Isosceles triangle and altitude 
$\triangle ABC$ is isosceles with altitude $CH$ and $\angle ACB=120 ^\circ$. $M$ lies on $AB$ such that $AM:MB=1:2$. I should show that $CM$ is the angle bisector of $\angle ACH$.


We can try to show that $\dfrac{AM}{MH}=\dfrac{AC}{CH}$. We have $\dfrac{AC}{CH}=\dfrac{2}{1}$ because $\triangle AHC$ is right-angled and $\angle CAH=30 ^\circ$. How can I show $\dfrac{AM}{MH}=\dfrac{2}{1}$?

Note that 
$$\sin \angle CMH = \frac{MH}{CM}
=\frac{\frac14MB}{MB\sin30}=\frac12$$
Then, $\angle CMH=30$. Thus, $CM$ is the angle bisector.
