Problem from a Math Competition

Let $O_1,O_2$ denote the centers of the two internal circles, which are tangent to circle $O$ at $S,T$ respectively. Note that $\triangle O_1SN, \triangle OST$ are similar isosceles triangles; therefore $NO_1||TO$ . Similarly we can prove $NO_2||SO$ . Hence $NO_2OO_1$ is a parallelogram, which means $O_1O_2$ bisects $ON$ . Since $O_1O_2$ perpendicularly bisects $MN$ , $O_1O_2$ is the midline of $OM$ in $\triangle OMN$ . Hence $OM||O_1O_2$ and $OM\perp MN\implies \angle OMN=90^{\circ}$