος ἐστὶν ὁ Α τοῦ Γ ἢ μέρη, τὸ αὐτὸ μέρος ἐστὶ καὶ ὁ Β τοῦ Δ ἢ τὰ αὐτὰ μέρη. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Β πρὸς τὸν Δ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 118|>","<|""VertexLabel"" -> ""7.14"", ""Text"" -> ""If there be as many numbers as we please, and others equal to them in multitude, which taken two and two are in the same ratio, they will also be in the same ratio ex aequali."", ""TextWordCount"" -> 36, ""GreekText"" -> ""ἐὰν ὦσιν ὁποσοιοῦν ἀριθμοὶ καὶ ἄλλοι αὐτοῖς ἴσοι τὸ πλῆθος σύνδυο λαμβανόμενοι καὶ ἐν τῷ αὐτῷ λόγῳ, καὶ δι᾽ ἴσου ἐν τῷ αὐτῷ λόγῳ ἔσονται."", ""GreekTextWordCount"" -> 26, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 13}}, ""Proof"" -> ""Let there be as many numbers as we please A, B, C, and others equal to them in multitude D, E, F, which taken two and two are in the same ratio, so that, as A is to B, so is D to E, and, as B is to C, so is E to F; I say that, ex aequali, as A is to C, so also is D to F. For, since, as A is to B, so is D to E, therefore, alternately, as A is to D, so is B to E. [VII. 13] Again, since, as B is to C, so is E to F, therefore, alternately, as B is to E, so is C to F. [VII. 13] But, as B is to E, so is A to D; therefore also, as A is to D, so is C to F. Therefore, alternately, as A is to C, so is D to F."", ""ProofWordCount"" -> 158, ""GreekProof"" -> ""ἔστωσαν ὁποσοιοῦν ἀριθμοὶ οἱ Α, Β, Γ καὶ ἄλλοι αὐτοῖς ἴσοι τὸ πλῆθος σύνδυο λαμβανόμενοι ἐν τῷ αὐτῷ λόγῳ οἱ Δ, Ε, Ζ, ὡς μὲν ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε, ὡς δὲ ὁ Β πρὸς τὸν Γ, οὕτως ὁ Ε πρὸς τὸν Ζ: λέγω, ὅτι καὶ δι᾽ ἴσου ἐστὶν ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ζ. ἐπεὶ γάρ ἐστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, οὕτως ὁ Β πρὸς τὸν Ε. πάλιν, ἐπεί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, οὕτως ὁ Ε πρὸς τὸν Ζ, ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Β πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν Ζ. ὡς δὲ ὁ Β πρὸς τὸν Ε, οὕτως ὁ Α πρὸς τὸν δ: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Δ, οὕτως ὁ Γ πρὸς τὸν Ζ: ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ζ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 173|>","<|""VertexLabel"" -> ""7.15"", ""Text"" -> ""If an unit measure any number, and another number measure any other number the same number of times, alternately also, the unit will measure the third number the same number of times that the second measures the fourth."", ""TextWordCount"" -> 38, ""GreekText"" -> ""ἐὰν μονὰς ἀριθμόν τινα μετρῇ, ἰσάκις δὲ ἕτερος ἀριθμὸς ἄλλον τινὰ ἀριθμὸν μετρῇ, καὶ ἐναλλὰξ ἰσάκις ἡ μονὰς τὸν τρίτον ἀριθμὸν μετρήσει καὶ ὁ δεύτερος τὸν τέταρτον."", ""GreekTextWordCount"" -> 27, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 12}}, ""Proof"" -> ""For let the unit A measure any number BC, and let another number D measure any other number EF the same number of times; I say that, alternately also, the unit A measures the number D the same number of times that BC measures EF. For, since the unit A measures the number BC the same number of times that D measures EF, therefore, as many units as there are in BC, so many numbers equal to D are there in EF also. Let BC be divided into the units in it, BG, GH, HC, and EF into the numbers EK, KL, LF equal to D. Thus the multitude of BG, GH, HC will be equal to the multitude of EK, KL, LF. And, since the units BG, GH, HC are equal to one another, and the numbers EK, KL, LF are also equal to one another, while the multitude of the units BG, GH, HC is equal to the multitude of the numbers EK, KL, LF, therefore, as the unit BG is to the number EK, so will the unit GH be to the number KL, and the unit HC to the number LF. Therefore also, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [VII. 12] therefore, as the unit BG is to the number EK, so is BC to EF. But the unit BG is equal to the unit A, and the number EK to the number D. Therefore, as the unit A is to the number D, so is BC to EF. Therefore the unit A measures the number D the same number of times that BC measures EF."", ""ProofWordCount"" -> 285, ""GreekProof"" -> ""μονὰς γὰρ ἡ Α ἀριθμόν τινα τὸν ΒΓ μετρείτω, ἰσάκις δὲ ἕτερος ἀριθμὸς ὁ Δ ἄλλον τινὰ ἀριθμὸν τὸν ΕΖ μετρείτω: λέγω, ὅτι καὶ ἐναλλὰξ ἰσάκις ἡ Α μονὰς τὸν Δ ἀριθμὸν μετρεῖ καὶ ὁ ΒΓ τὸν ΕΖ. ἐπεὶ γὰρ ἰσάκις ἡ Α μονὰς τὸν ΒΓ ἀριθμὸν μετρεῖ καὶ ὁ Δ τὸν ΕΖ, ὅσαι ἄρα εἰσὶν ἐν τῷ ΒΓ μονάδες, τοσοῦτοί εἰσι καὶ ἐν τῷ ΕΖ ἀριθμοὶ ἴσοι τῷ Δ. διῃρήσθω ὁ μὲν ΒΓ εἰς τὰς ἐν ἑαυτῷ μονάδας τὰς ΒΗ, ΗΘ, ΘΓ, ὁ δὲ ΕΖ εἰς τοὺς τῷ Δ ἴσους τοὺς ΕΚ, ΚΛ, ΛΖ. ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΘ, ΘΓ τῷ πλήθει τῶν ΕΚ, ΚΛ, ΛΖ. καὶ ἐπεὶ ἴσαι εἰσὶν αἱ ΒΗ, ΗΘ, ΘΓ μονάδες ἀλλήλαις, εἰσὶ δὲ καὶ οἱ ΕΚ, ΚΛ, ΛΖ ἀριθμοὶ ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ πλῆθος τῶν ΒΗ, ΗΘ, ΘΓ μονάδων τῷ πλήθει τῶν ΕΚ, ΚΛ, ΛΖ ἀριθμῶν, ἔσται ἄρα ὡς ἡ ΒΗ μονὰς πρὸς τὸν ΕΚ ἀριθμόν, οὕτως ἡ ΗΘ μονὰς πρὸς τὸν ΚΛ ἀριθμὸν καὶ ἡ ΘΓ μονὰς πρὸς τὸν ΛΖ ἀριθμόν. ἔσται ἄρα καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους: ἔστιν ἄρα ὡς ἡ ΒΗ μονὰς πρὸς τὸν ΕΚ ἀριθμόν, οὕτως ὁ ΒΓ πρὸς τὸν ΕΖ. ἴση δὲ ἡ ΒΗ μονὰς τῇ Α μονάδι, ὁ δὲ ΕΚ ἀριθμὸς τῷ Δ ἀριθμῷ. ἔστιν ἄρα ὡς ἡ Α μονὰς πρὸς τὸν Δ ἀριθμόν, οὕτως ὁ ΒΓ πρὸς τὸν ΕΖ. ἰσάκις ἄρα ἡ Α μονὰς τὸν Δ ἀριθμὸν μετρεῖ καὶ ὁ ΒΓ τὸν ΕΖ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 256|>","<|""VertexLabel"" -> ""7.16"", ""Text"" -> ""If two numbers by multiplying one another make certain numbers, the numbers so produced will be equal to one another."", ""TextWordCount"" -> 20, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινας, οἱ γενόμενοι ἐξ αὐτῶν ἴσοι ἀλλήλοις ἔσονται."", ""GreekTextWordCount"" -> 14, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 15}}, ""Proof"" -> ""Let A, B be two numbers, and let A by multiplying B make C, and B by multiplying A make D; I say that C is equal to D. For, since A by multiplying B has made C, therefore B measures C according to the units in A. But the unit E also measures the number A according to the units in it; therefore the unit E measures A the same number of times that B measures C. Therefore, alternately, the unit E measures the number B the same number of times that A measures C. [VII. 15] Again, since B by multiplying A has made D, therefore A measures D according to the units in B. But the unit E also measures B according to the units in it; therefore the unit E measures the number B the same number of times that A measures D. But the unit E measured the number B the same number of times that A measures C; therefore A measures each of the numbers C, D the same number of times. Therefore C is equal to D."", ""ProofWordCount"" -> 184, ""GreekProof"" -> ""ἔστωσαν δύο ἀριθμοὶ οἱ Α, Β, καὶ ὁ μὲν Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω, ὁ δὲ Β τὸν Α πολλαπλασιάσας τὸν Δ ποιείτω: λέγω, ὅτι ἴσος ἐστὶν ὁ Γ τῷ Δ. ἐπεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ πεποίηκεν, ὁ Β ἄρα τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεῖ δὲ καὶ ἡ Ε μονὰς τὸν Α ἀριθμὸν κατὰ τὰς ἐν αὐτῷ μονάδας: ἰσάκις ἄρα ἡ Ε μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Γ. ἐναλλὰξ ἄρα ἰσάκις ἡ Ε μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Α τὸν Γ. πάλιν, ἐπεὶ ὁ Β τὸν Α πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ Α ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Β μονάδας. μετρεῖ δὲ καὶ ἡ Ε μονὰς τὸν Β κατὰ τὰς ἐν αὐτῷ μονάδας: ἰσάκις ἄρα ἡ Ε μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Α τὸν Δ. ἰσάκις δὲ ἡ Ε μονὰς τὸν Β ἀριθμὸν ἐμέτρει καὶ ὁ Α τὸν Γ: ἰσάκις ἄρα ὁ Α ἑκάτερον τῶν Γ, Δ μετρεῖ. ἴσος ἄρα ἐστὶν ὁ Γ τῷ Δ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 180|>","<|""VertexLabel"" -> ""7.17"", ""Text"" -> ""If a number by multiplying two numbers make certain numbers, the numbers so produced will have the same ratio as the numbers multiplied."", ""TextWordCount"" -> 23, ""GreekText"" -> ""ἐὰν ἀριθμὸς δύο ἀριθμοὺς πολλαπλασιάσας ποιῇ τινας, οἱ γενόμενοι ἐξ αὐτῶν τὸν αὐτὸν ἕξουσι λόγον τοῖς πολλαπλασιασθεῖσιν."", ""GreekTextWordCount"" -> 17, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 20}, {""Book"" -> 7, ""Theorem"" -> 13}}, ""Proof"" -> ""For let the number A by multiplying the two numbers B, C make D, E; I say that, as B is to C, so is D to E. For, since A by multiplying B has made D, therefore B measures D according to the units in A. But the unit F also measures the number A according to the units in it; therefore the unit F measures the number A the same number of times that B measures D. Therefore, as the unit F is to the number A, so is B to D. [VII. Def. 20] For the same reason, as the unit F is to the number A, so also is C to E; therefore also, as B is to D, so is C to E. Therefore, alternately, as B is to C, so is D to E. [VII. 13]"", ""ProofWordCount"" -> 142, ""GreekProof"" -> ""ἀριθμὸς γὰρ ὁ Α δύο ἀριθμοὺς τοὺς Β, Γ πολλαπλασιάσας τοὺς Δ, Ε ποιείτω: λέγω, ὅτι ἐστὶν ὡς ὁ Β πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ε. ἐπεὶ γὰρ ὁ Α τὸν Β πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ Β ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Α μονάδας. μετρεῖ δὲ καὶ ἡ Ζ μονὰς τὸν Α ἀριθμὸν κατὰ τὰς ἐν αὐτῷ μονάδας: ἰσάκις ἄρα ἡ Ζ μονὰς τὸν Α ἀριθμὸν μετρεῖ καὶ ὁ Β τὸν Δ. ἔστιν ἄρα ὡς ἡ Ζ μονὰς πρὸς τὸν Α ἀριθμόν, οὕτως ὁ Β πρὸς τὸν Δ. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ Ζ μονὰς πρὸς τὸν Α ἀριθμόν, οὕτως ὁ Γ πρὸς τὸν Ε: καὶ ὡς ἄρα ὁ Β πρὸς τὸν Δ, οὕτως ὁ Γ πρὸς τὸν Ε. ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Β πρὸς τὸν Γ, οὕτως ὁ Δ πρὸς τὸν Ε: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 146|>","<|""VertexLabel"" -> ""7.18"", ""Text"" -> ""If two numbers by multiplying any number make certain numbers, the numbers so produced will have the same ratio as the multipliers."", ""TextWordCount"" -> 22, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ ἀριθμόν τινα πολλαπλασιάσαντες ποιῶσί τινας, οἱ γενόμενοι ἐξ αὐτῶν τὸν αὐτὸν ἕξουσι λόγον τοῖς πολλαπλασιάσασιν."", ""GreekTextWordCount"" -> 18, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 16}, {""Book"" -> 7, ""Theorem"" -> 17}}, ""Proof"" -> ""For let two numbers A, B by multiplying any number C make D, E; I say that, as A is to B, so is D to E. For, since A by multiplying C has made D, therefore also C by multiplying A has made D. [VII. 16] For the same reason also C by multiplying B has made E. Therefore the number C by multiplying the two numbers A, B has made D, E. Therefore, as A is to B, so is D to E. [VII. 17]"", ""ProofWordCount"" -> 87, ""GreekProof"" -> ""δύο γὰρ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν Γ πολλαπλασιάσαντες τοὺς Δ, Ε ποιείτωσαν: λέγω, ὅτι ἐστὶν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε. ἐπεὶ γὰρ ὁ Α τὸν Γ πολλαπλασιάσας τὸν Δ πεποίηκεν, καὶ ὁ Γ ἄρα τὸν Α πολλαπλασιάσας τὸν Δ πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Γ τὸν Β πολλαπλασιάσας τὸν Ε πεποίηκεν. ἀριθμὸς δὴ ὁ Γ δύο ἀριθμοὺς τοὺς Α, Β πολλαπλασιάσας τοὺς Δ, Ε πεποίηκεν. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν ε: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 94|>","<|""VertexLabel"" -> ""7.19"", ""Text"" -> ""If four numbers be proportional, the number produced from the first and fourth will be equal to the number produced from the second and third; and, if the number produced from the first and fourth be equal to that produced from the second and third, the four numbers will be proportional."", ""TextWordCount"" -> 51, ""GreekText"" -> ""ἐὰν τέσσαρες ἀριθμοὶ ἀνάλογον ὦσιν, ὁ ἐκ πρώτου καὶ τετάρτου γενόμενος ἀριθμὸς ἴσος ἔσται τῷ ἐκ δευτέρου καὶ τρίτου γενομένῳ ἀριθμῷ: καὶ ἐὰν ὁ ἐκ πρώτου καὶ τετάρτου γενόμενος ἀριθμὸς ἴσος ᾖ τῷ ἐκ δευτέρου καὶ τρίτου, οἱ τέσσαρες ἀριθμοὶ ἀνάλογον ἔσονται."", ""GreekTextWordCount"" -> 42, ""References"" -> {{""Book"" -> 5, ""Theorem"" -> 7}, {""Book"" -> 5, ""Theorem"" -> 9}, {""Book"" -> 7, ""Theorem"" -> 17}, {""Book"" -> 7, ""Theorem"" -> 18}}, ""Proof"" -> ""Let A, B, C, D be four numbers in proportion, so that, as A is to B, so is C to D; and let A by multiplying D make E, and let B by multiplying C make F; I say that E is equal to F. For let A by multiplying C make G. Since, then, A by multiplying C has made G, and by multiplying D has made E, the number A by multiplying the two numbers C, D has made G, E. Therefore, as C is to D, so is G to E. [VII. 17] But, as C is to D, so is A to B; therefore also, as A is to B, so is G to E. Again, since A by multiplying C has made G, but, further, B has also by multiplying C made F, the two numbers A, B by multiplying a certain number C have made G, F. Therefore, as A is to B, so is G to F. [VII. 18] But further, as A is to B, so is G to E also; therefore also, as G is to E, so is G to F. Therefore G has to each of the numbers E, F the same ratio; therefore E is equal to F. [cf. V. 9] Again, let E be equal to F; I say that, as A is to B, so is C to D. For, with the same construction, since E is equal to F, therefore, as G is to E, so is G to F. [cf. V. 7] But, as G is to E, so is C to D, [VII. 17] and, as G is to F, so is A to B. [VII. 18] Therefore also, as A is to B, so is C to D."", ""ProofWordCount"" -> 297, ""GreekProof"" -> ""ἔστωσαν τέσσαρες ἀριθμοὶ ἀνάλογον οἱ Α, Β, Γ, Δ, ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ, καὶ ὁ μὲν Α τὸν Δ πολλαπλασιάσας τὸν Ε ποιείτω, ὁ δὲ Β τὸν Γ πολλαπλασιάσας τὸν Ζ ποιείτω: λέγω, ὅτι ἴσος ἐστὶν ὁ Ε τῷ Ζ. ὁ γὰρ Α τὸν Γ πολλαπλασιάσας τὸν Η ποιείτω. ἐπεὶ οὖν ὁ Α τὸν Γ πολλαπλασιάσας τὸν Η πεποίηκεν, τὸν δὲ Δ πολλαπλασιάσας τὸν Ε πεποίηκεν, ἀριθμὸς δὴ ὁ Α δύο ἀριθμοὺς τοὺς Γ, Δ πολλαπλασιάσας τοὺς Η, Ε πεποίηκεν. ἔστιν ἄρα ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Η πρὸς τὸν Ε. ἀλλ᾽ ὡς ὁ Γ πρὸς τὸν Δ, οὕτως ὁ Α πρὸς τὸν Β: καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Ε. πάλιν, ἐπεὶ ὁ Α τὸν Γ πολλαπλασιάσας τὸν Η πεποίηκεν, ἀλλὰ μὴν καὶ ὁ Β τὸν Γ πολλαπλασιάσας τὸν Ζ πεποίηκεν, δύο δὴ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν Γ πολλαπλασιάσαντες τοὺς Η, Ζ πεποιήκασιν. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Ζ. ἀλλὰ μὴν καὶ ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Η πρὸς τὸν Ε: καὶ ὡς ἄρα ὁ Η πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν Ζ. ὁ Η ἄρα πρὸς ἑκάτερον τῶν Ε, Ζ τὸν αὐτὸν ἔχει λόγον: ἴσος ἄρα ἐστὶν ὁ Ε τῷ Ζ. ἔστω δὴ πάλιν ἴσος ὁ Ε τῷ Ζ: λέγω, ὅτι ἐστὶν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ ἴσος ἐστὶν ὁ Ε τῷ Ζ, ἔστιν ἄρα ὡς ὁ Η πρὸς τὸν Ε, οὕτως ὁ Η πρὸς τὸν Ζ. ἀλλ᾽ ὡς μὲν ὁ Η πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν δ, ὡς δὲ ὁ Η πρὸς τὸν Ζ, οὕτως ὁ Α πρὸς τὸν Β. καὶ ὡς ἄρα ὁ Α πρὸς τὸν Β, οὕτως ὁ Γ πρὸς τὸν Δ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 321|>","<|""VertexLabel"" -> ""7.20"", ""Text"" -> ""The least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less."", ""TextWordCount"" -> 33, ""GreekText"" -> ""οἱ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα."", ""GreekTextWordCount"" -> 26, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 20}, {""Book"" -> 7, ""Theorem"" -> 4}, {""Book"" -> 7, ""Theorem"" -> 12}, {""Book"" -> 7, ""Theorem"" -> 13}}, ""Proof"" -> ""For let CD, EF be the least numbers