Intersecting lines from the vertices of a square with an arbitrary interior point. This question is pretty hard to word without a picture, so I have attached one. 
I am wondering if there is a general way to find the area of the green or blue areas given the ordered pair of some interior point (a,b). These lines are going from each vertex of the square and intersecting the interior point. For simplicity, I am assuming the shape is a square with side lengths 1. Of course, finding the area of one of the areas allows you to automatically have the other, so is one of the areas easier to find than the other?
Also, sorry the picture is not the best. I tried to draw it in paint.
Some attempts at a solution have included using just the principles of similar triangles and have yielded some pretty nasty expressions. I've considered attempting the use the shoelace formula as well but I don't know if it will work here. Any help is appreciated. Thanks!

 A: Let the figure be a square with each side = s and the point be (a, b).

More than half of the co-ordinates are then known (in terms of s, a, b) except h, k, p and m but they can all be expressed in terms of s, a, and b.
For example, (m, 0) is a point on the line formed by (0, s) and (a, b). The corresponding equation is $\frac {y – b}{x – a} = \frac {s – b}{0 – a}$. (m, 0) lies on this line means  $\frac {[0] – b}{[m] – a} = \frac {s – b}{0 – a}$.
Note that it is not necessary to calculate all the triangles area. This is because the sum of the areas all the light green triangles (originally was dark green) is equal to half of the area of the square. Thus, we only need to find the areas of triangles XHK, XAM and XBP.
A: This is a continuation of my answer and it serves as a suggestion in replying the OP’ further query. [I don’t think I can include a picture in the comment.]
Judging from your finding, I suspect the attached can explain why it works only at certain occasions.

Depending on the relative values of a and b, the point X(a, b) will be located at different region. A 2 cases discussion may be necessary.
