I say that the triangle ABC is equiangular with the triangle DEF , the angle ABC will be equal to the angle DEF , and the remaining angle, namely the angle at C , equal to the remaining angle, the angle at F .

For, if the angle ABC is unequal to the angle DEF , one of them is greater.

Let the angle ABC be greater; and on the straight line AB , and at the point B on it, let the angle ABG be constructed equal to the angle DEF . [[I. 23](elem.1.23)]

Then, since the angle A is equal to D , and the angle ABG to the angle DEF , therefore the remaining angle AGB is equal to the remaining angle DFE . [[I. 32](elem.1.32)]

Therefore the triangle ABG is equiangular with the triangle DEF .

Therefore, as AB is to BG , so is DE to EF [[VI. 4](elem.6.4)]

But, as DE is to EF , so by hypothesis is AB to BC ; therefore AB has the same ratio to each of the straight lines BC , BG ; [[V. 11](elem.5.11)] therefore BC is equal to BG , [[V. 9](elem.5.9)] so that the angle at C is also equal to the angle BGC . [[I. 5](elem.1.5)]

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](elem.1.13)]

And it was proved equal to the angle at F ; therefore the angle at F is also greater than a right angle.

But it is by hypothesis less than a right angle : which is absurd.

Therefore the angle ABC is not unequal to the angle DEF ; therefore it is equal to it.

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](elem.1.32)]

Therefore the triangle ABC is equiangular with the triangle DEF .

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 .

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](elem.1.5)]

But the angle at C is not less than a right angle; therefore neither is the angle BGC less than a right angle.

Thus in the triangle BGC the two angles are not less than two right angles: which is impossible. [[I. 17](elem.1.17)]

Therefore, once more, the angle ABC is not unequal to the angle DEF ; therefore it is equal to it.

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](elem.1.32)]

Therefore the triangle ABC is equiangular with the triangle DEF .

Therefore etc. Q. E. D.