[14 points] In $\triangle A B C , \angle B A C = 2 \angle A C B$ and $0 ^ { \circ } < \angle B A C < 120 ^ { \circ }$. A point $M$ is chosen in the interior of $\triangle A B C$ such that $B A = B M$ and $M A = M C$. Prove that $\angle M C B = 30 ^ { \circ }$.
Hint (use this or your own method): Draw a suitable segment $C D$ of appropriate length making an appropriate angle with $C A$.