Next, let each of the angles at C and F be not less than a right angle: the triangle ABC shall be equiangular to the triangle DEF.
For, the same construction being made, it may be shewn in the same manner, that BC is equal to BG;
therefore the angle BCG is equal to the angle BGC. [I. 5.
But the angle BCG is not less than a right angle; [Hyp.
therefore the angle BCG is not less than a right angle; that is, two angles of the triangle BCG are together not less than two right angles; which is impossible. [I, 17.
Therefore the triangle ABC may be shewn to be equiangular to the triangle DEF, as in the first case.
Lastly, let one of the angles at C and F be a right angle, namely, the angle at C: the triangle ABC shall be equiangular to the triangle DEF.
For, if the triangle ABC be not equiangular to the triangle DEF, at the point B, in the straight line AB, make the angle ABG equal to the angle DEF. [I. 23.
Then it may be shewn, as in the first case, that BC is equal to BG; therefore the angle BCG is equal to the angle BGC [I. 5.
But the angle BCG is a right angle: [Hypothesis.
therefore the angle BGC is a right angle;
that is, two angles of the triangle BCG are together equal to two right angles; which is impossible. [I. 17.
Therefore the triangle ABC is equiangular to the triangle DEF.
Wherefore, if two triangles &c. q.e.d.
