Book 10
17
But CD is commensurable in length with CD, BF, for CD is equal to BF. [X. 6]
17
Therefore BC is also commensurable in length with BF, CD, [X. 12] so that BC is also commensurable in length with the remainder FD; [X. 15] therefore the square on BC is greater than the square on A by the square on a straight line commensurable with BC.
17
Next, let the square on BC be greater than the square on A by the square on a straight line commensurable with BC, let a parallelogram be applied to BC equal to the fourth part of the square on A and deficient by a square figure, and let it be the rectangle BD, DC.
17
It is to be proved that BD is commensurable in length with DC.
17
With the same construction, we can prove similarly that the square on BC is greater than the square on A by the square on FD.
17
But the square on BC is greater than the square on A by the square on a straight line commensurable with BC.
17
Therefore BC is commensurable in length with FD, so that BC is also commensurable in length with the remainder, the sum of BF, DC. [X. 15]
17
But the sum of BF, DC is commensurable with DC, [X. 6] so that BC is also commensurable in length with CD; [X. 12] and therefore, separando, BD is commensurable in length with DC. [X. 15]
