For let DG be the annex to AD;
therefore AG, GD are rational straight lines commensurable in square only,
neither of them is commensurable in length with the rational straight line AC set out,
and the square on the whole AG is greater than the square on the annex DG by the square on a straight line incommensurable in length with AG. [X. Deff. III. 6]
Since then the square on AG is greater than the square on GD by the square on a straight line incommensurable in length with AG,
therefore, if there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it will divide it into incommensurable parts. [X. 18]
Let then DG be bisected at E, let there be applied to AG a parallelogram equal to the square on EG and deficient by a square figure,
and let it be the rectangle AF, FG;
therefore AF is incommensurable in length with FG.
But, as AF is to FG, so is AI to FK, [VI. 1]
therefore AI is incommensurable with FK. [X. 11]
And, since AG, AC are rational straight lines commensurable in square only, AK is medial. [X. 21]
Again, since AC, DG are rational straight lines and incommensurable in length, DK is also medial. [X. 21]
Now, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD.
But, as AG is to GD, so is AK to KD; [VI. 1]
therefore AK is incommensurable with KD. [X. 11]