Extend $CO$ to a point $D$ such that $CO = OD$. Since $\angle BOC = 90^\circ$, the triangles $BOC$ and $BOD$ are congruent. Let $\omega_1$ be the circumcircle of the triangle $BOD$. Since $\angle BDO = \angle BCO = \angle BAO$, the point $A$ lies on $\omega_1$. As $M$, $O$, $N$ are the midpoints of $CA$, $CD$, $CB$ respectively, the homothety centered at $C$ with similitude ratio $\frac{1}{2}$ maps $\omega_1$ to the circumcircle $\omega_2$ of the triangle $MON$. As $DB$ is a diameter of $\omega_1$, $ON$ is a diameter of $\omega_2$. Thus $\angle OMN = 90^\circ$.