Denote the directed angle by the symbols $\measuredangle$ throughout the solution. Since $\measuredangle BAR=\measuredangle DRA-\measuredangle RBA=\measuredangle DOA-\measuredangle RBA=2\measuredangle DBA-\measuredangle RBA=\measuredangle RBA$, then $RA=RB$. Since $Y$ is a reflection of $A$ over $RS$, we have $RA=RY$. Hence $R$ is the circumcenter of $\triangle ABY$. Since $\measuredangle ABY=\measuredangle ARS=\measuredangle ADS=\measuredangle ABC$, and hence $Y$ lies on $BC$. Similarly, $Q$ is the circumcenter of $\triangle DCX$ and $X$ lies on $BC$. Let $(BPX)$ meet $(AOD)$ at $T\neq P$. Consider $$\measuredangle XTQ=\measuredangle XTP+\measuredangle PTQ=\measuredangle XBP+\measuredangle PAQ=\measuredangle CBA+\measuredangle BAC=\measuredangle BCA=\measuredangle XCQ$$and hence $X,C,Q,T$ are concyclic. Consider \begin{align*}\measuredangle RTB&=\measuredangle RTP+\measuredangle PTB=\measuredangle RAB+\measuredangle PXB=\measuredangle RAB+(\measuredangle PXQ+\measuredangle QXB)\\&=\measuredangle RAB+\measuredangle QDP+\measuredangle XCQ=\measuredangle RAB+\measuredangle CAB+\measuredangle BCA\\ &=\measuredangle ABR+\measuredangle CBA=\measuredangle CBR=\measuredangle RYB \end{align*}and hence $R,T,Y,B$ are concyclic. Finally, consider $$\measuredangle STY=\measuredangle STR+\measuredangle RTY=\measuredangle SDR+\measuredangle RBY=\measuredangle CDB+\measuredangle DBC=\measuredangle DCB=\measuredangle SCY$$and hence $S,T,Y,C$ are concyclic which means that $(AOD),(BPX)$ and $(CSY)$ meet at $T$.

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