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 $\dfrac{AM}{MH}=\dfrac{2}{1}$?
 A: Let $AC = 2x$. Since triangle $AHC$ is half of the equliateral triangle with side $2x$ (reflect $C$ across $AB$) we see that $CH =x$.
Let $y= AB/6$ then $AH =3y$ and $AM = 2y$ so $MH = y$ and thus $${AC\over CH} = {AM\over MH}$$
so by angle bisector theorem $CM$ is angle bisector for $\angle ACH$.
A: If we reflect $C$ across $AB$ to new point $D$ we see that $ACD$ is equilateral triangle and that $AH:HM = 2:1$. So $M$ is gravity center for $ACD$ which in equalateral triangle is also incenter (and orthocenter and circumcenter) of triangle. So $CM$ is angle bisector for $\angle ACD = \angle ACH$.
A: ACH is a right triangle ( 30,60,90 angles)  with
$$AC:CH:AH = 2:1:\sqrt{3}$$
as given by Pythagoras thm.
When 
$$ \frac{AC}{CH} =\frac{AM}{MH} =\frac{2}{1} $$
is satisfied the $CM$ line is a bisector of $\angle ACH$ by virtue of a converse of a theorem that states: A bisector divides adjacent sides and opposite side of a triangle in the same ratio.
A: Note that 
$$\sin \angle CMH = \frac{MH}{CM}
=\frac{\frac14MB}{MB\sin30}=\frac12$$
Then, $\angle CMH=30$. Thus, $CM$ is the angle bisector.
