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$ABCD$ 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.

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  • $\begingroup$ Are you sure that there is nothing else given , like type of quadrilateral etc? $\endgroup$ – The Dead Legend May 19 '17 at 8:58
  • $\begingroup$ @TheDeadLegend No... There is no other information. $\endgroup$ – user333900 May 19 '17 at 9:01
  • $\begingroup$ One hint, Hit the Coordinate System. It's easy to solve it by that system. $\endgroup$ – The Dead Legend May 19 '17 at 9:08
  • $\begingroup$ @TheDeadLegend Thanks for the hint. $\endgroup$ – user333900 May 19 '17 at 9:21
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Quadrilateral ABCD

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$_{PXQY}$.

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