splitting polygon in 4 equal parts I have a convex polygon and I want to divide into 4 equal parts using the two perpedicular splits. Like in a picture. I need s1 = s2 = s3 = s4;  I need to get coordinates of point where the lines cross and angle of rotation of polygon.
I think that firstly I need to divide polygon in 2 parts by 1st line, then by 2nd line. Then I get point of line cross and then I check s1=s2=s3=s4. If it is not then I need to rotate polygon at some angle.
Is there anything better for this purpose?

 A: This only applies for convex polygons.
We can use a continuity argument. Fix a direction $v\in S^1$: there exists a (only one) line $t_v$ parallel to $v$ that splits the area of the polygon in two equal parts, $L_v$ on the left of $t_v$ and $R_v$ on the right of $t_v$. For a similar reason, there exists a (only one) line $l_v$ orthogonal to $t_v$ that splits in two equal parts $L_v$, and a (only one) line $r_v$, orthogonal to $t_v$, that splits in two equal parts $R_v$. Call $P_L=t_v\cap l_v$ and $P_R=t_v\cap r_v$. Say that $v$ is "right-handed" if $P_R-P_L$ has the same direction of $v$, "left-handed" if $P_R-P_L$ has the opposite direction.
If $v$ is right-handed then $-v$ is left-handed, so somewhere between $v$ and $-v$ we must have a direction that is neither left- or right- handed, i.e. a solution for our splitting problem.
And yes, this is just a consequence of the Borsuk-Ulam theorem.
A: Edit: Only applies for symmetrical polygons, see comments
Lets take a physical reasoning. Assume that you try to balance the polygon on one line, if the line doesn't include the center of mass (centroid) of the polygon then it's obvious that the polygon will tip to one side and fall off. If it doesn't tip then it is in equilibrium and the mass (area) on either side of the polygon must be equal.
From the above reasoning it is obvious that any line you use to cut the polygon must go through the polygon's centroid. 
Under the constraint that both lines must be perpendicular you're left with determining the rotation of the polygon. You can do that by a linear search through $[0,\frac{\pi}{2}[$ due to symmetry of the cut.
