Cutting a square of area $A$ through It's mid point yields two polygons of area $A/2$ for arbitrary cuts? A square of area $A$ is cut by a straight line at It's mid point :

It consists now of two rectangles, with area $A/2$. I want to find a proof that I could rotate the line and the area of the two polygons is still $A/2$, for example:

And conclude that the area will be the same for arbitrary rotations (presuming It's true). It is visually obvious that it's true, but I don't know how to make a proof for it. The best I could achieve was this:

As we rotate right side of the line upwards, the length of the segment $\color{red}{G}$ will be the same of segment $\color{red}{H}$, and the length of segment $\color{orange}{I}$ will be the same of segment $\color{orange}{J}$ thus:
$$G=H\text{ and } I=J$$
$$(G+h)(I-h)=(G+h)(I-h)$$
I guess the argument applies to rotations of more than $45$ degrees, one needs only to do the same procedure with segments from the top and bottom of the square. I'm not sure if this would work as a proof. What do you think?
 A: It is not true in general if you pass a line through the centroid of a figure, then the line will cut the figure in two equal areas. The simplest example is an equilateral triangle, e.g. those
with vertices
$(-\frac{1}{\sqrt{3}},0), (\frac{1}{2\sqrt{3}}, \frac12 )\text{ and } (\frac{1}{2\sqrt{3}}, -\frac12 )$. The centroid is the origin and yet if you cut the triangle along the $y$-axis, the area of the left is only $\frac49$ of the whole triangle.
However, the statement is true when the figure is centrally symmetric (i.e figure has reflection symmetry with respect to a point).
Let's say the origin is the center of symmetry. No matter how you cut the figure using a line through the origin. For any area element at $\vec{x}$ on one side, there is equal area element at $-\vec{x}$ on the other side. In pure geometric term, if you rotate the portion of figure on one side of the line for $180^{\circ}$ with respect to the origin, it will match the portion of figure on the other side exactly.
Any regular polygon of even number of sides, in particular the square, are special cases of this.
A: Try triangulating the square using the center.  The proof should follow naturally by extending your argument above about the side lengths of certain triangles being equal, then noting that the area on either side is just the sum of the three triangles.  I've colored triangles that are equal in the decomposition.
This should solve the problem at hand, although I'm not certain about extending it to general polygons as most of the comments have mentioned.