of those which have the same ratio with A, B; I say that CD measures A the same number of times that EF measures B. Now CD is not parts of A. For, if possible, let it be so; therefore EF is also the same parts of B that CD is of A. [VII. 13 and Def. 20] Therefore, as many parts of A as there are in CD, so many parts of B are there also in EF. Let CD be divided into the parts of A, namely CG, GD, and EF into the parts of B, namely EH, HF; thus the multitude of CG, GD will be equal to the multitude of EH, HF. Now, since the numbers CG, GD are equal to one another, and the numbers EH, HF are also equal to one another, while the multitude of CG, GD is equal to the multitude of EH, HF, therefore, as CG is to EH, so is GD to HF. Therefore also, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents. [VII. 12] Therefore, as CG is to EH, so is CD to EF. Therefore CG, EH are in the same ratio with CD, EF, being less than they: which is impossible, for by hypothesis CD, EF are the least numbers of those which have the same ratio with them. Therefore CD is not parts of A; therefore it is a part of it. [VII. 4] And EF is the same part of B that CD is of A; [VII. 13 and Def. 20] therefore CD measures A the same number of times that EF measures B."", ""ProofWordCount"" -> 290, ""GreekProof"" -> ""ἔστωσαν γὰρ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β οἱ ΓΔ, ΕΖ: λέγω, ὅτι ἰσάκις ὁ ΓΔ τὸν Α μετρεῖ καὶ ὁ ΕΖ τὸν Β. ὁ ΓΔ γὰρ τοῦ Α οὔκ ἐστι μέρη. εἰ γὰρ δυνατόν, ἔστω: καὶ ὁ ΕΖ ἄρα τοῦ Β τὰ αὐτὰ μέρη ἐστίν, ἅπερ ὁ ΓΔ τοῦ Α. ὅσα ἄρα ἐστὶν ἐν τῷ ΓΔ μέρη τοῦ Α, τοσαῦτά ἐστι καὶ ἐν τῷ ΕΖ μέρη τοῦ Β. διῃρήσθω ὁ μὲν ΓΔ εἰς τὰ τοῦ Α μέρη τὰ ΓΗ, ΗΔ, ὁ δὲ ΕΖ εἰς τὰ τοῦ Β μέρη τὰ ΕΘ, ΘΖ: ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΓΗ, ΗΔ τῷ πλήθει τῶν ΕΘ, ΘΖ. καὶ ἐπεὶ ἴσοι εἰσὶν οἱ ΓΗ, ΗΔ ἀριθμοὶ ἀλλήλοις, εἰσὶ δὲ καὶ οἱ ΕΘ, ΘΖ ἀριθμοὶ ἴσοι ἀλλήλοις, καί ἐστιν ἴσον τὸ πλῆθος τῶν ΓΗ, ΗΔ τῷ πλήθει τῶν ΕΘ, ΘΖ, ἔστιν ἄρα ὡς ὁ ΓΗ πρὸς τὸν ΕΘ, οὕτως ὁ ΗΔ πρὸς τὸν ΘΖ. ἔσται ἄρα καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους. ἔστιν ἄρα ὡς ὁ ΓΗ πρὸς τὸν ΕΘ, οὕτως ὁ ΓΔ πρὸς τὸν ΕΖ: οἱ ΓΗ, ΕΘ ἄρα τοῖς ΓΔ, ΕΖ ἐν τῷ αὐτῷ λόγῳ εἰσὶν ἐλάσσονες ὄντες αὐτῶν: ὅπερ ἐστὶν ἀδύνατον: ὑπόκεινται γὰρ οἱ ΓΔ, ΕΖ ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. οὐκ ἄρα μέρη ἐστὶν ὁ ΓΔ τοῦ Α: μέρος ἄρα. καὶ ὁ ΕΖ τοῦ Β τὸ αὐτὸ μέρος ἐστίν, ὅπερ ὁ ΓΔ τοῦ Α: ἰσάκις ἄρα ὁ ΓΔ τὸν Α μετρεῖ καὶ ὁ ΕΖ τὸν Β: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 256|>","<|""VertexLabel"" -> ""7.21"", ""Text"" -> ""Numbers prime to one another are the least of those which have the same ratio with them."", ""TextWordCount"" -> 17, ""GreekText"" -> ""οἱ πρῶτοι πρὸς ἀλλήλους ἀριθμοὶ ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς."", ""GreekTextWordCount"" -> 13, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 12}, {""Book"" -> 7, ""Theorem"" -> 16}, {""Book"" -> 7, ""Theorem"" -> 20}}, ""Proof"" -> ""Let A, B be numbers prime to one another; I say that A, B are the least of those which have the same ratio with them. For, if not, there will be some numbers less than A, B which are in the same ratio with A, B. Let them be C, D. Since, then, the least numbers of those which have the same ratio measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent, [VII. 20] therefore C measures A the same number of times that D measures B. Now, as many times as C measures A, so many units let there be in E. Therefore D also measures B according to the units in E. And, since C measures A according to the units in E, therefore E also measures A according to the units in C. [VII. 16] For the same reason E also measures B according to the units in D. [VII. 16] Therefore E measures A, B which are prime to one another: which is impossible. [VII. Def. 12] Therefore there will be no numbers less than A, B which are in the same ratio with A, B. Therefore A, B are the least of those which have the same ratio with them."", ""ProofWordCount"" -> 229, ""GreekProof"" -> ""ἔστωσαν πρῶτοι πρὸς ἀλλήλους ἀριθμοὶ οἱ Α, Β: λέγω, ὅτι οἱ Α, Β ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. εἰ γὰρ μή, ἔσονταί τινες τῶν Α, Β ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β. ἔστωσαν οἱ Γ, Δ. ἐπεὶ οὖν οἱ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάττων τὸν ἐλάττονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον, ἰσάκις ἄρα ὁ Γ τὸν Α μετρεῖ καὶ ὁ Δ τὸν Β. ὁσάκις δὴ ὁ Γ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. καὶ ὁ Δ ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας. καὶ ἐπεὶ ὁ Γ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, καὶ ὁ Ε ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Γ μονάδας. διὰ τὰ αὐτὰ δὴ ὁ Ε καὶ τὸν Β μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας. ὁ Ε ἄρα τοὺς Α, Β μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα ἔσονταί τινες τῶν Α, Β ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β. οἱ Α, Β ἄρα ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 208|>","<|""VertexLabel"" -> ""7.22"", ""Text"" -> ""The least numbers of those which have the same ratio with them are prime to one another."", ""TextWordCount"" -> 17, ""GreekText"" -> ""οἱ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς πρῶτοι πρὸς ἀλλήλους εἰσίν."", ""GreekTextWordCount"" -> 13, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 15}, {""Book"" -> 7, ""Theorem"" -> 17}}, ""Proof"" -> ""Let A, B be the least numbers of those which have the same ratio with them; I say that A, B are prime to one another. For, if they are not prime to one another, some number will measure them. Let some number measure them, and let it be C. And, as many times as C measures A, so many units let there be in D, and, as many times as C measures B, so many units let there be in E Since C measures A according to the units in D, therefore C by multiplying D has made A. [VII. Def. 15] For the same reason also C by multiplying E has made B. Thus the number C by multiplying the two numbers D, E has made A, B; therefore, as D is to E, so is A to B; [VII. 17] therefore D, E are in the same ratio with A, B, being less than they: which is impossible. Therefore no number will measure the numbers A, B. Therefore A, B are prime to one another."", ""ProofWordCount"" -> 178, ""GreekProof"" -> ""ἔστωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς οἱ Α, Β: λέγω, ὅτι οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσι πρῶτοι πρὸς ἀλλήλους, μετρήσει τις αὐτοὺς ἀριθμός. μετρείτω, καὶ ἔστω ὁ Γ. καὶ ὁσάκις μὲν ὁ Γ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Δ, ὁσάκις δὲ ὁ Γ τὸν Β μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ ὁ Γ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας, ὁ Γ ἄρα τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Γ τὸν Ε πολλαπλασιάσας τὸν Β πεποίηκεν. ἀριθμὸς δὴ ὁ Γ δύο ἀριθμοὺς τοὺς Δ, Ε πολλαπλασιάσας τοὺς Α, Β πεποίηκεν: ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Ε, οὕτως ὁ Α πρὸς τὸν Β: οἱ Δ, Ε ἄρα τοῖς Α, Β ἐν τῷ αὐτῷ λόγῳ εἰσὶν ἐλάσσονες ὄντες αὐτῶν: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Α, Β ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ α, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 164|>","<|""VertexLabel"" -> ""7.23"", ""Text"" -> ""If two number be prime to one another, the number which measures the one of them will be prime to the remaining number."", ""TextWordCount"" -> 23, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ τὸν ἕνα αὐτῶν μετρῶν ἀριθμὸς πρὸς τὸν λοιπὸν πρῶτος ἔσται."", ""GreekTextWordCount"" -> 18, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 