Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

Let EH, FI, GK be drawn through the points E, F, G parallel to AC.

Therefore AF, FG are commensurable; therefore AI is also commensurable with FK. [VI. 1, X. 11]

And, since AF, FG are commensurable in length, therefore AG is also commensurable in length with each of the straight lines AF, FG. [X. 15]

But AG is rational and incommensurable in length with AC; so that AF, FG are so also. [X. 13]

Therefore each of the rectangles AI, FK is medial. [X. 21]

Again, 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]

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]

And, since AG, GD are commensurable in square only, therefore AG is incommensurable in length with GD.

But AG is commensurable in length with AF, and DG with EG; therefore AF is incommensurable in length with EG. [X. 13]

But, as AF is to EG, so is AI to EK; [VI. 1] therefore AI is incommensurable with EK. [X. 11]

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]

Let PR be their diameter, and let the figure be drawn.

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]