Bulgarian Spring Tournament

Let $AA_1$ be the height from point $A$ in $\triangle ABC$. Note that it lies on the continuation of $BC$, because this triangle is obtuse. The triangle $\triangle AA_1B$ is right-angled with angle $30^\circ$. Therefore, $AA_1 = AM = A_1M = BM = x$. Consequently, we directly find $\angle CMB = 45^\circ$, $\angle A_1MA = 60^\circ$, $\angle A_1MC = 75^\circ$, $\angle MCA_1 = 75^\circ$. So $\triangle MCA_1$ is isosceles. Then $CA_1 = x$ and $\triangle AA_1C$ is isosceles and right-angled, therefore $\angle ACA_1 = 45^\circ$ and therefore $\angle ACM = 30^\circ$.
We will calculate the area of $\triangle ACM$ in two different ways.

$S_{ACM} = \frac{AC \cdot CM}{4}$ because $\angle ACM = 30^\circ$ i.e. the height to $AC$ is equal to $\frac{MC}{2}$. But also $M$ is the middle of $AB$. So $S_{ACM} = \frac{S_{ABC}}{2} = \frac{BC \cdot AA_1}{4} = \frac{BC \cdot BM}{4}$ and we get what we are looking for. $\square$