12}}, ""Proof"" -> ""Let A, B be two numbers prime to one another, and let any number C measure A; I say that C, B are also prime to one another. For, if C, B are not prime to one another, some number will measure C, B. Let a number measure them, and let it be D. Since D measures C, and C measures A, therefore D also measures A. But it also measures B; therefore D measures A, B which are prime to one another: which is impossible. [VII. Def. 12] Therefore no number will measure the numbers C, B. Therefore C, B are prime to one another."", ""ProofWordCount"" -> 106, ""GreekProof"" -> ""ἔστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, τὸν δὲ Α μετρείτω τις ἀριθμὸς ὁ Γ: λέγω, ὅτι καὶ οἱ Γ, Β πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσιν οἱ Γ, Β πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς Γ, Β ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. ἐπεὶ ὁ Δ τὸν Γ μετρεῖ, ὁ δὲ Γ τὸν Α μετρεῖ, καὶ ὁ Δ ἄρα τὸν Α μετρεῖ. μετρεῖ δὲ καὶ τὸν Β: ὁ Δ ἄρα τοὺς Α, Β μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Γ, Β ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ Γ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 106|>","<|""VertexLabel"" -> ""7.24"", ""Text"" -> ""If two numbers be prime to any number, their product also will be prime to the same."", ""TextWordCount"" -> 17, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ πρός τινα ἀριθμὸν πρῶτοι ὦσιν, καὶ ὁ ἐξ αὐτῶν γενόμενος πρὸς τὸν αὐτὸν πρῶτος ἔσται."", ""GreekTextWordCount"" -> 18, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 12}, {""Book"" -> 7, ""Definition"" -> 15}, {""Book"" -> 7, ""Theorem"" -> 16}, {""Book"" -> 7, ""Theorem"" -> 19}, {""Book"" -> 7, ""Theorem"" -> 20}, {""Book"" -> 7, ""Theorem"" -> 21}, {""Book"" -> 7, ""Theorem"" -> 23}}, ""Proof"" -> ""For let the two numbers A, B be prime to any number C, and let A by multiplying B make D; I say that C, D are prime to one another. For, if C, D are not prime to one another, some number will measure C, D. Let a number measure them, and let it be E. Now, since C, A are prime to one another, and a certain number E measures C, therefore A, E are prime to one another. [VII. 23] As many times, then, as E measures D, so many units let there be in F; therefore F also measures D according to the units in E. [VII. 16] Therefore E by multiplying F has made D. [VII. Def. 15] But, further, A by multiplying B has also made D; therefore the product of E, F is equal to the product of A, B. But, if the product of the extremes be equal to that of the means, the four numbers are proportional; [VII. 19] therefore, as E is to A, so is B to F. But A, E are prime to one another, numbers which are prime to one another are also the least of those which have the same ratio, [VII. 21] and the least numbers of those which have the same ratio with them measure those which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore E measures B. But it also measures C; therefore E measures B, C which are prime to one another: which is impossible. [VII. Def. 12] Therefore no number will measure the numbers C, D. Therefore C, D are prime to one another."", ""ProofWordCount"" -> 297, ""GreekProof"" -> ""δύο γὰρ ἀριθμοὶ οἱ Α, Β πρός τινα ἀριθμὸν τὸν Γ πρῶτοι ἔστωσαν, καὶ ὁ α τὸν Β πολλαπλασιάσας τὸν Δ ποιείτω: λέγω, ὅτι οἱ Γ, Δ πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσιν οἱ Γ, Δ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς Γ, Δ ἀριθμός. μετρείτω, καὶ ἔστω ὁ Ε. καὶ ἐπεὶ οἱ Γ, Α πρῶτοι πρὸς ἀλλήλους εἰσίν, τὸν δὲ Γ μετρεῖ τις ἀριθμὸς ὁ Ε, οἱ Α, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. ὁσάκις δὴ ὁ Ε τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ζ: καὶ ὁ Ζ ἄρα τὸν Δ μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας. ὁ Ε ἄρα τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Δ πεποίηκεν: ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Ε, Ζ τῷ ἐκ τῶν Α, Β. ἐὰν δὲ ὁ ὑπὸ τῶν ἄκρων ἴσος ᾖ τῷ ὑπὸ τῶν μέσων, οἱ τέσσαρες ἀριθμοὶ ἀνάλογόν εἰσιν: ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Ζ. οἱ δὲ Α, Ε πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: ὁ Ε ἄρα τὸν Β μετρεῖ. μετρεῖ δὲ καὶ τὸν Γ: ὁ Ε ἄρα τοὺς Β, Γ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Γ, Δ ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ Γ, Δ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 260|>","<|""VertexLabel"" -> ""7.25"", ""Text"" -> ""If two numbers be prime to one another, the product of one of them into itself will be prime to the remaining one."", ""TextWordCount"" -> 23, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, ὁ ἐκ τοῦ ἑνὸς αὐτῶν γενόμενος πρὸς τὸν λοιπὸν πρῶτος ἔσται."", ""GreekTextWordCount"" -> 18, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 24}}, ""Proof"" -> ""Let A, B be two numbers prime to one another, and let A by multiplying itself make C: I say that B, C are prime to one another. For let D be made equal to A. Since A, B are prime to one another, and A is equal to D, therefore D, B are also prime to one another. Therefore each of the two numbers D, A is prime to B; therefore the product of D, A will also be prime to B. [VII. 24] But the number which is the product of D, A is C. Therefore C, B are prime to one another."", ""ProofWordCount"" -> 105, ""GreekProof"" -> ""ἔστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Γ ποιείτω: λέγω, ὅτι οἱ Β, Γ πρῶτοι πρὸς ἀλλήλους εἰσίν. κείσθω γὰρ τῷ Α ἴσος ὁ Δ. ἐπεὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, ἴσος δὲ ὁ Α τῷ Δ, καὶ οἱ Δ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. ἑκάτερος ἄρα τῶν Δ, Α πρὸς τὸν Β πρῶτός ἐστιν: καὶ ὁ ἐκ τῶν Δ, Α ἄρα γενόμενος πρὸς τὸν Β πρῶτος ἔσται. ὁ δὲ ἐκ τῶν Δ, Α γενόμενος ἀριθμός ἐστιν ὁ Γ. οἱ Γ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 101|>","<|""VertexLabel"" -> ""7.26"", ""Text"" -> ""If two numbers be prime to two numbers, both to each, their products also will be prime to one another."", ""TextWordCount"" -> 20, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ πρὸς δύο ἀριθμοὺς ἀμφότεροι πρὸς ἑκάτερον πρῶτοι ὦσιν, καὶ οἱ ἐξ αὐτῶν γενόμενοι πρῶτοι πρὸς ἀλλήλους ἔσονται."", ""GreekTextWordCount"" -> 20, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 24}}, ""Proof"" -> ""For let the two numbers A, B be prime to the two numbers C, D; both to each, and let A by multiplying B make E, and let C by multiplying D make F; I say that E, F are prime to one another. For, since each of the numbers A, B is prime to C, therefore the product of A, B will also be prime to C. [VII. 24] But the product of A, B is E; therefore E, C are prime to one another. For the same reason E, D are also prime to one another. Therefore each of the numbers C, D is prime to E. Therefore the product of C, D will also be prime to E. [VII. 24] But the product of C, D is F. Therefore E, F are prime to one another."", ""ProofWordCount"" -> 139, ""GreekProof"" -> ""δύο γὰρ ἀριθμοὶ οἱ Α, Β πρὸς δύο ἀριθμοὺς τοὺς Γ, Δ ἀμφότεροι πρὸς ἑκάτερον πρῶτοι ἔστωσαν, καὶ ὁ μὲν Α τὸν Β πολλαπλασιάσας τὸν Ε ποιείτω, ὁ δὲ Γ τὸν Δ πολλαπλασιάσας τὸν Ζ ποιείτω: λέγω, ὅτι οἱ Ε, Ζ πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐπεὶ γὰρ ἑκάτερος τῶν Α, Β πρὸς τὸν Γ πρῶτός ἐστιν, καὶ ὁ ἐκ τῶν Α, Β ἄρα γενόμενος πρὸς τὸν Γ πρῶτος ἔσται. ὁ δὲ ἐκ τῶν Α, Β γενόμενός ἐστιν ὁ Ε: οἱ Ε, Γ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. διὰ τὰ αὐτὰ δὴ καὶ οἱ Δ, Ε πρῶτοι πρὸς ἀλλήλους εἰσίν. ἑκάτερος ἄρα τῶν Γ, Δ πρὸς τὸν Ε πρῶτός ἐστιν. καὶ ὁ ἐκ τῶν Γ, Δ ἄρα γενόμενος πρὸς τὸν Ε πρῶτος ἔσται. ὁ δὲ ἐκ τῶν Γ, Δ γενόμενός ἐστιν ὁ Ζ. οἱ Ε, Ζ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 