Bisection Theorem I don't know whether this is true or false But I did try to prove it as true using similar arguments as in the Bisection theorem Statement: Given a simple bounded region in $\mathbb{R}^2$ there exist at least two straight lines where each bisect it into two parts of equal area. But this one I feel it is false but again I cannot find an example, Given as above there exist a point in $\mathbb{R}^2$ such that any line passing through it will bisect the area. Some help please.

There exists such a bisector with any given slope $m$. The proof is pretty much the same as that of the pancake theorem. Let $X$ denote the given bounded region, and assume without loss of generality (by taking a suitable rotation of the plane) that $m = 0$; that is, the required bisector is horizontal. Consider the half-planes $P_t = \\{(x, y): y \leq t\\}$, and let $f(t)$ denote the area of $P_t \cap X$. Assuming $X$ is measurable, $f(t)$ is well-defined and continuous by the fact that $X$ is bounded. (Explicitly, if $X$ lies in the cube $[-N, N]^2$, then $f(t + \epsilon) - f(t) \leq 2N \epsilon$ for $\epsilon > 0$.) Since $f(t) = 0$ for small $t$ and $f(t) = \operatorname{area}(X)$ for $t$ large, it follows that there exists some $t_0$ with $f(t_0) = \frac{1}{2} \operatorname{area}(X)$.