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Rioplatense Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it Argentina

Let $ABCD$ be a convex quadrilateral with $AB > AD$ and $\angle B = \angle D = 90^\circ$ . Let $P$ be the point on side $AB$ such that $AP = AD$ . Lines $PD$ and $BC$ intersect at $Q$ . The perpendicular line to $AC$ through $Q$ intersects $AB$ at $R$ . Let $S$ be the foot of the perpendicular from $D$ to $AC$ . Prove that $\angle PSQ = \angle RCP$ .

Solution

Let $T = QR \cap AC$ and $V = QS \cap AB$ . First, note that from $AP = AD$ , we have $\angle APD = \angle ADP$ . So
$\angle CQP = 90^\circ - \angle QPB = 90^\circ - \angle APD = 90^\circ - \angle ADP = \angle QDC,$
and hence $CQ = CD$ . Now, by metric relations in the right triangle $DAC$ , we have $AD^2 = AS \cdot AC$ , then $AP^2 = AS \cdot AC$ and thus $\triangle APS \sim \triangle ACP$ , where $\angle APS = \angle ACP$ . In a similar fashion we get $CQ^2 = CD^2 = CS \cdot CA$ , which implies $\angle CQS = \angle CAQ$ .
Observe that $QATB$ is a cyclic quadrilateral because $\angle QBA = \angle QTA = 90^\circ$ . Hence $\angle CQS = \angle CAQ = \angle CBT$ , i.e., $SQ \parallel TB$ . Therefore, $\angle AVS = \angle ABT$ . But $\angle AVS = \angle APS + \angle PSQ$ , and $\angle ABT = \angle RBT = \angle RCT$ (cyclic $RBCT$ ). Putting all together we get
$\angle APS + \angle PSQ = \angle RCT = \angle RCP + \angle ACP \Rightarrow \angle PSQ = \angle RCP.$

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