143|>","<|""VertexLabel"" -> ""7.27"", ""Text"" -> ""If two numbers be prime to one another, and each by multiplying itself make a certain number, the products will be prime to one another; and, if the original numbers by multiplying the products make certain numbers, the latter will also be prime to one another [and this is always the case with the extremes]."", ""TextWordCount"" -> 55, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ πολλαπλασιάσας ἑκάτερος ἑαυτὸν ποιῇ τινα, οἱ γενόμενοι ἐξ αὐτῶν πρῶτοι πρὸς ἀλλήλους ἔσονται, κἂν οἱ ἐξ ἀρχῆς τοὺς γενομένους πολλαπλασιάσαντες ποιῶσί τινας, κἀκεῖνοι πρῶτοι πρὸς ἀλλήλους ἔσονται καὶ ἀεὶ περὶ τοὺς ἄκρους τοῦτο συμβαίνει."", ""GreekTextWordCount"" -> 42, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 25}, {""Book"" -> 7, ""Theorem"" -> 26}}, ""Proof"" -> ""Let A, B be two numbers prime to one another, let A by multiplying itself make C, and by multiplying C make D, and let B by multiplying itself make E, and by multiplying E make F; I say that both C, E and D, F are prime to one another. For, since A, B are prime to one another, and A by multiplying itself has made C, therefore C, B are prime to one another. [VII. 25] Since then C, B are prime to one another, and B by multiplying itself has made E, therefore C, E are prime to one another. [id.] Again, since A, B are prime to one another, and B by multiplying itself has made E, therefore A, E are prime to one another. [id.] Since then the two numbers A, C are prime to the two numbers B, E, both to each, therefore also the product of A, C is prime to the product of B, E. [VII. 26] And the product of A, C is D, and the product of B, E is F. Therefore D, F are prime to one another."", ""ProofWordCount"" -> 189, ""GreekProof"" -> ""ἔστωσαν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ Α, Β, καὶ ὁ Α ἑαυτὸν μὲν πολλαπλασιάσας τὸν Γ ποιείτω, τὸν δὲ Γ πολλαπλασιάσας τὸν Δ ποιείτω, ὁ δὲ Β ἑαυτὸν μὲν πολλαπλασιάσας τὸν Ε ποιείτω, τὸν δὲ Ε πολλαπλασιάσας τὸν Ζ ποιείτω: λέγω, ὅτι οἵ τε Γ, Ε καὶ οἱ Δ, Ζ πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐπεὶ γὰρ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Α ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν, οἱ Γ, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐπεὶ οὖν οἱ Γ, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, οἱ Γ, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. πάλιν, ἐπεὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους εἰσίν, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Ε πεποίηκεν, οἱ Α, Ε ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. ἐπεὶ οὖν δύο ἀριθμοὶ οἱ Α, Γ πρὸς δύο ἀριθμοὺς τοὺς Β, Ε ἀμφότεροι πρὸς ἑκάτερον πρῶτοί εἰσιν, καὶ ὁ ἐκ τῶν Α, Γ ἄρα γενόμενος πρὸς τὸν ἐκ τῶν Β, Ε πρῶτός ἐστιν. καί ἐστιν ὁ μὲν ἐκ τῶν Α, Γ ὁ Δ, ὁ δὲ ἐκ τῶν Β, Ε ὁ Ζ. οἱ Δ, Ζ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 193|>","<|""VertexLabel"" -> ""7.28"", ""Text"" -> ""If two numbers be prime to one another, the sum will also be prime to each of them; and, if the sum of two numbers be prime to any one of them, the original numbers will also be prime to one another."", ""TextWordCount"" -> 42, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, καὶ συναμφότερος πρὸς ἑκάτερον αὐτῶν πρῶτος ἔσται: καὶ ἐὰν συναμφότερος πρὸς ἕνα τινὰ αὐτῶν πρῶτος ᾖ, καὶ οἱ ἐξ ἀρχῆς ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ἔσονται."", ""GreekTextWordCount"" -> 32, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 12}}, ""Proof"" -> ""For let two numbers AB, BC prime to one another be added; I say that the sum AC is also prime to each of the numbers AB, BC. For, if CA, AB are not prime to one another, some number will measure CA, AB. Let a number measure them, and let it be D. Since then D measures CA, AB, therefore it will also measure the remainder BC. But it also measures BA; therefore D measures AB, BC which are prime to one another: which is impossible. [VII. Def. 12] Therefore no number will measure the numbers CA, AB; therefore CA, AB are prime to one another. For the same reason AC, CB are also prime to one another. Therefore CA is prime to each of the numbers AB, BC. Again, let CA, AB be prime to one another; I say that AB, BC are also prime to one another. For, if AB, BC are not prime to one another, some number will measure AB, BC. Let a number measure them, and let it be D. Now, since D measures each of the numbers AB, BC, it will also measure the whole CA. But it also measures AB; therefore D measures CA, AB which are prime to one another: which is impossible. [VII. Def. 12] Therefore no number will measure the numbers AB, BC. Therefore AB, BC are prime to one another."", ""ProofWordCount"" -> 232, ""GreekProof"" -> ""Συγκείσθωσαν γὰρ δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους οἱ ΑΒ, ΒΓ: λέγω, ὅτι καὶ συναμφότερος ὁ ΑΓ πρὸς ἑκάτερον τῶν ΑΒ, ΒΓ πρῶτός ἐστιν. εἰ γὰρ μή εἰσιν οἱ ΓΑ, ΑΒ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς ΓΑ, ΑΒ ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. ἐπεὶ οὖν ὁ Δ τοὺς ΓΑ, ΑΒ μετρεῖ, καὶ λοιπὸν ἄρα τὸν ΒΓ μετρήσει. μετρεῖ δὲ καὶ τὸν ΒΑ: ὁ Δ ἄρα τοὺς ΑΒ, ΒΓ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς ΓΑ, ΑΒ ἀριθμοὺς ἀριθμός τις μετρήσει: οἱ ΓΑ, ΑΒ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. διὰ τὰ αὐτὰ δὴ καὶ οἱ ΑΓ, ΓΒ πρῶτοι πρὸς ἀλλήλους εἰσίν. ὁ ΓΑ ἄρα πρὸς ἑκάτερον τῶν ΑΒ, ΒΓ πρῶτός ἐστιν. ἔστωσαν δὴ πάλιν οἱ ΓΑ, ΑΒ πρῶτοι πρὸς ἀλλήλους: λέγω, ὅτι καὶ οἱ ΑΒ, ΒΓ πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσιν οἱ ΑΒ, ΒΓ πρῶτοι πρὸς ἀλλήλους, μετρήσει τις τοὺς ΑΒ, ΒΓ ἀριθμός. μετρείτω, καὶ ἔστω ὁ Δ. καὶ ἐπεὶ ὁ Δ ἑκάτερον τῶν ΑΒ, ΒΓ μετρεῖ, καὶ ὅλον ἄρα τὸν ΓΑ μετρήσει. μετρεῖ δὲ καὶ τὸν ΑΒ: ὁ Δ ἄρα τοὺς ΓΑ, ΑΒ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς ΑΒ, ΒΓ ἀριθμοὺς ἀριθμός τις μετρήσει. οἱ ΑΒ, ΒΓ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 210|>","<|""VertexLabel"" -> ""7.29"", ""Text"" -> ""Any prime number is prime to any number which it does not measure."", ""TextWordCount"" -> 13, ""GreekText"" -> ""ἅπας πρῶτος ἀριθμὸς πρὸς ἅπαντα ἀριθμόν, ὃν μὴ μετρεῖ, πρῶτός ἐστιν."", ""GreekTextWordCount"" -> 11, ""References"" -> {}, ""Proof"" -> ""Let A be a prime number, and let it not measure B; I say that B, A are prime to one another. For, if B, A are not prime to one another, some number will measure them. Let C measure them. Since C measures B, and A does not measure B, therefore C is not the same with A. Now, since C measures B, A, therefore it also measures A which is prime, though it is not the same with it: which is impossible. Therefore no number will measure B, A. Therefore A, B are prime to one another."", ""ProofWordCount"" -> 99, ""GreekProof"" -> ""ἔστω πρῶτος ἀριθμὸς ὁ Α καὶ τὸν Β μὴ μετρείτω: λέγω, ὅτι οἱ Β, Α πρῶτοι πρὸς ἀλλήλους εἰσίν. εἰ γὰρ μή εἰσιν οἱ Β, Α πρῶτοι πρὸς ἀλλήλους, μετρήσει τις αὐτοὺς ἀριθμός. μετρείτω ὁ Γ. ἐπεὶ ὁ Γ τὸν Β μετρεῖ, ὁ δὲ Α τὸν Β οὐ μετρεῖ, ὁ Γ ἄρα τῷ Α οὔκ ἐστιν ὁ αὐτός. καὶ ἐπεὶ ὁ Γ τοὺς Β, Α μετρεῖ, καὶ τὸν Α ἄρα μετρεῖ πρῶτον ὄντα μὴ ὢν αὐτῷ ὁ αὐτός: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα τοὺς Β, Α μετρήσει τις ἀριθμός. οἱ Α, Β ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 100|>","<|""VertexLabel"" -> ""7.30"", ""Text"" -> ""If two numbers by multiplying one another make some number, and any prime number measure the product, it will also measure one of the original numbers."", ""TextWordCount"" -> 26, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ πολλαπλασιάσαντες ἀλλήλους ποιῶσί τινα, τὸν δὲ γενόμενον ἐξ αὐτῶν μετρῇ τις πρῶτος