Mathematical correct division of a piece of property This is real life calling.
My parents own a piece of property in which we would like to divide equally in two parts. There are two buildings on the property - an old house (H) from 1935 and an annex (A) to it's west side. I need a way to calculate this correctly, the trial-and-error ways of doing it have not yielded results my geeky nature will accept easily (I'm always off by at least one square meter).
The division must take into account that we cannot cut through any of the buildings but rather have to draw the division line (x) at the line where the buildings touch each other, then divide the rest of the property from the endpoint of that line on so that in the end we have two equal parts in size.

(x) in this picture is parallel to e, this is just my bad drawing skills.
The variables I see here are:
n,s,e,w as the length of the sides of the property.
x as the length of the straight line between the buildings.
s1,s2 as the lengths of the lines at which the divison line x starts.
The result should show the length of n1,n2 as where the angled, calculated division line will meet n.
As I am quite at lost here I was hoping for your help. Thank you!
 A: OK, here's the solution. The distance of the corner (that is the point where the dotted lines meet) to the side $e$ is $n\frac{s_1}{s}$, while to the side $w$ it is $n\frac{s_2}{s}$. Also, $e-w=\sqrt{s^2-n^2}$ by the Pythagorean theorem. 
This allows us to compute the surface of four trapezia determined by that corner and the corners of the entire terrain. The area of the bottom left trapezium is
$$s_1\frac{n}{s}\left(x+\frac{s_1}{2s}(e-w)\right) \; .$$
The area of the bottom right trapezium is 
$$s_2\frac{n}{s}\left(x-\frac{s_2}{2s}(e-w)\right) \; .$$
The area of the top left trapezium is
$$\frac{1}{2}\left(e-x-\frac{s_1}{s}(e-w)\right)\left(n_1+s_1\frac{n}{s}\right) \; .$$
Finally, the area of the top right trapezium is
$$\frac{1}{2}\left(w-x+\frac{s_2}{s}(e-w)\right)\left(n_2+s_2\frac{n}{s}\right) \; .$$
Our condition is that the sum of the left areas equals the sum of the right areas. This gives after some algebra the following relation
$$x\frac{s_1-s_2}{s}+\frac{s_1^2+s_2^2}{2s^2}(e-w)=\frac{1}{2}\left(\frac{es_1+ws_2}{s}-x\right)\left(\frac{n_2-n_1}{n}+\frac{s_2-s_1}{s}\right) \; .$$
