For let DG be the annex to AD;
therefore AG, GD are rational straight lines commensurable in square only, and neither of the straight lines AG, GD is commensurable in length with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex DG by the square on a straight line commensurable with AG. [X. Deff. III. 3]
Since then the square on AG is greater than the square on GD by the square on a straight line commensurable 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 commensurable parts. [X. 17]
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.
Let EH, FI, GK be drawn through the points E, F, G parallel to AC.
Therefore AF, FG are commensurable;
therefore AI is also commensurable with FK. [VI. 1, X. 11]
And, since AF, FG are commensurable in length,
therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]
But AG is rational and incommensurable in length with AC;
so that AF, FG are so also. [X. 13]
Therefore each of the rectangles AI, FK is medial. [X. 21]
Again, since DE is commensurable in length with EG,
therefore DG is also commensurable in length with each of the straight lines DE, EG. [X. 15]
But GD is rational and incommensurable in length with AC;
therefore each of the straight lines DE, EG is also rational and incommensurable in length with AC; [X. 13]
therefore each of the rectangles DH, EK is medial. [X. 21]
And, since AG, GD are commensurable in square only,
therefore AG is incommensurable in length with GD.
But AG is commensurable in length with AF, and DG with EG;
therefore AF is incommensurable in length with EG. [X. 13]
But, as AF is to EG, so is AI to EK; [VI. 1]
therefore AI is incommensurable with EK. [X. 11]