Book 10
92
For let the area AB be contained by the rational straight line AC and the second apotome AD; I say that the side of the area AB is a first apotome of a medial straight line.
92
For let DG be the annex to AD; therefore AG, GD are rational straight lines commensurable in square only, [X. 73] and the annex DG is commensurable with the rational straight line AC set out, while the square on the whole AG is greater than the square on the annex GD by the square on a straight line commensurable in length with AG. [X. Deff. III. 2]
92
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 GD and deficient by a square figure, it divides it into commensurable parts. [X. 17]
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.
