I didn't solve the problem, but I am just wondering if my work can lead to a solution. Work: Extend the parallel to $BC$ passing through $A$ to intersect the circumcircle of $\triangle ABC$ at $C'$. Then $-1=(M,N;O,P_{\infty})=(B,C;A',C')$.Therefore $A'$ lies on the symmedian of $C'$ in $\triangle BC'C$. I was wondering if there was a way to make a connection between the symmedian of $C'$ and the fact that $O$ is the midpoint of $MN$ to prove the problem.

If we have the cercumcircle of ABC then AO meets circle on A'.we know that AO=OA' and MO=ON so AMA'N is a parallelogram.and AN=MA' also AN||MA' so A'MB=60.A'M:MB=2