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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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    @draks: quadrilateral $\neq$ rectangle. Otherwise it looks okay.2012-03-01

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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.