Question 32
Two triangles $ABC$ and $C'B'A$ are congruent. $M$ is the midpoint of $AB$ and the extension of $CB'$ beyond $B'$ passes through $M$. Point $F$ lies on the arc $MC$ (which does not contain $B$) of the circumcircle of $BMC$ in such a way that $\angle FB'A = 90^\circ$, and $\angle C'CB' \neq 90^\circ$. Prove that $\angle B'C'C = \angle CAF$.