<s>[INST] 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!

 [/INST] 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.
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