AoPS - Problem

Let the midpoint of $BO$ be $G$, let line $BO$ intersect $AC,XY$ at $F,E$ respectively. Let $XY$ intersect $AC$ at $D$. Firstly, we note that $(C,A;D,F)=-1$ so $MF\cdot MD=MA^2 \Longrightarrow AM^2=DM(DM-DF) $ $\Longrightarrow DM\cdot DF=DM^2-AM^2 $ $\Longrightarrow DM\cdot DF= (DM-AM)(DM+CM) $ $\Longrightarrow DA\cdot DC=DM\cdot DF$. Similarly, $(X,Y;D,E)=-1$ so $DN\cdot DE=DX\cdot DY$. Combined, we have $DN\cdot DE=DX\cdot DY=DA\cdot DC=DM\cdot DF$ so $E,F,M,N$ are concyclic. Next, consider that $G,N,M$ are collinear since it is the Newton-Gauss line of complete quadrilateral $BXCOAY$. Hence, $GN\cdot GM=GE\cdot GF$. But $(B,O;F,E)=-1$ implies that $GB^2=GO^2=GE\cdot GF$. Hence $GM\cdot GN =GO^2$ and we are done.