In the right triangle $\triangle ABC$, Assume $AB = EM$ , $BM = MC$ and $\angle BME = 50^{\circ}$. Find the angle $\angle ACB$.
Here is my approach:
In order to find the angle $\angle C$, I drew a perpendicular from $M$ to $AC$ and assume its intersation point with $AC$ is $D$. Since $M$ is the midpoint of $AB$ and $MD \parallel AB$ we can conclude $MD = \dfrac{AB}2$ and in ritght triangle $\triangle EMD$, since $MD = \dfrac{EM}2$, $\angle E = 30^{\circ}$ and hence $x = 20^{\circ}$.
Although this approach is straightforward, I would like to see alternative approaches to solve this problem.
In△ABC:
$$\sin x=\frac{AB}{2MC}$$
In△MCE:
$$\frac{\sin x}{sin(50^\circ-x)}=\frac{ME}{MC}$$
As $AB=ME$ \begin{align*} \frac{\sin x}{\sin (50^\circ-x)} &= 2\sin x \\ \sin (50^\circ-x) &= \frac{1}{2} \\ 50^\circ-x&= 30^\circ\\ x&=20^\circ \end{align*}