have the angle ABC equal to the angle DEF, and the angle at C equal to the angle at F.
For, if the angles ABC, DEF be not equal, one of them must be greater than the other.
Let ABC be the greater, and at the point B, in the straight line AB, make the angle ABC equal to the angle DEF. [I, 23.
Then, because the angle at A is equal to the angle at D, [Hyp.
and the angle ABG is equal to the angle DEF, [Constr.
therefore the remaining angle AGB is equal to the remaining angle DFE;
therefore the triangle ABG is equiangular to the triangle DEF.
Therefore AB is to BG as DE is to EF. [VI. 4.
But AB is to BG as DE is to EF; [Hypothesis.
therefore AB is to BC as AB is to BG; [V. 11.
therefore BC is equal to BG; [V. 9.
and therefore the angle BCG is equal to the angle BGC. [I. 5.
But the angle BCG is less than a right angle; [Hyp.
therefore the angle BGC is less than a right angle;
and therefore the adjacent angle AGB must be greater than a right angle. [I. 13.
But the angle AGB was shewn to be equal to the angle at F;
therefore the angle at F is greater than a right angle.
But the angle at F is less than a right angle; [Hypothesis.
which is absurd.
Therefore the angles ABC and DEF are not unequal; that is, they are equal.
And the angle at A is equal to the angle at D; [Hypothesis.
therefore the remaining angle at C is equal to the remaining angle at F;
therefore the triangle ABC is equiangular to the triangle DEF.
