Problem

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Tags: geometry, circumcircle, geometric transformation, reflection, IMO Shortlist



Let $ABC$ be an acute triangle with $D, E, F$ the feet of the altitudes lying on $BC, CA, AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P.$ The lines $BP$ and $DF$ meet at point $Q.$ Prove that $AP = AQ.$ Proposed by Christopher Bradley, United Kingdom