Let $ABC$ be a triangle and $A'$ be the reflection of $A$ about $BC$. Let $M$ be the mid-point of $BC$ and $I$ be the point of intersection of $A'M$ with the circumcircle of triangle $ABC$ (See figure).
To prove that: $\angle ABC=\angle MIB$.
I am not able to make any progress.
Let $AM$ cuts circle second time at $J$. Then by symmetry with respect to perpendicular bisector of $BC$, the line $AM$ goes to line $MA'$ and circle goes to it self (also $B$ to $C$ and vice versa, and $M$ to it self), so $I$ goes to $J$. Thus $$\angle ABC = \angle AJC = \angle MJC = \angle MIB$$