Napoleon's theorem

Idea. The centre of each small triangle is a complex-linear combination of two vertices of $ABC$; a root-of-unity identity finishes.

Third vertices. Write $\varepsilon=e^{i\pi/3}$. For counter-clockwise $ABC$, the outer vertex across $BC$ is
$$A'=B+(C-B)\varepsilon^{-1}=\varepsilon B+\varepsilon^{-1}C.$$
The centre of $BCA'$ is
$$G_a=\frac{B+C+A'}3=\frac{uB+\bar uC}{\sqrt3},\qquad u=e^{i\pi/6}.$$
By cyclic symmetry,
$$G_b=\frac{uC+\bar uA}{\sqrt3},\qquad G_c=\frac{uA+\bar uB}{\sqrt3}.$$

The key identity. With $\omega=e^{2i\pi/3}$, each of the three brackets in
$$\sqrt3\,(G_a+\omega G_b+\omega^2G_c)=A(\omega\bar u+\omega^2u)+B(u+\omega^2\bar u)+C(\bar u+\omega u)$$
vanishes (e.g. $\omega\bar u=i$, $\omega^2u=-i$). So $G_a+\omega G_b+\omega^2G_c=0$.

Conclusion. Using $1+\omega+\omega^2=0$, $(G_a-G_c)+\omega(G_b-G_c)=0$, i.e. $G_a-G_c=e^{-i\pi/3}(G_b-G_c)$. So $G_aG_bG_c$ is equilateral. $\blacksquare$

Remark. Holds with inward triangles too. The centre of the Napoleon triangle is the centroid of $ABC$.

Lesson. Rotations by $\pm60^\circ$ and $\pm90^\circ$ are multiplications by fixed unit complex numbers; equilateral triangles are detected by $z_1+\omega z_2+\omega^2z_3=0$.