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.
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.