Kitap 6
7
But, by hypothesis, the angle at C is less than a right angle; therefore the angle BGC is also less than a right angle; so that the angle AGB adjacent to it is greater than a right angle. [I. 13]
7
And it was proved equal to the angle at F; therefore the angle at F is also greater than a right angle.
7
But it is by hypothesis less than a right angle : which is absurd.
7
Therefore the angle ABC is not unequal to the angle DEF; therefore it is equal to it.
7
But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]
7
Therefore the triangle ABC is equiangular with the triangle DEF.
7
But, again, let each of the angles at C, F be supposed not less than a right angle; I say again that, in this case too, the triangle ABC is equiangular with the triangle DEF.
7
For, with the same construction, we can prove similarly that BC is equal to BG; so that the angle at C is also equal to the angle BGC. [I. 5]
7
But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle.
7
Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [I. 17]
7
Therefore, once more, the angle ABC is not unequal to the angle DEF; therefore it is equal to it.
7
But the angle at A is also equal to the angle at D; therefore the remaining angle at C is equal to the remaining angle at F. [I. 32]
7
Therefore the triangle ABC is equiangular with the triangle DEF.
7
Therefore etc. Q. E. D.
