Austrian MO 2020/Advanced/p1 wrote: Let $ABCD$ be a convex cyclic quadrilateral with the diagonal intersection $S$. Let further be $P$ the circumcenter of the triangle $ABS$ and $Q$ the circumcenter of the triangle $BCS$. The parallel to $AD$ through $P$ and the parallel to $CD$ through $Q$ intersect at point $R$. Prove that $R$ is on $BD$. Karl Czakler It's clear that $CD\perp PS$ and so $PS\perp QR$ and analogously $QS\perp PR$ . Due to fact that $BS$ is the radical axis of $(ABS)$ , $(BSC)$ , we have $\overline{BSD}\perp PQ$ , and since $\{P,S,Q,R\}$ form an orthocentric system , $RS\perp PQ$ . Hence $R\in BD$ . $\textbf{Q.E.D.}$