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Geometry Difficulty 4.4 AIME Prove it Saudi Arabia

Let $ABCD$ be a square inscribed in circle $(O)$ . Let $P$ be a point lies on minor arc $CD$ of $(O)$ . Line $PB$ intersects $AC$ at $E$ . Line $PA$ intersects $DB$ at $F$ . Circumcircle of triangle $PEF$ cuts $(O)$ again at $Q$ . Prove that $PQ$ is parallel to $CD$ .

Solution

Let $K$ be the circumcenter of triangle $PEF$ . Note that
$\angle APB = \angle ADB = 45^\circ$
so triangle $KEF$ is right isosceles at $K$ . Thus, quadrilateral $OEKF$ is cyclic.
Hence, $\angle FOK = \angle FEK = 45^\circ$ , this means $OK$ is bisector of angle $DOC$ , it is also perpendicular bisector of $CD$ . Since $KP = KQ$ , $OP = OQ$ , we have $OK \perp PQ$ .
Therefore, $PQ$ and $CD$ are parallel. $\square$

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