Show that MNPQ and ABCD have the same centroid. If ABCD is a quadrilateral and M,N,P,Q are exterior points such that ABM, BNC, CPD, DQA are equilateral triangles. Show that MNPQ and ABCD have the same centroid.
I tried to solve it by vectors but I don't know how can I use that the triangles are equilateral.

Let $\varepsilon = e^{-i{\pi\over 3}} = \cos {\pi\over 3} -
i\cdot \sin{\pi\over 3}$ and let $G$, $G'$ be gravity centers of $ABCD$
and $MNPQ$, so
$$ G ={1\over 4}(A+B+C+D)$$ and $$G'= {1\over 4}(M+N+P+Q)$$
Now since we get the vector $\vec{AM}$ with rotation of $\vec{AB}$ around $A$ for angle $-{\pi\over 3}$ we have (and similary for others):
\begin{eqnarray}
% \nonumber to remove numbering (before each equation)
  M-A &=& \varepsilon(B-A) \\
 N-B &=& \varepsilon(C-B) \\
P-C &=& \varepsilon(D-C) \\
  Q-D &=& \varepsilon(A-D)
\end{eqnarray}
If we sun these equation and divide it with 4 we get:
$$ G'-G = 0$$
and we are done.
