Question 15
On the sides of an acute triangle $ABC$ three triangles $A'BC$, $AB'C$, $ABC'$ are constructed outside with $\angle ABC' = \angle A'BC = \angle B'AC = 30^\circ$ and $\angle BAC' = \angle AB'C = \angle A'CB = 90^\circ$. Prove that $A'C' \perp B'M$, where $M$ is the midpoint of $BC$.