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Brazilian Math Olympiad

Geometry Difficulty 4.5 AIME Prove it Brazil

Let $ABC$ be a triangle. The internal bisector of $\angle B$ meets $AC$ in $P$ and $I$ is the incenter of $ABC$ . Prove that if $AP + AB = CB$ , then $API$ is an isosceles triangle.

Solution

Draw $PP'$ parallel to $IA$ so that $P'$ is on line $AB$ . Then $\triangle PAP'$ is isosceles, which implies that $BC = AB + AP = AB + AP' = BP'$ . This then implies that $\triangle P'BC$ is isosceles, which in turn implies that, since $P$ is on the angle bisector of $\angle B$ , $P'PC$ is also isosceles, with $PP' = PC$ . It then follows, using similarity of triangles and the angle bisector theorem, that
$\frac{IA}{PP'} = \frac{BA}{BP'} = \frac{BA}{BC} = \frac{AP}{PC} = \frac{AP}{PP'}$
from which $IA = AP$ .

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