Kitap 2
10
Again, since the angle EGF is half a right angle, and the angle at F is right, for it is equal to the opposite angle, the angle at C, [I. 34] the remaining angle FEG is half a right angle; [I. 32] therefore the angle EGF is equal to the angle FEG, so that the side GF is also equal to the side EF. [I. 6]
10
Now, since the square on EC is equal to the square on CA, the squares on EC, CA are double of the square on CA.
10
But the square on EA is equal to the squares on EC, CA; [I. 47] therefore the square on EA is double of the square on AC. [C. N. 1]
10
Again, since FG is equal to EF, the square on FG is also equal to the square on FE; therefore the squares on GF, FE are double of the square on EF.
10
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 EF.
10
And EF is equal to CD; [I. 34] therefore the square on EG is double of the square on CD. But the square on EA was also proved double of the square on AC; therefore the squares on AE, EG are double of the squares on AC, CD.
10
And 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.
10
And DG is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD.
10
Therefore etc. Q. E. D.
Proposition 11.
