I'm trying to solve the following question using complex numbers:
Let $ABCD$ be a convex quadrilateral with the equilateral triangles $ABE,BFC,CGD,DHA$ constructed externally on its sides. Prove that the line joining the centers (centroids) of triangles $DHA$ and $BFC$ is perpendicular to the line joining $G$ and $E$.
My try: Let $x$ and $y$ be the centers of the triangles $DHA$ and $BFC$, respectively.
Showing that $(E-G)i=k(x-y)$, where $k$ is a constant real number will end the proof.
Let $\omega=e^{\tfrac{\pi i}{6}}$. Than we have all the following:
First of all, we notice that $i=\omega^3$. Now I'll use $\omega$ to express all the vertices:
$E=A+(B-A)\omega^2$
$G=C+(D-C)\omega^2$
$H=D+(A-D)\omega^2$
$G=B+(C-B)\omega^2$
$x=\frac{1}{3}(A+D+H)$
$y=\frac{1}{3}(B+C+F)$
We can also express $x$ and $y$ like this:
$x=D+\frac{1}{2}(A+D)\omega$
$y=B+\frac{1}{2}(B+C)\omega$
I was trying to combine some or all of these identities in various ways, but IT didn't yield the expected result.
Any help will be greatly appreciated.