For let DG be the annex to AD;
therefore AG, GD are rational straight lines commensurable in square only, the annex GD 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 with AG. [X. Deff. III. 5]
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 the point 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.
Now, since AG is incommensurable in length with CA, and both are rational, therefore AK is medial. [X. 21]
Again, since DG is rational and commensurable in length with AC, DK is rational. [X. 19]