Regular hexagon divided into triangles Problem: Give an regular hexagon and an interior point of this, join this point with each vertex. The hexagon is divided in $6$ triangles, paint the triangles alternately. Show that the sum of the areas of the painted triangles is equal to that of the unpainted triangles
The same problem is proposed with an square and is easy prove this, because the sum of heights of opposite triangles is $l$ (the side of the square). But in this case I can't. I try to prove that the sum of heights of painted triangles is $\frac{3\sqrt3\cdot l}{2}$. I tried to test this but I was not able, what do you suggest or how else can I do it?
 A: Let $ABCDEF$ be the hexagon.  Extend the alternating sides $AB,CD,EF$ until pairs of them meet at points $G,H,I$, where $\triangle GHI$ is equilateral with sides measuring $l+l+l=3l$.  Then the sum of the distances from these sides to any interior point is twice the area divided by each side of the triangle, which you should be able to render as $3\sqrt3l/2$.
A: Imagine that the center of the hexagon is at the origin, and so placed that the perpendiculars from the center to $3$ alternate sides are $-\hat j$, $\frac{\sqrt3}2\hat i+ \frac12\hat j$, and $-\frac{\sqrt3}2\hat i+ \frac12\hat j$ These are the altitudes of three like-colored triangles.

The statement is clearly true when the chosen point is the center.  When the point moves from the center to a point $x\hat i+y\hat j$, the sides of the triangle don't change, so the change in the triangle's area is proportional to the change in the triangle's height.  This change is the projection of the displacement vector on the altitude of the triangle, which is just the dot product of the displacement and the altitude, since the altitude has unit length.
The total change in area is the dot product of the displacement vector and the sum of the altitudes, and as the sum of the altitudes is $0$, the sum of the areas of the three triangles doesn't change.
