Solution:

If $M$ and $O$ coincide, then $B C$ is a diameter. Then $\angle A$ is right (this follows from the well-known theorem that an angle inscribed in a semicircle is a right angle).

If $M$ and $O$ do not coincide, then line $M O$ must be the perpendicular bisector of $B C$ since $M$ and $O$ are both equidistant from $B$ and $C$. We are given that $A$ lies on $M O$. So $A$ is equidistant from $B$ and $C$, i.e. $\triangle A B C$ is isosceles.