Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.5 AIME Prove it Hong Kong

$PQRS$ is a cyclic quadrilateral with $\angle PSR = 90^\circ$ . $H$ , $K$ are the feet of the perpendiculars from $Q$ to $PR$ , $PS$ (suitably extended if necessary) respectively. Show that $HK$ bisects $QS$ .

Solution

Let $HK$ meet $QS$ at $A$ . Since $\angle PHQ = \angle PKQ = 90^\circ$ , the points $P$ , $Q$ , $H$ , $K$ are concyclic. As both $QK$ and $RS$ are perpendicular to $PS$ , they are parallel. Therefore, we have
$\angle QKA = \angle QKH = \angle QPH = \angle QPR = \angle QSR = \angle SQK = \angle AQK.$
Thus, $AK = AQ$ . As $\angle QKS = 90^\circ$ , $A$ is the midpoint of $QS$ . This means $HK$ bisects $QS$ .

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