Prove that any line passing through the intersection of two bisectors is also a bisector. Given an arbitrary closed shape $F$, a line, $H$, that bisects $F$ horizontally, and a line, $V$, that bisects $F$ vertically, is it true that any line that passes through the intersection of $H$ and $V$ also bisects $F$?
I know that any given line $L$ through the intersection must divide $F$ into 6 (possibly empty) regions.
It looks kinda like this:
 L\   V
  ┌\──┼───┐
  │ \B│ C │
  │A \│   │
H─┼───\───┼──
  │   │\ D│
  │ F │E\ │
  └───┼──\┘
      │   \

So I have $$A+B+C = C+E+D = D+E+F = F+A+B$$
and $$A+B = C = D+E = F.$$
I just need to prove that $A = D$ or $B = E$, but no matter how I manipulate the given equations, I can't seem to isolate $A$ and $B$ or $D$ and $E$.
I think I should be using the fact that $L$ is a straight line somehow, since the given equations could still be true even if $L$ were curved.
My idea is that, any closed shape should have a centroid, like the center of mass, such that every line that passes through it should bisect the shape. Since there can only one such centroid, it must be the intersection of any two bisectors.

The lines which bisect the area of a triangle form an envelope as shown in this picture

The blue medians intersect in the centroid of the triangle, but no other lines through the centroid bisect the area of the triangle; none of the green bisectors of the area of the triangle pass through the centroid.
It is easiest to consider the horizontal green line parallel to the base.  It must be $\frac1{\sqrt{2}}$ of the way down to bisect the area of the triangle, rather than two-thirds of the way down to pass through the centroid.
