For, with the same construction, we can prove, similarly, that AB is double of AF, and CD of CG.
And, since AE is equal to CE, the square on AE is equal to the square on CE.
But the squares on EF, FA are equal to the square on AE,
and the squares on EG, GC equal to the square on CE. [I. 47]
Therefore 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 AB is equal to CD.