Let $ABCDEF$ be a hexagon such that the diagonals $AD,BE,CF$ intersect at point $O$, and the area of the triangle formed by any three adjacent points is $2$. Find the area of the hexagon.
What I Tried: Here is a picture :-
I drew some kind of an irregular hexagon for convenience, but I am thinking that is this true only for a certain regular hexagon? In this case the triangles are definitely don't have equal area as $2$, is it possible only for a regular hexagon, along with diagonals $AD,BE,CF$ concurring at $O$? If it is true (which I am not sure), then it is great, but I have found no way to prove it.
As another way, by adding and subtracting areas, I was only able to conclude that $[\Delta BFD] = [\Delta AEC]$, other than that I have no idea.
Can anyone help me here? Thank You.
Best Answer
Given that the area of the triangle formed by any three adjacent points are equal, there are a lot of parallel lines in this hexagon. For example:
$$[\triangle ABC] = [\triangle BCD] \implies BC//AD \implies [\triangle BOC] = [\triangle ABC] = [\triangle BCD]$$
Using this property for each side of the hexagon, the area is $2 \times 6$.
EDIT: Found a squished hexagon with concurrent diagonals which are parallel to the sides: