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the squares on EF, FA are equal to the squares on EG, GC, of which the square on EF is equal to the square on EG, for EF is equal to EG;
therefore
the square on AF which remains is equal to the square on CG;
therefore
AF is equal to CG.
And AB is double of AF, and CD double of CG;
therefore
Being what it was required to prove.
