Prove that $\frac{PQ}{MN} = \frac{|[BCE] - [ADE]|}{[ABCD]}$ in a quadrilateral ABCD where P and Q are related to the diagonals I've recently been given a few challenge problems that I really want to find out. But for the most part, I just can't figure out how to completely prove the problems. Now one of the problems goes something a little like this.
Let's say we are given a convex quadrilateral $ABCD$. We can begin by making a few constructions to it, starting with denoting point $E$ as the intersection of the diagonals of $ABCD$. Furthermore, let's say points $M$ and $N$ are the midpoints of sides $AB$ and $CD$, respectively. And continuing on segment $MN$, we are able to find that it meets our diagonals $AC$ and $BC$, which we can label the points that it meets the diagonals at as points $P$ and $Q$, respectively.
And we're given the task to prove that $\frac{PQ}{MN} = \frac{|[BCE] - [ADE]|}{[ABCD]}$. Now for the most part, I've been able to understand what this question is asking, and I've been able to construct a diagram online. I've put a screenshot of it below. Now the part that's throwing me off is that we need to relate the length of two segments to the area of a few figures. I've recognized that the length of the segments do influence the triangles in the numerator, but I'm not exactly sure how I can make a concrete connection between them. Does anyone have an idea how we can do this?

 A: Let $S_{\Delta EPN}=a$, $S_{\Delta EPQ}=b$ and $S_{\Delta EQM}=c$.
Thus, $$S_{QMB}\cdot b=S_{\Delta QPB}\cdot c,$$
which gives $$S_{\Delta QPB}=\frac{bS_{\Delta QMB}}{c}$$ and since $$S_{\Delta PAM}=S_{\Delta PBM},$$ we obtain:
$$b+c+c+S_{\Delta QMB}=\frac{bS_{\Delta QMB}}{c}+S_{\Delta QMB},$$ which gives $$S_{\Delta QMB}=\frac{c(2c+b)}{b},$$
$$S_{\Delta AEB}=2S_{\Delta AEM}=2\left(c+\frac{c(2c+b)}{b}\right)=\frac{4c(b+c)}{b}.$$
By the same way we obtain:
$$S_{\Delta PNC}=\frac{a(2a+b)}{b}$$ and $$S_{\Delta DEC}=\frac{4a(a+b)}{b}.$$
Also, $$S_{\Delta QPB}=\frac{bS_{\Delta QMB}}{c}=\frac{b}{c}\cdot\frac{c(2c+b)}{b}=2c+b,$$ which gives
$$\frac{S_{\Delta PBC}}{b+2c+b}=\frac{\frac{a(2a+b)}{b}}{a}$$ or
$$S_{\Delta PBC}=\frac{2(2a+b)(b+c)}{b}$$ and $$S_{\Delta EBC}=b+2c+b+\frac{2(2a+b)(b+c)}{b}=\frac{4(a+b)(b+c)}{b}.$$
Thus, $$S_{\Delta ADE}=\frac{S_{\Delta DEC}S_{\Delta AEB}}{S_{\Delta EBC}}=\frac{4ac}{b}.$$
Id est, $$\frac{|S_{\Delta BCE}-S_{\Delta ADE}|}{S_{ABCD}}=\frac{\frac{4(a+b)(b+c)}{b}-\frac{4ac}{b}}{\frac{4(a+b)(b+c)}{b}+\frac{4ac}{b}+\frac{4c(b+c)}{b}+\frac{4a(a+b)}{b}}=$$
$$=\frac{b(a+b+c)}{(a+b+c)^2}=\frac{b}{a+b+c}=\frac{PQ}{MN}.$$
A: 
We use the fact that in quadrilaterals with opposite sides parallel(figures a, b and c) diagonals and the line that connect the midpoints of parallel sides cross through one point.Now consider figure c, where vertex G transformed to D and trapezoid  ABCG is transformed to ABCD and triangle PEQ has taken shape, or point H is transformed to triangle PEQ. In shape ABCG triangles CHB and GHA are equal so their difference is zero , so is the area of FEQ and measure of PQ. Therefore the fraction holds. Now consider figure e where D is coincident on A and equilateral ABCD is transformed to triangle ABC. Now area of triangle DEA is zero and area of equilateral ABCD is  equal to the area of triangle ABC and again the fraction holds. Hence by induction one may conclude that the fraction holds wherever D locates.
