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Iberoamerican Mathematical Olympiad

Geometry Difficulty 4.9 AIME Prove it Ibero-American Mathematical Olympiad

Problem:

$ABC$ is a triangle. The incircle has center $I$ and touches the sides $BC$ , $CA$ , $AB$ at $D$ , $E$ , $F$ respectively. The rays $BI$ and $CI$ meet the line $EF$ at $P$ and $Q$ respectively. Show that if $DPQ$ is isosceles, then $ABC$ is isosceles.

Solution

Solution:

$AF = AE$ , so $\angle AFE = 90^{\circ} - A / 2$ . Hence $\angle BFP = 90^{\circ} + A / 2$ . But $\angle FBP = B / 2$ , so $\angle FPB = C / 2$ . But $BFP$ and $BDP$ are congruent ( $BF = BD$ , $BP$ common, $\angle FBP = \angle FDP$ ), so $\angle DPB = C / 2$ and $\angle DPQ = C$ . Similarly, $\angle DQP = B$ . Hence $\angle PDQ = A$ . So $DQP$ and $ABC$ are similar. So if one is isosceles, so is the other.

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