Equilateral triangle in a circle

Suppose you have a circle and consider three disjoint $60$ degree arcs $A,B,C$ in the circle. (i.e the arcs $A,B,C$ are separated by three arcs $x,y,z$ (with $x+y+z=180$ degrees and $x,y,z>0$)). Now take the chords on $x,y,z$. Call them $X,Y,Z$ respectively. Prove that the triangle that has its vertices at the midpoints of $X,Y,Z$ is an equilateral triangle.

You have three points $x,y,z$ on the unit circle, and their images under a $60°$ rotation, which is multiplication by $\omega=e^{\frac{2i\pi}{6}}$. The three midpoints are $$m=\frac{z+\omega y}{2},\quad, m'=\frac{x+\omega z}{2},\quad m''=\frac{y+\omega x}{2}$$ and the claim is that they form an equilateral triangle, i.e., the three sides $$\begin{eqnarray} m'-m&=\frac{x-z}{2}+\omega\,\frac{z-y}{2}\\ m''-m'&=\frac{y-x}{2}+\omega\,\frac{x-z}{2}\\ m-m''&=\frac{z-y}{2}+\omega\,\frac{y-x}{2} \end{eqnarray}$$ have the same modulus. But they are of the form $$\begin{eqnarray} A+\omega B\\ C+\omega A\\ B+\omega C \end{eqnarray}$$ with $A+B+C=0$, and therefore must have the same modulus. Indeed, $$|A+\omega B|=|-(B+C)+\omega B|=|(1-\omega)B+C|=|B+\omega C|$$ since $1-\omega=\overline{\omega}=\omega^{-1}$ and $|\omega|=1$. The other equality is derived similarly. Hence $$|m'-m|=|m''-m'|=|m-m''|$$ and the triangle is equilateral.

It can be done (even if it shouldn't be done!) by a mindless calculation.

If $\alpha$ is a cube root of $-1$, other than $-1$, then $1 - \alpha + \alpha^2 = 0.$

Let $u, v, w$ be any three complex numbers. \begin{gather*} 0 = (1 - \alpha + \alpha^2)u = (1 - \alpha + \alpha^2)w \\ \therefore\ (-1 - \alpha^2)w + \alpha^2u + v = \alpha{u} + v - (\alpha{w} + u) \\ \text{i.e. } \alpha^2[(\alpha - 1)w +(u - \alpha{v})] = (\alpha{u} + v) - (\alpha{w} + u) \\ \text{i.e. } \alpha^2[(\alpha{w} + u) - (\alpha{v} + w)] = (\alpha{u} + v) - (\alpha{w} + u) \\ \end{gather*}

Dividing by 2, and taking $\alpha = (1 + \sqrt{3}i)/2$, so that the beginnings and ends of three 60-degree arcs of the unit circle in the complex plane are represented by $u$, $\alpha{u}$, $v$, $\alpha{v}$, $w$, $\alpha{w}$, we see that the complex numbers representing the oriented line segments joining the midpoints $(\alpha{w} + u)/2$, $(\alpha{u} + v)/2$, $(\alpha{v} + w)/2$, are each equal to $\alpha^2$ times the preceding one in cyclic order, i.e. they form an equilateral triangle.

Disjointness doesn't seem to play any part, so long as you take the 6 endpoints in the right order.