The 14th Thailand Mathematical Olympiad

First we show that $PORG$ is a parallelogram. Since $R$ and $O$ lie on the perpendicular bisector of chord $CD$, we have $OR$ is perpendicular to $CD$. Let $L$ be the intersection of $PG$ and $CD$, and $X$ be the midpoint of $BG$. Since $\angle (LP,PX)=\angle (GP,PX)=\angle (GA,AB)=\angle (CA,AB)=\angle (CD,DB)=\angle (LD,DX)$, the points $L,P,D,X$ are concyclic. From $PX\perp XG$, we deduce $LP\perp CD$ and thus $GP\parallel OR$.

A similar argument shows that $GR\parallel OP$. Thus, $PORG$ is a parallelogram. Since $PQ$ and $RS$ are perpendicular bisectors of segments $BG$ and $DG$ respectively, we have $PQ\parallel RS$ (since both are perpendicular to $BD$). Similarly we have $QR\parallel PS$, and hence $PQRS$ is a parallelogram. Since the diagonals of a parallelogram bisect each other, from the parallelogram $PQRS$ we deduce that $M$ is the midpoint of $PR$. On the other hand, the parallelogram $PORG$ yields that $M$ is the midpoint of $GO$ as required.