Kitap 10
18
Let A, BC be two unequal straight lines, of which BC is the greater, and to BC let there be applied a parallelogram equal to the fourth part of the square on the less, A, and deficient by a square figure. Let this be the rectangle BD, DC, [cf. Lemma before X. 17] and let BD be incommensurable in length with DC; I say that the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.
18
For, with the same construction as before, we can prove similarly that the square on BC is greater than the square on A by the square on FD.
18
It is to be proved that BC is incommensurable in length with DF.
18
Since BD is incommensurable in length with DC, therefore BC is also incommensurable in length with CD. [X. 16]
18
But DC is commensurable with the sum of BF, DC; [X. 6] therefore BC is also incommensurable with the sum of BF, DC; [X. 13] so that BC is also incommensurable in length with the remainder FD. [X. 16]
18
And the square on BC is greater than the square on A by the square on FD; therefore the square on BC is greater than the square on A by the square on a straight line incommensurable with BC.
18
Again, let the square on BC be greater than the square on A by the square on a straight line incommensurable with BC, and let there be applied to BC a parallelogram equal to the fourth part of the square on A and deficient by a square figure. Let this be the rectangle BD, DC.
18
It is to be proved that BD is incommensurable in length with DC.
18
For, 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.
18
But the square on BC is greater than the square on A by the square on a straight line incommensurable with BC; therefore BC is incommensurable in length with FD. so that BC is also incommensurable with the remainder, the sum of BF, DC. [X. 16]
