$ABCD$ is a quadrilateral and $P,Q$ are midpoints of $CD, AB.$... 
$ABCD$ is a quadrilateral and $P,Q$ are midpoints of $CD, AB.$ $AP$ and $DQ$ meet at $X, BP$ and $CQ$ meet at $Y.$ Prove that $$|ADX|+|BCY|=|PXQY|$$ (here $|N|$ means area of the shape $N$) 

I have absolutely no idea how to solve this problem. 
Any help will be appreciated. 
 A: 
Let's S$_{N}$ be the area of N.
First let's look at $_\triangle$ABP: Q - midpoint $\Rightarrow$ PQ - median. From PQ - median $\Rightarrow$ S$_{AQP}$ = S$_{BQP}$ $\Rightarrow$ S$_2$ + S$_3$ = S$_5$ + S$_8$  (1).
Now let's look at $_\triangle$DQC: P - midpoint $\Rightarrow$ QP - median. From QP - median $\Rightarrow$ S$_{DPQ}$ = S$_{QPC}$ $\Rightarrow$ S$_4$ + S$_3$ = S$_7$ + S$_8$  (2).
Adding equation (1) and (2) together gives us S$_2$ + S$_7$ = S$_4$ + S$_5$ (3).
Now we look at $_\triangle$ABC: CP - median (Q-midpoint) $\Rightarrow$ S$_{AQC}$ = S$_5$ + S$_6$ (4). In $_\triangle$CAD $\rightarrow$ AP - median $\Rightarrow$ S$_{ACP}$ = S$_1$ + S$_4$ (5).
Adding (4) and (5) together we get S$_{AQCP}$ = S$_1$ + S$_4$ + S$_5$ + S$_6$ (6).
Let's look now at AQPC and DQBP: 
S$_{AQCP}$ = S$_2$ + S$_3$ + S$_8$ + S$_7$ (7).
S$_{DQBP}$ = S$_5$ + S$_3$ + S$_8$ + S$_4$ (8).
From (3), (7) and (8) $\Rightarrow$ S$_{AQCP}$ = S$_{DQBP}$ (9).
From (9) and (6) we get S$_{BQBP}$ = S$_1$ + S$_4$ + S$_5$ + S$_6$, or S$_5$ + S$_3$ + S$_8$ + S$_4$ = S$_1$ + S$_4$ + S$_5$ + S$_6$ $\Rightarrow$
$\Rightarrow$ S$_3$ + S$_8$ = S$_1$ + S$_6$ $\Rightarrow$ S$_{ADX}$ + S$_{BCY}$ = S$_{PXQY}$.
