Prove equilateral triangle in another triangle We got triangle $ABC$ ($AC = BC$). ∠$ACB = 120^\circ$. $AM = MN = NB = \frac 13AB$.
Prove that CNM is equilateral triangle.
Any ideas how to do that? I'd be grateful if someone helps me!
P.S.I am sorry that the drawing is not very accurate!


Let us use complex numbers. There is no loss of generality in assuming $C$ as the origin, $X$ axis along $CA$ and $A$ is the point $(1,0)$. Then $B$ is represented by the complex number $e^{\frac{2\pi i }{3}} = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$. $M$ and $N$ are respectively
$$m = \frac{1 \cdot e^{\frac{2\pi i }{3}} +2 \cdot 1}{3} = \frac{\frac{3}{2} + \frac{i\sqrt{3}}{2}}{3}$$
and 
$$n = \frac{2 \cdot e^{\frac{2\pi i }{3}} +1 \cdot 1}{3} = \frac{i\sqrt{3}}{3}$$
Since
$$m = \left(\frac{1}{2}-\frac{i\sqrt{3}}{2}\right)n$$
It follows that $M$ can be obtained by rotating $CN$ clockwise by $60^\circ$ and consequently, the triangle $CMN$ is equilateral.
