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.
 A: Given a bounded figure, and some line, there is only one line parallel to the given that will bisect the area by the intermediate value theorem (just examine the area to one side of the line as it is moving from one end to the other). Given a bisecting line, we must have that the point lies on it. Else, we could potentially make two parallel lines that both bisect the area, which cannot happen. Using this, we can construct a counterexample quite easily.
Consider a 5 by 5 square with a 1 by 1 corner cut off. The line of symmetry along the diagonal is one such line so the point must lie on it, as shown by the very bad diagram.
Also, there is another bisecting line parallel to the other diagonal. Note that the area is 24 so the area of the lower triangle is 12, and each side length is $2\sqrt6$. This implies that the intersection point has height $\sqrt6$. However, once you draw your line parallel to the side, you can see that in order for it to bisect the area, it must be $\frac{12}{5}$ units above the side. Thus, there is no such point in this example. 
This theorem probably holds if and only if there's a nontrivial rotational symmetry of order 2n. I haven't verified this though.
A: 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 $f(t_0) = \frac{1}{2} \operatorname{area}(X)$.
