Book 10
94
For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, AG 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. 4]
94
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]
94
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.
94
Let EH, FI, GK be drawn through E, F, G parallel to AC, BD.
94
Since then AG is rational and commensurable in length with AC, therefore the whole AK is rational. [X. 19]
94
Again, since DG is incommensurable in length with AC, and both are rational, therefore DK is medial. [X. 21]
94
Again, since AF is incommensurable in length with FG, therefore AI is also incommensurable with FK. [VI. 1, X. 11]
94
Now let the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and about the same angle, the angle LPM.
94
Therefore the squares LM, NO are about the same diameter. [VI. 26]