ἀριθμός, καὶ ἕνα τῶν ἐξ ἀρχῆς μετρήσει."", ""GreekTextWordCount"" -> 22, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 15}, {""Book"" -> 7, ""Theorem"" -> 19}, {""Book"" -> 7, ""Theorem"" -> 20}, {""Book"" -> 7, ""Theorem"" -> 21}, {""Book"" -> 7, ""Theorem"" -> 29}}, ""Proof"" -> ""For let the two numbers A, B by multiplying one another make C, and let any prime number D measure C; I say that D measures one of the numbers A, B. For let it not measure A. Now D is prime; therefore A, D are prime to one another. [VII. 29] And, as many times as D measures C, so many units let there be in E. Since then D measures C according to the units in E, therefore D by multiplying E has made C. [VII. Def. 15] Further, A by multiplying B has also made C; therefore the product of D, E is equal to the product of A, B. Therefore, as D is to A, so is B to E. [VII. 19] But D, A are prime to one another, primes are also least, [VII. 21] and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less, that is, the antecedent the antecedent and the consequent the consequent; [VII. 20] therefore D measures B. Similarly we can also show that, if D do not measure B, it will measure A. Therefore D measures one of the numbers A, B."", ""ProofWordCount"" -> 207, ""GreekProof"" -> ""δύο γὰρ ἀριθμοὶ οἱ Α, Β πολλαπλασιάσαντες ἀλλήλους τὸν Γ ποιείτωσαν, τὸν δὲ Γ μετρείτω τις πρῶτος ἀριθμὸς ὁ Δ: λέγω, ὅτι ὁ Δ ἕνα τῶν Α, Β μετρεῖ. τὸν γὰρ Α μὴ μετρείτω: καί ἐστι πρῶτος ὁ Δ: οἱ Α, Δ ἄρα πρῶτοι πρὸς ἀλλήλους εἰσίν. καὶ ὁσάκις ὁ Δ τὸν Γ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε. ἐπεὶ οὖν ὁ Δ τὸν Γ μετρεῖ κατὰ τὰς ἐν τῷ Ε μονάδας, ὁ Δ ἄρα τὸν Ε πολλαπλασιάσας τὸν Γ πεποίηκεν. ἀλλὰ μὴν καὶ ὁ Α τὸν β πολλαπλασιάσας τὸν Γ πεποίηκεν: ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Δ, Ε τῷ ἐκ τῶν Α, Β. ἔστιν ἄρα ὡς ὁ Δ πρὸς τὸν Α, οὕτως ὁ Β πρὸς τὸν Ε. οἱ δὲ Δ, Α πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα, τουτέστιν ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον: ὁ Δ ἄρα τὸν Β μετρεῖ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἐὰν τὸν Β μὴ μετρῇ, τὸν Α μετρήσει. ὁ Δ ἄρα ἕνα τῶν Α, Β μετρεῖ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 191|>","<|""VertexLabel"" -> ""7.31"", ""Text"" -> ""Any composite number is measured by some prime number."", ""TextWordCount"" -> 9, ""GreekText"" -> ""ἅπας σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται."", ""GreekTextWordCount"" -> 8, ""References"" -> {}, ""Proof"" -> ""Let A be a composite number; I say that A is measured by some prime number. For, since A is composite,some number will measure it. Let a number measure it, and let it be B. Now, if B is prime, what was enjoined will have been done. But if it is composite, some number will measure it. Let a number measure it, and let it be C. Then, since C measures B, and B measures A, therefore C also measures A. And, if C is prime, what was enjoined will have been done. But if it is composite, some number will measure it. Thus, if the investigation be continued in this way, some prime number will be found which will measure the numberbefore it, which will also measure A. For, if it is not found, an infinite series of numbers will measure the number A, each of which is less than the other: which is impossible in numbers. Therefore some prime number will be found which willmeasure the one before it, which will also measure A. Therefore any composite number is measured by some prime number."", ""ProofWordCount"" -> 187, ""GreekProof"" -> ""ἔστω σύνθετος ἀριθμὸς ὁ Α: λέγω, ὅτι ὁ Α ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. ᾿επεὶ γὰρ σύνθετός ἐστιν ὁ Α, μετρήσει τις αὐτὸν ἀριθμός. μετρείτω, καὶ ἔστω ὁ Β. καὶ εἰ μὲν πρῶτός ἐστιν ὁ Β, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει τις αὐτὸν ἀριθμός. μετρείτω, καὶ ἔστω ὁ Γ. καὶ ἐπεὶ ὁ Γ τὸν Β μετρεῖ, ὁ δὲ Β τὸν Α μετρεῖ, καὶ ὁ Γ ἄρα τὸν Α μετρεῖ. καὶ εἰ μὲν πρῶτός ἐστιν ὁ Γ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει τις αὐτὸν ἀριθμός. τοιαύτης δὴ γινομένης ἐπισκέψεως ληφθήσεταί τις πρῶτος ἀριθμός, ὃς μετρήσει. εἰ γὰρ οὐ ληφθήσεται, μετρήσουσι τὸν Α ἀριθμὸν ἄπειροι ἀριθμοί, ὧν ἕτερος ἑτέρου ἐλάσσων ἐστίν: ὅπερ ἐστὶν ἀδύνατον ἐν ἀριθμοῖς. ληφθήσεταί τις ἄρα πρῶτος ἀριθμός, ὃς μετρήσει τὸν πρὸ ἑαυτοῦ, ὃς καὶ τὸν Α μετρήσει. ἅπας ἄρα σύνθετος ἀριθμὸς ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 150|>","<|""VertexLabel"" -> ""7.32"", ""Text"" -> ""Any number either is prime or is measured by some prime number."", ""TextWordCount"" -> 12, ""GreekText"" -> ""ἅπας ἀριθμὸς ἤτοι πρῶτός ἐστιν ἢ ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται."", ""GreekTextWordCount"" -> 11, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 31}}, ""Proof"" -> ""Let A be a number; I say that A either is prime or is measured by some prime number. If now A is prime, that which was enjoined will have been done. But if it is composite, some prime number will measure it. [VII. 31] Therefore any number either is prime or is measured by some prime number."", ""ProofWordCount"" -> 58, ""GreekProof"" -> ""ἔστω ἀριθμὸς ὁ Α: λέγω, ὅτι ὁ Α ἤτοι πρῶτός ἐστιν ἢ ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται. εἰ μὲν οὖν πρῶτός ἐστιν ὁ Α, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. εἰ δὲ σύνθετος, μετρήσει τις αὐτὸν πρῶτος ἀριθμός. ἅπας ἄρα ἀριθμὸς ἤτοι πρῶτός ἐστιν ἢ ὑπὸ πρώτου τινὸς ἀριθμοῦ μετρεῖται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 52|>","<|""VertexLabel"" -> ""7.33"", ""Text"" -> ""Given as many numbers as we please, to find the least of those which have the same ratio with them."", ""TextWordCount"" -> 20, ""GreekText"" -> ""ἀριθμῶν δοθέντων ὁποσωνοῦν εὑρεῖν τοὺς ἐλαχίστους τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς."", ""GreekTextWordCount"" -> 12, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 15}, {""Book"" -> 7, ""Definition"" -> 20}, {""Book"" -> 7, ""Theorem"" -> 3}, {""Book"" -> 7, ""Theorem"" -> 16}, {""Book"" -> 7, ""Theorem"" -> 19}, {""Book"" -> 7, ""Theorem"" -> 21}}, ""Proof"" -> ""Let A, B, C be the given numbers, as many as we please; thus it is required to find the least ofthose which have the same ratio with A, B, C. A, B, C are either prime to one another or not. Now, if A, B, C are prime to oneanother, they are the least of those which have the same ratio with them. [VII. 21] But, if not, let D the greatest common measure of A, B, C be taken, [VII. 3] and, as many times as D measures the numbers A, B, C respectively, so many units let there be in the numbers E, F, G respectively. Therefore the numbers E, F, G measure the numbers A, B, C respectively according to the units in D. [VII. 16] Therefore E, F, G measure A, B, C the same number oftimes; therefore E, F, G are in the same ratio with A, B, C. [VII. Def. 20] I say next that they are the least that are in that ratio. For, if E, F, G are not the least of those which have the same ratio with A, B, C,there will be numbers less than E, F, G which are in the same ratio with A, B, C. Let them be H, K, L; therefore H measures A the same number of times that the numbers K, L measure the numbers B, C respectively. Now, as many times as H measures A, so many units let there be in M; therefore the numbers K, L also measure the numbers B, C respectively according to the units in M. And, since H measures A according to the units in M,therefore M also measures A according to the units in H. [VII. 16] For the same reason M also measures the numbers B, C according to the units in the numbers K, L respectively; Therefore M measures A, B, C. Now, since H measures A