Leonard Giugiuc
9 October 2016

Let $\Delta ABC\;$ be positively oriented, with $A=a,\;$ $B=b\;$ and $C=c.\;$ Introduce $\displaystyle u=-\frac{1}{2}+i\frac{\sqrt{3}}{2}.\;$ Let's remark that $u^2+u+1=0\;$ and $u^3=1.$

$\Delta DCB\;$ is equilateral and positively oriented, implying $D+Cu+Bu^2=0\;$ so that $D=-cu-bu^2.\;$ Similarly, $E=-au-cu^2\;$ and $F=-bu-au^2.$

Let $G,H,I\;$ be the centers of triangles $DCB,\;$ $EAC,\;$ $FBA,\;$ respectively. We have:

$\displaystyle\begin{align} G&=\frac{1}{3} [(1-u)c+(1-u^2 )b],\\ H&=\frac{1}{3} [(1-u)a+(1-u^2 )c],\\ I&=\frac{1}{3} [(1-u)b+(1-u^2 )a]. \end{align}$

From here,clearly, $G+Hu+Iu^2=0,\;$ making $\Delta GHI\;$ equilateral.