Book 10
113
But the square on A is rational; therefore the rectangle BC, G is also rational.
113
And it has been applied to the rational straight line BC; therefore G is rational and commensurable in length with BC. [X. 20]
113
Since now the rectangle BC, G is equal to the rectangle BD, KH, therefore, proportionally, as CB is to BD, so is KH to G. [VI. 16]
113
But BC is greater than BD; therefore KH is also greater than G. [V. 16, V. 14]
113
Let KE be made equal to G; therefore KE is commensurable in length with BC.
113
And since, as CB is to BD, so is HK to KE, therefore, convertendo, as BC is to CD, so is KH to HE. [V. 19, Por.]
113
Let it be contrived that, as KH is to HE, so is HF to FE; therefore also the remainder KF is to FH as KH is to HE, that is, as BC is to CD. [V. 19]
113
But BC, CD are commensurable in square only; therefore KF, FH are also commensurable in square only. [X. 11]
113
And since, as KH is to HE, so is KF to FH, while, as KH is to HE, so is HF to FE, therefore also, as KF is to FH, so is HF to FE, [V. 11] so that also, as the first is to the third, so is the square on the first to the square on the second; [V. Def. 9] therefore also, as KF is to FE, so is the square on KF to the square on FH.
113
But the square on KF is commensurable with the square on FH, for KF, FH are commensurable in square; therefore KF is also commensurable in length with FE, [X. 11] so that KF is also commensurable in length with KE. [X. 15]
113
But KE is rational and commensurable in length with BC; therefore KF is also rational and commensurable in length with BC. [X. 12]
113
And, since, as BC is to CD, so is KF to FH, alternately, as BC is to KF, so is DC to FH. [V. 16]
113
But BC is commensurable with KF; therefore FH is also commensurable in length with CD. [X. 11]
113
But BC, CD are rational straight lines commensurable in square only; therefore KF, FH are also rational straight lines [X. Def. 3] commensurable in square only; therefore KH is binomial. [X. 36]
