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Taiwan IMO Selection Camp

Geometry Difficulty 4.5 AIME Prove it Taiwan

Let circle $O$ be the circumcircle of $\Delta ABC$ . The angle bisector of $\angle A$ meets circle $O$ again at $P$ , and meets $BC$ at $D$ . From $D$ , drop a perpendicular to $AB$ , meeting $AB$ at $E$ , and meeting circle $O$ at $Q$ , such that $E$ lies on segment $DQ$ . Then connect $PQ$ to meet $BC$ at $F$ , and connect $AF$ to meet $CQ$ at $G$ . Prove that: $EG$ is parallel to $FC$ .

Solution

Solution: First, since
$\angle PDB = \frac{1}{2}(\widehat{CA} + \widehat{BP}) = \frac{1}{2}(\widehat{CA} + \widehat{CP}) = \angle AQP,$
therefore points $A$ , $D$ , $F$ , $Q$ are concyclic. Hence, $\angle AQE = \angle AQD = \angle AFD = \angle GFC$ .
$\angle EAQ = \angle BAQ = \angle BCQ = \angle FCG,$
thus $\angle CGF = \angle AEQ = 90^\circ$ , and therefore points $A$ , $G$ , $E$ , $Q$ are concyclic, from which $\angle QGE = \angle QAE = \angle QCB$ . Hence, $EG$ is parallel to $FC$ . Q.E.D.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.