Book 10
92
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 commensurable in length with FG.
92
Therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]
92
But AG is rational and incommensurable in length with AC; therefore each of the straight lines AF, FG is also rational and incommensurable in length with AC; [X. 13] therefore each of the rectangles AI, FK is medial. [X. 21]
92
Again, since DE is commensurable with EG, therefore DG is also commensurable with each of the straight lines DE, EG. [X. 15]
92
But DG is commensurable in length with AC.
92
Therefore each of the rectangles DH, EK is rational. [X. 19]
92
Let then the square LM be constructed equal to AI, and let there be subtracted NO equal to FK and being about the same angle with LM, namely the angle LPM; therefore the squares LM, NO are about the same diameter. [VI. 26]
92
Let PR be their diameter, and let the figure be drawn.
92
Since then AI, FK are medial and are equal to the squares on LP, PN, the squares on LP, PN are also medial; therefore LP, PN are also medial straight lines commensurable in square only.
92
And, since the rectangle AF, FG is equal to the square on EG, therefore, as AF is to EG, so is EG to FG, [VI. 17] while, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore EK is a mean proportional between AI, FK. [V. 11]
