Cyclic hexagon gives equilateral triangle 
Suppose that $ABCDEF$ is a cyclic hexagon and the radius of a circumcircle is $r$. Suppose further that $|AB|=|CD|=|EF|=r$. Let the midpoints of $BC, DE$ and $FA$ be $M_1$, $M_2$ and $M_3$ respectively. Is the triangle $M_1M_2M_3$ necessary equilateral?

I tried to use the law of cosines to prove that, but it was hard to prove that those system of equations has a solution only is three variable are equal. I also tried to rotate the triangle but could not found anything useful. I thing it has something to do with the fact that the equal length of hexagon sides is used to perform angle equalities. 
 A: It's true because
$$R_{60^{\circ}}\left(\vec{M_3M_2}\right)=R_{60^{\circ}}\left(\frac{1}{2}\left(\vec{AD}+\vec{FE}\right)\right)=$$
$$=R_{60^{\circ}}\left(\frac{1}{2}\left(\vec{AO}+\vec{OD}+\vec{FE}\right)\right)=$$
$$=\frac{1}{2}\left(\vec{AB}+\vec{OC}+\vec{FO}\right)=$$
$$=\frac{1}{2}\left(\vec{AB}+\vec{FC}\right)=\vec{M_3M_1}.$$
Done!
A: The answer is YES.
WOLOG, consider the case $r = 1$, the circumcenter is located at origin and the vertices are arranged in counterclockwise manner along the circumcircle.
Identify the euclidean plane $\mathbb{R}^2$ with the complex plane $\mathbb{C}$. We will abuse the notation and use same symbol to denote a point in the euclidean plane and the corresponding number in complex plane.
Let $\omega = e^{i\pi/3}$ be the primitive root of unity with order $6$.
Since $|AB| = |CD| = |EF| = r = 1$ and $A, \cdots, F$ are arranged counterclockwisely along the circumcircle, we have
$$
\begin{cases}
B = \omega A,\\
D = \omega C,\\
F = \omega E\\
\end{cases}
\implies 
\begin{cases}
M_1 = \frac12(B+C) = \frac12( \omega A + C)\\
M_2 = \frac12(D+E) = \frac12( \omega C + E)\\
M_3 = \frac12(F+A) = \frac12( \omega E + A)
\end{cases}
$$
Notice following three complex numbers are multiple of each other upto some powers of $\omega^2$.
$$\begin{align}
M_2 - M_1 &= \frac12 ((\omega - 1) C + E - \omega A)
= \frac12 ( E + \omega^2 C + \omega^4 A),\\
M_3 - M_2 &= \frac12((\omega - 1) E + A - \omega C)
= \frac12 ( A + \omega^2 E + \omega^4 C) = \omega^2 (M_2 - M_1)\\
M_1 - M_3 &= \frac12((\omega - 1) A + C  - \omega E)
= \frac12 ( C + \omega^2 A + \omega^4 E ) = \omega^2 (M_3 - M_2)
\end{align}
$$
Translate these relations back to Euclidean plane.
We can obtain the line segment $M_2M_3$ from $M_1M_2$ by a counterclockwise rotation of $120^\circ$ followed by a translation to move $M_1$ into $M_2$. Similarly, we can obtain the line segment $M_3M_1$ from $M_2M_3$ by a counterclockwise rotation of $120^\circ$ followed by a translation to move $M_2$ into $M_3$.
As a result, $\triangle M_1M_2M_3$ is an equilateral triangle.
A: Let $N_1, N_2, N_3$ be the midpoints of segments $EF, AB, CD$ respectively and let $O$ be the center of the circle. The three triangles $ABO, CDO, EFO$ are equilateral so if you perform $60^{\circ}$ rotation around $O$, the points $A, C, E$ are mapped to $B, D, F$ respectively, so triangle $ACE$ is mapped to triangle $BDF$ which means that segments $AC, CE, EA$ are mapped to segments $BD, DF, FB$ respectively. Thus, $|AC| = |BD|$ and the angle between them is $60^{\circ}$, $|CE| = |DF|$ and the angle between them is $60^{\circ}$,  $|EA| = |FB|$ and the angle between them is $60^{\circ}$. 
Now, the segment $N_2M_1$ is a mid-segment of triangle $ABC$, so $N_2M_1 \, || \, AC$ and $|N_2M_1| = \frac{1}{2}|AC|$. Furthermore, segment $M_1N_3$ is a mid-segment of triangle $BCD$, so $M_1N_3 \, || \, BD$ and $|M_1N_3| = \frac{1}{2}|ND|$. Therefore, $\angle \, N_2M_1N_3 = 120^{\circ}$ and $|N_2M_1| = |M_1N_3|$.  Analogously,  $\angle \, N_3M_2N_1 = 120^{\circ}$ and $|N_3M_2| = |M_2N_1|$ as well as  $\angle \, N_2M_1N_3 = 120^{\circ}$ and $|N_2M_1| = |M_1N_3|$. These observations show that if one constructs the three external equilateral triangles on the three sides of the triangle $N_2N_3N_1$ then the points $M_1, M_2, M_3$ are the centroids of the said three equilateral triangles. Since this is the configuration of Napoleon's Theorem, it follows immediately that triangle $M_1M_2M_3$ is equilateral.   
