Complex Number in Geometry: equilateral triangle Question states:
"Let $ABCDEF$ be a hexagon in a circle of radius $r$. Show that if $AB = CD = EF = r$, then the midpoints of $BC$, $DE$, and $FA$ are the vertices of an equilateral triangle."
Let $G$ be the midpoint of $BC$, $H$ be midpoint of $DE$, and $I$ be midpoint of $FA$. Also, Let $O$ the center of the circle.
We can use complex number representation.
If we construct a geometric representation of the problem, it becomes evident that triangles $ABO$, $CDO$, and $EFO$ are equilateral triangles. Let $a,b,c,d,e,f,o, g,h,i$ be affixes of $A,B,C,D,E,F,O, G,H$, and $I$, respectively. Then $a$ can be represented as $-ow-bw^2$, $c$ can be represented as $-ow-dw^2$, and $e$ can be represented as $-ow-fw^2$, where $w$ is the cubic root of unity (i.e. $w^3 = 1$).
Also, $g = \frac{b+c}{2}, h = \frac{d+e}{2}$, and $i = \frac{f+a}{2}$. 
After this, I can show that the triangle $GHI$ is equilateral by showing that $g = -hw-iw^2$. However, I cannot seem to prove that, however much I try. Is my approach correct? and if so, could you how the calculations?
If anyone has another approach (i.e. without complex number representation), please let me know.
 A: Choose the origin of the complex plane at $O$ so that its affix is $o=0$. Let $\omega = \text{cis}\,\frac{\pi}{3}$ be the $60^\circ$ rotation, and note that $\omega^3 = -1$ so that $\omega^2 - \omega + 1 = 0$. The problem gives that $b = a \omega$, $d = c \omega$, $f=e \omega$. Then:
$$
\begin{cases}
2(g-h) = b+c-d-e = a \omega + c - c \omega - e = a \omega + c(1-\omega) -e \\
2(i-g) = f+a-b-c = e \omega + a - a \omega - c = a(1- \omega) - c + e \omega
\end{cases}
$$
Multiplying the first equation by $\omega$ and adding the two gives:
$$
\require{cancel}
\begin{align}
2\big(\omega(g-h) + (i-g)\big) &= a(\omega^2 + 1 - \omega) + c\big(\omega(1-w) - 1\big) + \cancel{e( -\omega + \omega)} \\
 & = a(\omega^2 - \omega +1) + c(-\omega^2 + \omega - 1) \\
 & = 0
\end{align}
$$
Therefore $g-i = \omega(g-h$) so the two sides $GI, GH$ have the same length and the angle between them is $60^\circ$, thus $\triangle GHI$ is equilateral.

[ EDIT ] $\;$ Alternate proof, prompted by a very similar question.
With the same notations, the midpoints $\,g=\frac{1}{2}(\omega a + c)$, $h=\frac{1}{2}(\omega c + e)$, $i = \frac{1}{2}(\omega e + a)\,$, then:
$$
\require{cancel}
2\left(g + \omega^2 h + \omega^4 i\right) \,=\, \left(\cancel{\omega a} + \bcancel{c}\right) + \omega^2\left(\bcancel{\omega c} + \xcancel{e}\right) + \omega^4\left(\xcancel{\omega e} + \cancel{a}\right) \,=\, 0
$$
It is known (or easily proved) that $g + \omega^2 h + \omega^4 i = 0$ iff $\triangle GHI$ is equilateral, see e.g. here.
