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!

 A: 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 
$\tan(∠CMN=∠CNM)=CH/MH=\frac{l\sqrt{3}/2}{l/2}=\sqrt{3} \Rightarrow ∠CMN=∠CNM=60°$
A: Look at that mirrored triangle

since $\angle ABC = 120^o$ then $\angle ABD = {1 \over 2}\angle ABC =60^o$
also $\angle BAC = \angle ACB = {1 \over 2}(180^o-\angle ABC) = 30^o$
therefore $\angle BAD = 2 \cdot\angle BAD = 60^o$ as well
so $\triangle ABD$ is equilateral, let it's edge be of length $2l$
Solution 1: Pythagoras
and let $|AM|=|MN|=|NC|=2x$
consider $\triangle ABS$, we know that:

*

*$|AB|=2l$, $|BS|=l$, $|AS|=3x$

*$\angle BAS=30^o$, $\angle ABS=60^o$
from the pythagorean equation we get
$$9x^2+l^2=4l^2$$
after simplifying
$$3x^2=l^2$$
now lets divide both sides by $3xl$, we get
$${x \over l}={l \over 3x}$$
and that proportion is actually
$${|MS| \over |BS|}={|BS| \over |AS|}$$
and since $\triangle ABS$ and $\triangle MBS$ share an angle at $S$ they are similar hence $$\angle BMS=\angle ABS = 60^o$$ which proves that $\triangle MBN$ is equilateral
Solution 2: Geometry
if we know some properties of triangles we don't need to calculate anything.
We need following facts:

*

*All 3 medians in every triangle intersect each other in one point (called centroid). What's more that point splits each of them into 2:1 ratio. (The distance between a vertex and the centroid is ${2 \over 3}$ and the distance between centroid and the opposite edge is ${1 \over 3}$.)

*In equilateral triangles medians, altitudes, angle bisectors and perpendicular bisectors are the same segments. (For example in $\triangle MBN$ segment $BS$ is a median of edge MN, its altitude, its bisector as well as it's bisector of $\angle MBS$)

Knowing that much we just need to look at the $\triangle ABD$ and note that CS is a median and N is a centroid. Since $\triangle ABD$ is equilateral all 3 altitudes (and therefore medians) have equal length, and both $|BM|$ and $|AM|$ is ${2 \over 3}$ of that length so $|BM|=|AM|$
which proves that $$|BM|=|MN|=|BN|$$
A: 
Rotate the given picture by $120^\circ$ and $240^\circ$ to get the equilateral triangle $ABB_1$. Note that $$AM = MN=NB=BM_1=M_1N_1=N_1B_1=B_1M_2=M_2N_2=N_2A$$ Thus $MN_1$ is parallel to $AB_1$. Hence $\angle NMC = 60^\circ$. Similarly $\angle MNC = 60^\circ$ and the triangle $MNC$ is equilateral.
A: 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.
A: 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).

