# Areas of quadrilateral and parallelograms and triangles.

The diagonals of a quadrilateral $ABCD$ meet at $P$.

Prove that $ar(APB)*ar(DPC)=ar(ADP)*ar(BPC)$

Please solve this question. I have tried a lot on this question. Please do not use trigonometry, but if you want you can use trigonometry.

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I think it would be wise for you to consider how helpful answers would be to you if they put in as much effort into preparing their response as you did into your question. –  Cam McLeman Mar 1 '12 at 14:29
I edited your question. Please check, that everything is still what you menat. BTW: Did you draw a picture? –  draks ... Mar 1 '12 at 15:21
@draks: quadrilateral $\neq$ rectangle. Otherwise it looks okay. –  Willie Wong Mar 1 '12 at 15:22

Take points $S,T$ on $BD$ such that $AS\perp BD$ and $CT\perp BD$.

Then

$$ar(ABP) = \frac12 AS\times BP\qquad ar(DCP) = \frac12 CT\times DP$$

similarly

$$ar(BCP) = \frac12 CT\times BP\qquad ar(ADP) = \frac12 AS\times DP$$

It is then easy to see that the area products you listed are equal to each other.

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