As discussed in comments of this question I am opening a new question about this specific problem:
Given a closed polygon $P$, of area $S(P)$ find a point $x\in \mathbb{R}^2$ such that every line through $x$ cuts $P$ into two parts of area at least $S(P)/3$. As suggested by Misha Lavrov I have tried to emulate the proof of centerpoint theorem but I am not sure if I did it right.
We begin by considering halfspaces $\gamma$ such that $S(\gamma \cap P)>2/3S(P)$. Now we move to convex sets $C(\gamma \cap P)= C_{\gamma}$ in order to apply Helly theorem. We notice that for $\gamma_1, \gamma_2, \gamma_3$ we will have $C_{\gamma_1}\cap C_{\gamma_2}\cap C_{\gamma_3} \neq \emptyset$ by the surface condition imposed on our sets. Thus by Helly theorem, intersection of all such $C_{\gamma}$'s contains a point $x$. We will show that this $x$ satisfies the theorem. Assume that it doesn't, then we can find a line through $x$ defining a closed halfplace $\beta$ such that $S(\beta \cap P)< S(P)/3) \implies S(\beta^c \cap P) > 2S(P)/3 \implies x\in \beta^c$. But then $x\in \beta \cap \beta^c$ giving us a contradiction.
I am wondering if this is correct? Mainly I am not sure if my logic about the final contradiction is sound? I am also not sure if I am missing some obvious counterexample to $C_{\gamma_2}\cap C_{\gamma_3} \neq \emptyset$. I think I should be okay as otherwise the surface area of their respective intersections with $P$ would be bigger than $S(P)$ which can't happen.
Anyways I am happy to get any critique and advice!