Indija TS 2012

Let $H$ be the radical center of the circles $ABCD$, $ABOP$ and $CDPO$. Then the radical axes of any two of these circles, i.e. the lines $AB$, $CD$ and $OP$, pass through $H$. Since $P$ lies on the shorter arcs $AO$ and $DO$, it follows that $H$ lies on the extension of $OP$ beyond $P$. The radical center satisfies $$HA\cdot HB=HC\cdot HD=HO\cdot HP.$$ (1) Since the quadrilateral $ABOP$ is cyclic, $$\begin{array}{rl}\angle QAB+\angle BRQ& =(\angle PAB+\angle QAP)+(\angle BOP-\angle OBR)\\ & =(\angle PAB+\angle BOP)+(\angle QAP-\angle OBR)\\ & =180^{\circ }+(\angle QAP-\angle OBR).\end{array}$$ Therefore, $\angle QAP=\angle OBR$ holds if and only if the quadrilateral $ABRQ$ is cyclic, which is equivalent to $HQ\cdot HR=HA\cdot HB$. Similarly, $\angle QAP=\angle OBR$ holds if and only if $HQ\cdot HR=HC\cdot HD$. Combining with (1), $$\angle QAP=\angle OBR\iff HQ\cdot HR=HA\cdot HB\iff HQ\cdot HR=HC\cdot HD\iff \angle PDQ=\angle RCO.$$