according to the units in M, therefore H by multiplying M has made A. [VII. Def. 15] For the same reason also E by multiplying D has made A. Therefore the product of E, D is equal to the product ofH, M. Therefore, as E is to H, so is M to D. [VII. 19] But E is greater than H; therefore M is also greater than D. And it measures A, B, C:which is impossible, for by hypothesis D is the greatest common measure of A, B, C. Therefore there cannot be any numbers less than E, F, G which are in the same ratio with A, B, C. Therefore E, F, G are the least of those which have thesame ratio with A, B, C."", ""ProofWordCount"" -> 454, ""GreekProof"" -> ""ἔστωσαν οἱ δοθέντες ὁποσοιοῦν ἀριθμοὶ οἱ Α, Β, Γ: δεῖ δὴ εὑρεῖν τοὺς ἐλαχίστους τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ. οἱ Α, Β, Γ γὰρ ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν ἢ οὔ. εἰ μὲν οὖν οἱ Α, Β, Γ πρῶτοι πρὸς ἀλλήλους εἰσίν, ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων αὐτοῖς. εἰ δὲ οὔ, εἰλήφθω τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον ὁ Δ, καὶ ὁσάκις ὁ Δ ἕκαστον τῶν Α, Β, Γ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν ἑκάστῳ τῶν Ε, Ζ, Η. καὶ ἕκαστος ἄρα τῶν Ε, Ζ, Η ἕκαστον τῶν Α, Β, Γ μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας. οἱ Ε, Ζ, Η ἄρα τοὺς Α, Β, Γ ἰσάκις μετροῦσιν: οἱ Ε, Ζ, Η ἄρα τοῖς Α, Β, Γ ἐν τῷ αὐτῷ λόγῳ εἰσίν. λέγω δή, ὅτι καὶ ἐλάχιστοι. εἰ γὰρ μή εἰσιν οἱ Ε, Ζ, Η ἐλάχιστοι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ, ἔσονται τινες τῶν Ε, Ζ, Η ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β, Γ. ἔστωσαν οἱ Θ, Κ, Λ: ἰσάκις ἄρα ὁ Θ τὸν Α μετρεῖ καὶ ἑκάτερος τῶν Κ, Λ ἑκάτερον τῶν Β, Γ. ὁσάκις δὲ ὁ Θ τὸν Α μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Μ: καὶ ἑκάτερος ἄρα τῶν Κ, Λ ἑκάτερον τῶν Β, Γ μετρεῖ κατὰ τὰς ἐν τῷ Μ μονάδας. καὶ ἐπεὶ ὁ Θ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Μ μονάδας, καὶ ὁ Μ ἄρα τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Θ μονάδας. διὰ τὰ αὐτὰ δὴ ὁ Μ καὶ ἑκάτερον τῶν Β, Γ μετρεῖ κατὰ τὰς ἐν ἑκατέρῳ τῶν Κ, Λ μονάδας: ὁ Μ ἄρα τοὺς Α, Β, Γ μετρεῖ. καὶ ἐπεὶ ὁ Θ τὸν Α μετρεῖ κατὰ τὰς ἐν τῷ Μ μονάδας, ὁ Θ ἄρα τὸν Μ πολλαπλασιάσας τὸν Α πεποίηκεν. διὰ τὰ αὐτὰ δὴ καὶ ὁ Ε τὸν Δ πολλαπλασιάσας τὸν Α πεποίηκεν. ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Ε, Δ τῷ ἐκ τῶν Θ, Μ. ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Θ, οὕτως ὁ Μ πρὸς τὸν Δ. μείζων δὲ ὁ Ε τοῦ Θ: μείζων ἄρα καὶ ὁ Μ τοῦ Δ. καὶ μετρεῖ τοὺς Α, Β, Γ: ὅπερ ἐστὶν ἀδύνατον: ὑπόκειται γὰρ ὁ Δ τῶν Α, Β, Γ τὸ μέγιστον κοινὸν μέτρον. οὐκ ἄρα ἔσονταί τινες τῶν Ε, Ζ, Η ἐλάσσονες ἀριθμοὶ ἐν τῷ αὐτῷ λόγῳ ὄντες τοῖς Α, Β, Γ. οἱ Ε, Ζ, Η ἄρα ἐλάχιστοί εἰσι τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β, Γ: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 410|>","<|""VertexLabel"" -> ""7.34"", ""Text"" -> ""Given two numbers, to find the least number which they measure."", ""TextWordCount"" -> 11, ""GreekText"" -> ""δύο ἀριθμῶν δοθέντων εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν."", ""GreekTextWordCount"" -> 8, ""References"" -> {{""Book"" -> 7, ""Definition"" -> 15}, {""Book"" -> 7, ""Theorem"" -> 16}, {""Book"" -> 7, ""Theorem"" -> 17}, {""Book"" -> 7, ""Theorem"" -> 19}, {""Book"" -> 7, ""Theorem"" -> 20}, {""Book"" -> 7, ""Theorem"" -> 21}, {""Book"" -> 7, ""Theorem"" -> 33}}, ""Proof"" -> ""Let A, B be the two given numbers; thus it is required to find the least number which they measure. Now A, B are either prime to one another or not. First, let A, B be prime to one another, and let A by multiplying B make C; therefore also B by multiplying A has made C. [VII. 16] Therefore A, B measure C I say next that it is also the least number they measure. For, if not, A, B will measure some number which is less than C. Let them measure D. Then, as many times as A measures D, so many units let there be in E, and, as many times as B measures D, so many units let there be in F; therefore A by multiplying E has made D, and B by multiplying F has made D; [VII. Def. 15] therefore the product of A, E is equal to the product of B, F. Therefore, as A is to B, so is F E. [VII. 19] But A, B are prime, primes are also least, [VII. 21] and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore B measures E, as consequent consequent. And, since A by multiplying B, E has made C, D, therefore, as B is to E, so is C to D. [VII. 17] But B measures E; therefore C also measures D, the greater the less: which is impossible. Therefore A, B do not measure any number less than C; therefore C is the least that is measured by A, B. Next, let A, B not be prime to one another, and let F, E, the least numbers of those which have the same ratio with A, B, be taken; [VII. 33] therefore the product of A, E is equal to the product of B, F. [VII. 19] And let A by multiplying E make C; therefore also B by multiplying F has made C; therefore A, B measure C. I say next that it is also the least number that they measure. For, if not, A, B will measure some number which is less than C. Let them measure D. And, as many times as A measures D, so many units let there be in G, and, as many times as B measures D, so many units let there be in H. Therefore A by multiplying G has made D, and B by multiplying H has made D. Therefore the product of A, G is equal to the product of B, H; therefore, as A is to B, so is H to G. [VII. 19] But, as A is to B, so is F to E. Therefore also, as F is to E, so is H to G. But F, E are least, and the least measure the numbers which have the same ratio the same number of times, the greater the greater and the less the less; [VII. 20] therefore E measures G. And, since A by multiplying E, G has made C, D, therefore, as E is to G, so is C to D. [VII. 17] But E measures G; therefore C also measures D, the greater the less: which is impossible. Therefore A, B will not measure any number which is less than C. Therefore C is the least that is measured by A, B."", ""ProofWordCount"" -> 576, ""GreekProof"" -> ""ἔστωσαν οἱ δοθέντες δύο ἀριθμοὶ οἱ Α, Β: δεῖ δὴ εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν. οἱ Α, Β γὰρ ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν ἢ οὔ. ἔστωσαν πρότερον οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, καὶ ὁ Α τὸν Β πολλαπλασιάσας τὸν Γ ποιείτω: καὶ ὁ Β ἄρα τὸν Α πολλαπλασιάσας τὸν Γ πεποίηκεν. οἱ Α, Β ἄρα τὸν Γ μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσί τινα ἀριθμὸν οἱ Α, Β ἐλάσσονα ὄντα τοῦ Γ. μετρείτωσαν τὸν Δ. καὶ ὁσάκις ὁ Α τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ε, ὁσάκις δὲ ὁ Β τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Ζ: ὁ μὲν Α ἄρα τὸν Ε πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ δὲ Β τὸν Ζ πολλαπλασιάσας τὸν Δ πεποίηκεν: ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Α, Ε τῷ ἐκ τῶν Β, Ζ. