Question 26
Two circles $\omega_1$ and $\omega_2$ with the same radius meet at $P$ and $Q$. Points $B$ and $C$ lie on the circles $\omega_1$ and $\omega_2$, respectively. Points $X$ and $Y$ different from $P$ are on $\omega_1$ and $\omega_2$, respectively, such that $\measuredangle CXQ = \measuredangle QPC$ and $\measuredangle BYQ = \measuredangle QPB$ (directed angles modulo $180^\circ$; in the configuration of the figure, where the triangles $PCQ$ and $PBQ$ are acute, this means $\angle CXQ + \angle CPQ = 180^\circ$ and $\angle BYQ + \angle BPQ = 180^\circ$). Denote by $S$ the second intersection (other than $P$) of the circumcircles of $XPC$ and $YPB$. Prove that $XY$, $BC$, and $QS$ are concurrent.