Kitap 10
91
If an area be contained by a rational straight line and a first apotome, the side of the area is an apotome.
91
For let the area AB be contained by the rational straight line AC and the first apotome AD;
91
I say that the side of the area AB is an apotome.
91
For, since AD is a first apotome, let DG be its annex; therefore AG, GD are rational straight lines commensurable in square only. [X. 73]
91
And the whole AG is commensurable with the rational straight line AC set out, and the square on AG is greater than the square on GD by the square on a straight line commensurable in length with AG; [X. Deff. III. 1] if therefore there be applied to AG a parallelogram equal to the fourth part of the square on DG and deficient by a square figure, it divides it into commensurable parts. [X. 17]
91
Let 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 with FG.
91
And through the points E, F, G let EH, FI, GK be drawn parallel to AC.
91
Now, since AF is commensurable in length with FG, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]
91
But AG is commensurable with AC; therefore each of the straight lines AF, FG is commensurable in length with AC. [X. 12]
91
And AC is rational; therefore each of the straight lines AF, FG is also rational, so that each of the rectangles AI, FK is also rational. [X. 19]
91
Now, 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]
91
But DG 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]
