Rioplatense Mathematical Olympiad

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 $A{D}^2=AS\cdot AC$, then $A{P}^2=AS\cdot AC$ and thus $△APS\sim △ACP$, where $\angle APS=\angle ACP$. In a similar fashion we get $C{Q}^2=C{D}^2=CS\cdot CA$, 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 $$\angle APS+\angle PSQ=\angle RCT=\angle RCP+\angle ACP\Rightarrow \angle PSQ=\angle RCP.$$