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Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 4.9 AIME Prove it Singapore

In the cyclic quadrilateral $ABCD$ , the sides $AB$ , $DC$ meet at $Q$ , the sides $AD$ , $BC$ meet at $P$ , $M$ is the midpoint of $BD$ . If $\angle APQ = 90^\circ$ , prove that $PM$ is perpendicular to $AB$ .

Solution

Drop perpendicular $DE$ from $D$ onto $AB$ . Join $PE$ . In the cyclic quadrilaterals $DEQP$ and $ABCD$ , we have $\angle PEB = \angle PDQ = \angle PBE$ . Thus $\triangle PBE$ is isosceles. Let $F$ be the midpoint of $BE$ . Then both $PF$ and $MF$ are perpendicular to $AB$ . Thus $P$ , $M$ , $E$ are collinear and $PM$ is perpendicular to $AB$ .

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