I've been trying to solve this problem:
Let $P\subset\mathbb{R}^2$ be a closed polygon with area $A(P)$, not necessarily convex. Prove that there is a point $x\in\mathbb{R}^2$ such that every line that contains $x$ divides $P$ into two parts, each with area at least $\frac{A(P)}{3}$.
My only idea so far was to use the Shoelace formula, and possibly the Centerpoint theorem to find this point, but nothing has given me the wanted result. I'm not sure what I can use to show this.
Thanks in advance for any suggestions on how to approach this.