Example of polygon. I wonder how to find a point within $n$-sides polygonal shape that makes all sub-triangle areas equal. In an attached picture, how to find point $C$ that makes area $A_1=A_2=A_3=A_4=A_5$?
Thanks.
Another picture for clarifying. Example 2
It is relatively easy to demonstrate that in a triangle $ABC$ the locus of points $X$ such that $[AXB]=[AXC]$ is the median $AA'$ (where $A'$ is the midpoint of $BC$).
With this observation, the construction of the point in question in an arbitrary polygon is the following. For any triangle consisting of three adjacent vertices of the polygon $A_{i-1}A_iA_{i+1}$ construct the median $A_iA'_i$ (where $A'_i$ is the midpoint of $A_{i-1}A_{i+1}$). If all $n$ such "trimedians" intersect in a common point, this is the point in question. If on the other hand they do not intersect in a common point (what is rather a rule for $n>3$), such a point does not exist.
Update:
Below is an example of a pentagon where the point in question (denoted as $F$) do exists. $ABCD$ and $AFDE$ are parallelograms. It is seen that (different from the cases of triangle and quadrilateral) the point $F$ coincides neither with the area centroid $Z$ nor with the vertex centroid $V$.