Let $P$ be an interior point of a circle $\Gamma$ centered at $O$ where $P \ne O$. Let $A$ and $B$ be distinct points on $\Gamma$. Lines $AP$ and $BP$ meet $\Gamma$ again at $C$ and $D$, respectively. Let $S$ be any interior point on line segment $PC$. The circumcircle of $\vartriangle ABS$ intersects line segment $PD$ at $T$. The line through $S$ perpendicular to $AC$ intersects $\Gamma$ at $U$ and $V$ . The line through $T$ perpendicular to $BD$ intersects $\Gamma$ at $X$ and $Y$ . Let $M$ and $N$ be the midpoints of $UV$ and $XY$ , respectively. Let $AM$ and $BN$ meet at $Q$. Suppose that $AB$ is not parallel to $CD$. Show that $P, Q$, and $O$ are collinear if and only if $S$ is the midpoint of $PC$.