Question 98
Let $O$ be the circumcenter of triangle $ABC$, and $\omega$ be the circumcircle of triangle $BOC$. Line $AO$ intersects with circle $\omega$ again at the point $G$. Let $M$ be the midpoint of side $BC$, and the perpendicular bisector of $BC$ meets circle $\omega$ at the points $O$ and $N$.
Prove that the midpoint of the segment $AN$ lies on the radical axis of the circumcircle of triangle $OMG$, and the circle whose diameter is $AO$.
Prove that the midpoint of the segment $AN$ lies on the radical axis of the circumcircle of triangle $OMG$, and the circle whose diameter is $AO$.