Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

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]

But BC is greater than BD; therefore KH is also greater than G. [V. 16, V. 14]

Let KE be made equal to G; therefore KE is commensurable in length with BC.

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.]

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]

But BC, CD are commensurable in square only; therefore KF, FH are also commensurable in square only. [X. 11]

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.

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]

But KE is rational and commensurable in length with BC; therefore KF is also rational and commensurable in length with BC. [X. 12]

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]

But BC is commensurable with KF; therefore FH is also commensurable in length with CD. [X. 11]