Proving a triangle is equilateral, under some conditions, using vectors. I'm teaching myself about the applications of vectors in geometry and I came across this problem.
Question:
In a circle, three non-intersecting chords $AB$, $CD$, and $EF$ are given, each congruent to the radius of the circle, and the midpoints of the segments $BC$, $DE$, and $FA$ are connected. Prove that the resulting triangle is equilateral.
My attempt: Right now, since this question is intended to be solved through vectors, I've done some messing around and I think I have an idea how to approach it, however, I can't seem to get anywhere with it. I know that certain centres of triangles can be represented quite nicely with vectors, so I then looked into working with those geometrically. After doodling a few cases, it appears as though the triangle in question and the triangle formed by the midpoints of the congruent chords may share the same centroid. Perhaps if the centres of the triangles align, it may be equilateral? All this is new to me, so I'm not entirely sure.
Any help or guidance would be greatly appreciated!
 A: Just in case you are interested, here is a very straightforward proof using complex numbers (and vectors).
Let the centre of the circle be the origin, and points $A$, $B$, and so on, be given by complex numbers $a$, $b$ and so on.
Let $P$ be the midpoint of $AF$, denoted by complex number $p$, and likewise for $Q$ and $R$ the midpoints of $BC$ and $DE$ respectively.
Let $\omega$ be a cube root of unity so that $\omega^3=1$ and $1+\omega+\omega^2=0$.
Multiplying a complex number by $\omega$ rotates it anti-clockwise by $120^o$.
To prove that triangle $PQR$ is equilateral we just have to show that $$\omega\overrightarrow{RP}=\overrightarrow{PQ}$$
Firstly, $$\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{AB}$$
$$\implies b=a+a\omega=-\omega^2a$$
Likewise, $$d=-\omega^2c$$ and $$f=-\omega^2e$$
Now $$p=\frac12(a+f)=\frac12(a-\omega^2e)$$
Similarly, $$q=\frac12(b+c)=\frac12(-\omega^2a+c)$$
and
$$r=\frac12(d+e)=\frac12(-\omega^2c+e)$$
Then we have
$$\overrightarrow{PQ}=q-p=\frac12(-\omega^2a+c-a+\omega^2e)=\frac12(\omega a+c+\omega^2e)$$
And also
$$\overrightarrow{RP}=p-r=\frac12(a-\omega^2e+\omega^2c-e)=\frac12(a+\omega^2c+\omega e)$$
Then, sure enough, $$\omega\overrightarrow{RP}=\frac12(a\omega+c+\omega^2 e)=\overrightarrow{PQ} $$
Hence proved.
