Let $ABC$ be a triangle with circumcenter $O$. Let $\omega (O_1)$ be the circumference which passes through $A$ and $B$ and is tangent to $BC$ at $B$. $\omega (O_2)$ the circle that passes through $A$ and $C$ and is tangent to $BC$ at $C$. Let $M$ the midpoint of $O_1O_2$ and $D$ the symmetric point of $O$ with respect to $A$. Prove that $\angle O_1DM = \angle ODO_2$.