Kitap 10
93
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]
93
And, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD.
93
But AG is commensurable in length with AF, and DG with EG; therefore AF is incommensurable in length with EG. [X. 13]
93
But, as AF is to EG, so is AI to EK; [VI. 1] therefore AI is incommensurable with EK. [X. 11]
93
Now let 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; therefore LM, NO are about the same diameter. [VI. 26]
93
Let PR be their diameter, and let the figure be drawn.
93
Now, 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]
93
But, as AF is to EG, so is AI to EK, and, as EG is to FG, so is EK to FK; [VI. 1] therefore also, as AI is to EK, so is EK to FK; [V. 11] therefore EK is a mean proportional between AI, FK.
93
But MN is also a mean proportional between the squares LM, NO, and AI is equal to LM, and FK to NO; therefore EK is also equal to MN.
93
But MN is equal to LO, and EK equal to DH; therefore the whole DK is also equal to the gnomon UVW and NO.
93
But AK is also equal to LM, NO; therefore the remainder AB is equal to ST, that is, to the square on LN; therefore LN is the side of the area AB.
