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.
Take points $S,T$ on $BD$ such that $AS\perp BD$ and $CT\perp BD$.
$$ ar(ABP) = \frac12 AS\times BP\qquad ar(DCP) = \frac12 CT\times DP $$
$$ ar(BCP) = \frac12 CT\times BP\qquad ar(ADP) = \frac12 AS\times DP $$
It is then easy to see that the area products you listed are equal to each other.