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!
Since $∠ACB = 120°$ and $AC=BC$ then $∠CAB=∠CBA=30°$ and let be $CH$ the altitude of triangle $ABC$ and $AB=3l$.

$\tan(∠CAB=30°)=\sqrt{3}/3=CH/AH=CH/(l+l/2) \Rightarrow CH=l\sqrt{3}/2$. But $CH$ is also the altitude of the triangle $CMN$ and then 

$\tan(∠CMN=∠CNM)=CH/MH=\frac{l\sqrt{3}/2}{l/2}=\sqrt{3} \Rightarrow ∠CMN=∠CNM=60°$
given


3AM=3MN=3NB=AB
$\angle ACM$=$\angle MCN$=$\angle NCB$(from 1)


So,  $\angle ACM$=$\angle MCN$=$\angle NCB$=60 ($\angle ACB=\angle ACM$+$\angle MCN$+$\angle NCB$)


also given


$\angle ACB=120$

SINCE it is equilateral triangle,


$\angle CAB$=$\angle CBA$=30(using sum of all $ang$ of $\triangle$ is 180)
ALSO
$\angle MCN$=$\angle CMN$=$\angle CMN$=60
(By solving triangle ACM and BCN).