Dividing a polygon in desired way I have a bunch of polygons. I want to divide them with a line that will pass through the center and divide the polygon into two equal pieces with respect to area. 
I cannot seem to find a place to start with. 
Hopefully clearer explanation of the problem: 
What i have is a bunch of polygons in 3D. Like below. They are more or less flat polygons suspended in 3 dimensions. 
I have Cartesian coordinates of each of their vertices. Now what i want is to divide them into two equal pieces. The line can be in any direction and has to pass by the center of the polygon while dividing it into two equal pieces, areawise.  


As an example above polygon with blue outline, if is divided as i want, would look something like this 

With dividing (black) line passing through the center (in red) and dividing the polygon in two equal halves in terms of area.
About how the polygons are shaped.
1. Each of the polygon is flat or the vertices of each polygon is co-planar.
2. However, different polygons are on different plane, think of these polygons as tiles tilling a surface
3. The area is calculated for each polygon based on the its vertices that all lie on the same plane.
You can see these properties of the polygons on the following shape.
For the division, each polygon is to be divided with separate lines. The dividing line are not the same


As the comments indicate, your phrasing of the problem is not quite precise.
There has been considerable work on computing the "center of area" of a
single convex polygon (a point $x$ such that any line through $x$ bisects the area), including this:

Braß, Peter, Laura Heinrich-Litan, and Pat Morin. "Computing the center of area of a convex polygon." International Journal of Computational Geometry & Applications 13.05 (2003): 439-445.
  Journal link.

They improved on my earlier work with a (then) student:

Diaz, Matthew, and Joseph O'Rourke. "Algorithms for computing the center of area of a convex polygon." The Visual Computer 10.8 (1994): 432-442.





Diaz-O'Rourke Fig.2: 5% contours.
The innermost contour ranges over $[45,46)$ percent.


