# 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? 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, $$\frac{|S_{\Delta BCE}-S_{\Delta ADE}|}{S_{ABCD}}=\frac{\frac{4(a+b)(b+c)}{b}-\frac{4ac}{b}}{\frac{4(a+b)(b+c)}{b}+\frac{4ac}{b}+\frac{4c(b+c)}{b}+\frac{4a(a+b)}{b}}=$$ $$=\frac{b(a+b+c)}{(a+b+c)^2}=\frac{b}{a+b+c}=\frac{PQ}{MN}.$$ 