Question 14
Let $ABCDEF$ be a convex hexagon with $\angle A \equiv \angle C \equiv \angle E$ and $\angle B \equiv \angle D \equiv \angle F$, such that there is a point $P$ in its interior that is equidistant from the sides $AB$, $CD$ and $EF$. If $G_1 \neq G_2$ are the centroids of triangles $ACE$ and $BDF$, respectively, prove that $\angle G_1 P G_2 = 60^\circ$.