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Taiwan IMO Selection Camp

Geometry Difficulty 4.5 AIME Prove it Taiwan

Let $O$ be the circumcenter of acute triangle $ABC$ , and let $AO$ meet $BC$ at point $D$ . From point $D$ , draw two rays perpendicular to $AB$ and $AC$ respectively, meeting $AB$ , $AC$ at $E$ , $F$ respectively, and meeting the circumcircle of $ABC$ at $K$ , $L$ . Prove: if $K$ , $E$ , $F$ , $L$ are concyclic, then $AB = AC$ .

Solution

Since $\angle EKL = \angle DFE = \angle DAE$ , $KL \perp AD$ . Hence $AO$ bisects $KL$ , so $K$ , $L$ are symmetric with respect to $AO$ . Hence $\angle ADK = \angle ADL$ , $\angle BAD = \angle CAD$ . Hence $AB = AC$ .

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