Converse of Napoleon's Theorem Consider a fixed scalene triangle $ABC$ together with a point $X$ in the semiplane determined by $BC$ opposite to $A$, a point $Y$ in the semiplane determined by $AC$ opposite to $B$ and a point $Z$ in the semiplane determined by $AB$ oppocite to $C$. Suppose that $\Delta AYC$, $\Delta CXB$ and $\Delta BZA$ are similar triangles (in that order) and moreover $\Delta XYZ$ is equilateral.
Is it true that the angles of $\Delta CXB$ are $30°$, $120°$ and $30°$?
 A: Here's a brute force derivation in the complex plane ...
Without loss of generality, take the vertices of $\triangle ABC$ to be $A=0$, $B=1+0i$, $C=c$. Then we can write
$$\begin{align}
X &= B + \rho (C-B) = 1+\rho(c-1) \\
Y &= C + \rho (A-C) = c(1-\rho) \\
Z &= A + \rho (B-A) = \rho
\end{align} \tag1$$
for some complex $\rho$. (Eg, we dilate $C$ relative to $B$ by scale factor $|\rho|$ then rotate the image about $B$ by $\arg \rho$ to get $X$; likewise for other vertices, thereby erecting identically-oriented similar triangles on each edge of $\triangle ABC$.)
As $\triangle XYZ$ is equilateral, we can get $Z$ from rotating $Y$ about $X$ by $60^\circ$, so
$$Z = X + \omega(Y-X) \quad\to\quad 
c (\rho + \omega(1-2\rho) ) = 
\omega\rho +(\omega -1)(1-2\rho)
\tag2$$
where $\omega := \cos60^\circ\pm i\sin60^\circ$. (The $\pm$ accounts for whether the rotation is clockwise or counter-.) The curious grouping in $(2)$ anticipates invoking the observation that $\omega-1=\omega^2$, which allows us to write
$$c (\rho + \omega(1-2\rho) ) = 
\omega\left(\rho +\omega(1-2\rho)\right) \quad\to\quad (c-\omega)(\rho+\omega(1-2\rho)) = 0 \tag{3}$$
Consequently, either $c = \omega$ (making $\triangle ABC$ equilateral), or else
$$\rho = \frac{\omega}{2\omega-1}=
\frac{\frac12\pm i\frac{\sqrt{3}}{2}}{\pm i\sqrt3}=\frac1{\sqrt3} (\cos30^\circ\mp i\sin30^\circ) \tag4$$
so that $\rho$ effects the exact dilation-and-rotation to erect $30^\circ$-$30^\circ$-$120^\circ$ triangles on the edges of $\triangle ABC$ via $(1)$. (The $\mp$ distinguishes between erecting the triangles externally or internally; which is which depends upon the orientation of the vertices of $\triangle ABC$.) $\square$
