and is therefore equal to the angle DBG;
therefore also the side BD is equal to the side DG. [I. 6.
Again, because the angle EGF is half a right angle, and the angle at F a right angle, for it is equal to the opposite angle ECD; [I. 34.
therefore the remaining angle FEG is half a right angle, [I. 32.
and is therefore equal to the angle EGF;
therefore also the side GF is equal to the side FE. [I. 6.
And because EC is equal to CA, the square on EC is equal to the square on CA;
therefore the squares on EC,CA are double of the square on CA.
But the square on AE is equal to the squares on EC, CA. [I. 47.
Therefore the square on AE is double of the square on AC.
Again, because GF is equal to FE, the square on GF is equal to the square on FE;
therefore the squares on GF, FE are double of the square on FE.
But the square on EG is equal to the squares on GF, FE. [I. 47.
Therefore the square on EG is double of the square on FE. And FE is equal to CD; [I. 34.
Therefore the square on EG is double of the square on CD.
But it has been shewn that the square on AE is double of the square on AC.
Therefore the squares on AE, EG are double of the squares on AC, CD.
But the square on AG is equal to the squares on AE,EG. [I. 47.
Therefore the square on AG is double of the squares on AC, CD.
But the squares on AD, DG are equal to the square on AG. [I. 47.
Therefore the squares on AD, DG are double of the squares on AC, CD.
And DG equal to DB;
therefore the squares on AD, DB are double of the squares on AC,CD.
Wherefore, if a straight line &c. Q.E.D.
