angle FDB a right angle, for it is equal to the interior and opposite angle ECB; [I. 29.
therefore the remaining angle BFD is half a right angle.
Therefore the angle at B is equal to the angle BFD and the side DF is equal to the side DB. [I. 6.
And because AC is equal to CE [Construction.
the square on AC is equal to the square on CE;
therefore the squares on AC, CE are double of the square on AC.
But the square on AE is equal to the squares on AC, CE, because the angle ACE is a right angle; [I. 47.
therefore the square on AE is double of the square on AC.
Again, because EG is equal to GF, [Construction.
the square on EG is equal to the square on GF;
therefore the squares on EG, GF are double of the square on GF.
But the square on EF is equal to the squares on EG, GF, because the angle EGF is a right angle; [I. 47.
therefore the square on EF is double of the square on GF. And GF is equal to CD; [I. 34.
therefore the square on EF is double of the square on CD.
But it has been shewn that the square on AE is also double of the square on AC.
Therefore the squares on AE, EF are double of the squares on AC, CD.
But the square on AF is equal to the squares on AE, EF, because the angle AEF is a right angle, [I. 47.
Therefore the square on AF is double of the squares on AC, CD.
But the squares on AD, DF are equal to the square on AF, because the angle ADF is a right angle. [I. 47.
Therefore the squares on AD, DF are double of the squares on AC, CD.
And DF is 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.
