TITLE: Diagonals dividing an hexagon into regions of the same area.
QUESTION [7 upvotes]: Let's say that a diagonal of a convex polygon with an even number of vertices is a line segment connecting opposite vertices.
I found the following question in an old exercise list that a teacher gave me:

If each of the three diagonals of a convex hexagon divides it into regions of the same area, show that the three diagonals meet at the same point.


In the image below I marked the areas determined by the diagonals

My first impulse was to try to show that $P = 0$. For this I set up the following system:
$\begin{array}{ll}
\Omega/2 &= A+B+Y\\
\Omega/2 &= B+C+Z\\
\Omega/2 &= C+A+X\\
\Omega/2 &= X+Y+A+P\\
\Omega/2 &= Y+Z+B+P\\
\Omega/2 &= Z+X+C+P\\
\end{array}$
Where $\Omega = A+B+C+X+Y+Z+P$ is the hexagon's total area. The system gives:
$\begin{array}{ll}
A &= Z+P\\
B &= X+P\\
C &= Y+P\\
\end{array}$
I don't see how to proceed from that point. Also, I wonder if the same statement is true by replacing the hexagon with a $2n-$agon. Of course, in this case, writing a system with the areas would become unfeasible, so a smarter approach would be needed.

REPLY [5 votes]: The area of a triangle $\triangle ABC$ is $\frac12 ab\sin \widehat{C}$.
In your figure, triangles $Z+P$ and $A$ share an angle, so the product of corresponding sides are equal. The same with $Y+P$ and $C$, and with $X+P$ and $B$. So the product of six sides of $X+P$, $Y+P$ and $Z+P$ equal six sides of $A$, $B$ and $C$.  But the sides of the former are longer than the latter, unless $P$ is a point.