The answer is no. There are two positions for $C$.
If $M$ and $C$ are on the same side of $AB$, $\angle ACM = \frac{1}{2}\angle C$ and $\angle CMA = \pi - \frac{1}{2} \angle AMB = \pi - \angle C$ and finally $\angle CAM = \pi - \angle ACM - \angle CMA = \frac{1}{2}\angle C$ and thus $\triangle AMC$ is isosceles and $CM = AM$ follows.
However, $C$ and $M$ could be on opposite sides of $AB$ (so that $AMBC$ is shaped like a kite). Then $CM = AM + 2MH$ where $MH$ is the height of $\triangle AMB$ as seen by folding $AMBC$ along $BC$ to get the first case.