South-Afrika, 2011

Drop a perpendicular from $C_1$ onto $AB$ meeting $AB$ at $C_2$. Similarly define $B_2$ such that $B_1B_2 \perp AC$. Let $M_c$ be on $AC$ such that $MM_c \perp AC$ and $M_b$ on $AB$ such that $MM_b \perp AB$. Without loss of generality, assume that the points on the angle bisector of $\angle A$ are in the order as shown:
$$\text{Now } \triangle AB_1B_2 \sim \triangle AMM_c \sim \triangle AC_1C$$
and $\triangle AC_1C_2 \sim \triangle AMM_b \sim \triangle AB_1B$.

So by the Midpoint Theorem, $B_2M_c = M_cC$ and $C_2M_b = M_bB$.
This means that $B_2M = CM$ and $C_2M = MB$.
Notice that $\angle MB_1B_2 = \angle BB_1M$ and by symmetry, $BB_1 = B_1B_2$.
This means that $\triangle MB_1B_2 \equiv \triangle MB_1B$. Hence $BM = MB_2$.
Thus $BM = MB_2 = CM$.