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Ζ πρὸς τὸν Ε. οἱ δὲ Α, Β πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα: ὁ Β ἄρα τὸν Ε μετρεῖ, ὡς ἑπόμενος ἑπόμενον. καὶ ἐπεὶ ὁ Α τοὺς Β, Ε πολλαπλασιάσας τοὺς Γ, Δ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Β πρὸς τὸν Ε, οὕτως ὁ Γ πρὸς τὸν Δ. μετρεῖ δὲ ὁ Β τὸν Ε: μετρεῖ ἄρα καὶ ὁ Γ τὸν Δ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β μετροῦσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Γ. ὁ Γ ἄρα ἐλάχιστος ὢν ὑπὸ τῶν Α, Β μετρεῖται. μὴ ἔστωσαν δὴ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, καὶ εἰλήφθωσαν ἐλάχιστοι ἀριθμοὶ τῶν τὸν αὐτὸν λόγον ἐχόντων τοῖς Α, Β οἱ Ζ, Ε: ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Α, Ε τῷ ἐκ τῶν Β, Ζ. καὶ ὁ Α τὸν Ε πολλαπλασιάσας τὸν Γ ποιείτω: καὶ ὁ Β ἄρα τὸν Ζ πολλαπλασιάσας τὸν Γ πεποίηκεν: οἱ Α, Β ἄρα τὸν Γ μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσί τινα ἀριθμὸν οἱ Α, Β ἐλάσσονα ὄντα τοῦ Γ. μετρείτωσαν τὸν Δ. καὶ ὁσάκις μὲν ὁ Α τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Η, ὁσάκις δὲ ὁ Β τὸν Δ μετρεῖ, τοσαῦται μονάδες ἔστωσαν ἐν τῷ Θ. ὁ μὲν Α ἄρα τὸν Η πολλαπλασιάσας τὸν Δ πεποίηκεν, ὁ δὲ Β τὸν Θ πολλαπλασιάσας τὸν Δ πεποίηκεν. ἴσος ἄρα ἐστὶν ὁ ἐκ τῶν Α, Η τῷ ἐκ τῶν Β, Θ: ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Θ πρὸς τὸν Η. ὡς δὲ ὁ Α πρὸς τὸν Β, οὕτως ὁ Ζ πρὸς τὸν Ε: καὶ ὡς ἄρα ὁ Ζ πρὸς τὸν Ε, οὕτως ὁ Θ πρὸς τὸν Η. οἱ δὲ Ζ, Ε ἐλάχιστοι, οἱ δὲ ἐλάχιστοι μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε μείζων τὸν μείζονα καὶ ὁ ἐλάσσων τὸν ἐλάσσονα: ὁ Ε ἄρα τὸν Η μετρεῖ. καὶ ἐπεὶ ὁ Α τοὺς Ε, Η πολλαπλασιάσας τοὺς Γ, Δ πεποίηκεν, ἔστιν ἄρα ὡς ὁ Ε πρὸς τὸν Η, οὕτως ὁ Γ πρὸς τὸν Δ. ὁ δὲ Ε τὸν Η μετρεῖ: καὶ ὁ Γ ἄρα τὸν Δ μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Γ. ὁ Γ ἄρα ἐλάχιστος ὢν ὑπὸ τῶν Α, Β μετρεῖται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 546|>","<|""VertexLabel"" -> ""7.35"", ""Text"" -> ""If two numbers measure any number, the least number measured by them will also measure the same."", ""TextWordCount"" -> 17, ""GreekText"" -> ""ἐὰν δύο ἀριθμοὶ ἀριθμόν τινα μετρῶσιν, καὶ ὁ ἐλάχιστος ὑπ᾽ αὐτῶν μετρούμενος τὸν αὐτὸν μετρήσει."", ""GreekTextWordCount"" -> 16, ""References"" -> {}, ""Proof"" -> ""For let the two numbers A, B measure any number CD, and let E be the least that they measure; I say that E also measures CD. For, if E does not measure CD, let E, measuring DF, leave CF less than itself. Now, since A, B measure E, and E measures DF, therefore A, B will also measure DF. But they also measure the whole CD; therefore they will also measure the remainder CF which is less than E: which is impossible. Therefore E cannot fail to measure CD; therefore it measures it."", ""ProofWordCount"" -> 94, ""GreekProof"" -> ""δύο γὰρ ἀριθμοὶ οἱ Α, Β ἀριθμόν τινα τὸν ΓΔ μετρείτωσαν, ἐλάχιστον δὲ τὸν Ε: λέγω, ὅτι καὶ ὁ Ε τὸν ΓΔ μετρεῖ. εἰ γὰρ οὐ μετρεῖ ὁ Ε τὸν ΓΔ, ὁ Ε τὸν ΔΖ μετρῶν λειπέτω ἑαυτοῦ ἐλάσσονα τὸν ΓΖ. καὶ ἐπεὶ οἱ Α, Β τὸν Ε μετροῦσιν, ὁ δὲ Ε τὸν ΔΖ μετρεῖ, καὶ οἱ Α, Β ἄρα τὸν ΔΖ μετρήσουσιν. μετροῦσι δὲ καὶ ὅλον τὸν ΓΔ: καὶ λοιπὸν ἄρα τὸν ΓΖ μετρήσουσιν ἐλάσσονα ὄντα τοῦ Ε: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οὐ μετρεῖ ὁ Ε τὸν ΓΔ: μετρεῖ ἄρα: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 95|>","<|""VertexLabel"" -> ""7.36"", ""Text"" -> ""Given three numbers, to find the least number which they measure."", ""TextWordCount"" -> 11, ""GreekText"" -> ""τριῶν ἀριθμῶν δοθέντων εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν."", ""GreekTextWordCount"" -> 8, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 34}, {""Book"" -> 7, ""Theorem"" -> 35}}, ""Proof"" -> ""Let A, B, C be the three given numbers; thus it is required to find the least number which they measure. Let D, the least number measured by the two numbers A, B, be taken. [VII. 34] Then C either measures, or does not measure, D. First, let it measure it. But A, B also measure D; therefore A, B, C measure D. I say next that it is also the least that they measure. For, if not, A, B, C will measure some number which is less than D. Let them measure E. Since A, B, C measure E, therefore also A, B measure E. Therefore the least number measured by A, B will also measure E. [VII. 35] But D is the least number measured by A, B; therefore D will measure E, the greater the less: which is impossible. Therefore A, B, C will not measure any number which is less than D; therefore D is the least that A, B, C measure. Again, let C not measure D, and let E, the least number measured by C, D, be taken. [VII. 34] Since A, B measure D, and D measures E, therefore also A, B measure E. But C also measures E; therefore also A, B, C measure E. I say next that it is also the least that they measure. For, if not, A, B, C will measure some number which is less than E. Let them measure F. Since A, B, C measure F, therefore also A, B measure F; therefore the least number measured by A, B will also measure F. [VII. 35] But D is the least number measured by A, B; therefore D measures F. But C also measures F; therefore D, C measure F, so that the least number measured by D, C will also measure F. But E is the least number measured by C, D; therefore E measures F, the greater the less: which is impossible. Therefore A, B, C will not measure any number which is less than E. Therefore E is the least that is measured by A, B, C."", ""ProofWordCount"" -> 354, ""GreekProof"" -> ""ἔστωσαν οἱ δοθέντες τρεῖς ἀριθμοὶ οἱ Α, Β, Γ: δεῖ δὴ εὑρεῖν, ὃν ἐλάχιστον μετροῦσιν ἀριθμόν. εἰλήφθω γὰρ ὑπὸ δύο τῶν Α, Β ἐλάχιστος μετρούμενος ὁ Δ. ὁ δὴ Γ τὸν Δ ἤτοι μετρεῖ ἢ οὐ μετρεῖ. μετρείτω πρότερον. μετροῦσι δὲ καὶ οἱ Α, Β τὸν Δ: οἱ Α, Β, Γ ἄρα τὸν Δ μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσιν τινα ἀριθμὸν οἱ Α, Β, Γ ἐλάσσονα ὄντα τοῦ Δ. μετρείτωσαν τὸν Ε. ἐπεὶ οἱ Α, Β, Γ τὸν Ε μετροῦσιν, καὶ οἱ Α, Β ἄρα τὸν Ε μετροῦσιν. καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Α, Β μετρούμενος τὸν Ε μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Α, Β μετρούμενός ἐστιν ὁ Δ: ὁ Δ ἄρα τὸν Ε μετρήσει ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β, Γ μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Δ: οἱ Α, Β, Γ ἄρα ἐλάχιστον τὸν Δ μετροῦσιν. μὴ μετρείτω δὴ πάλιν ὁ Γ τὸν Δ, καὶ εἰλήφθω ὑπὸ τῶν Γ, Δ ἐλάχιστος μετρούμενος ἀριθμὸς ὁ Ε. ἐπεὶ οἱ Α, Β τὸν Δ μετροῦσιν, ὁ δὲ Δ τὸν Ε μετρεῖ, καὶ οἱ Α, Β ἄρα τὸν Ε μετροῦσιν. μετρεῖ δὲ καὶ ὁ Γ τὸν Ε: καὶ οἱ Α, Β, Γ ἄρα τὸν Ε μετροῦσιν. λέγω δή, ὅτι καὶ ἐλάχιστον. εἰ γὰρ μή, μετρήσουσί τινα οἱ Α, Β, Γ ἐλάσσονα ὄντα τοῦ Ε. μετρείτωσαν τὸν Ζ. ἐπεὶ οἱ Α, Β, Γ τὸν Ζ μετροῦσιν, καὶ οἱ Α, Β ἄρα τὸν Ζ μετροῦσιν: καὶ ὁ ἐλάχιστος ἄρα ὑπὸ τῶν Α, Β μετρούμενος τὸν Ζ μετρήσει. ἐλάχιστος δὲ ὑπὸ τῶν Α, Β μετρούμενός ἐστιν ὁ Δ: ὁ Δ ἄρα τὸν Ζ μετρεῖ. μετρεῖ δὲ καὶ ὁ Γ τὸν Ζ: οἱ Δ, Γ ἄρα τὸν Ζ μετροῦσιν: ὥστε καὶ ὁ ἐλάχιστος ὑπὸ τῶν Δ, Γ μετρούμενος τὸν Ζ μετρήσει. ὁ δὲ ἐλάχιστος ὑπὸ τῶν Γ, Δ μετρούμενός ἐστιν ὁ Ε: ὁ Ε ἄρα τὸν Ζ μετρεῖ ὁ μείζων τὸν ἐλάσσονα: ὅπερ ἐστὶν ἀδύνατον. οὐκ ἄρα οἱ Α, Β, Γ μετρήσουσί τινα ἀριθμὸν ἐλάσσονα ὄντα τοῦ Ε. ὁ Ε ἄρα ἐλάχιστος ὢν ὑπὸ τῶν Α, Β, Γ μετρεῖται: ὅπερ ἔδει δεῖξαι."", ""GreekProofWordCount"" -> 348|>","<|""VertexLabel"" -> ""7.37"", ""Text"" -> ""If a number be measured by any number, the number which is measured will have a part called by the same name as the measuring number."", ""TextWordCount"" -> 26, ""GreekText"" -> ""ἐὰν ἀριθμὸς ὑπό τινος ἀριθμοῦ μετρῆται, ὁ μετρούμενος ὁμώνυμον μέρος ἕξει τῷ μετροῦντι."", ""GreekTextWordCount"" -> 13, ""References"" -> {{""Book"" -> 7, ""Theorem"" -> 15}}, ""Proof"" -> ""For let the number A be measured by any number B; I say that A has a part called by the same name as B. For, as many times as B measures A, so many units let there be in C. Since B measures A according to the units in C, and the unit D also measures the number C according to the units in it, therefore the unit D measures the number C the same number of times as B measures A. Therefore, alternately, the unit D measures the number B the same number of times as C measures A; [VII. 15] therefore, whatever part the unit D is of the number B, the same part is C of A also. But the unit D is a part of the number B called by the same name as it; therefore C is also a part of A called by the same name as B, so that A has a part C which is called by the same name as B."", ""ProofWordCount"" -> 171, ""GreekPro