From chapter areas of parallelogram and triangles

Q. Diagonals AC and BD of a quadrilateral intersect at P. show that-
ar(APB)$\text{×}$ ar(CPD)=ar(APD)$\text{×}$ ar(BPC).


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Given, ABCD is a quadrilateral in which diagonals AC and BD of a quadrilateral ABCD intersect each other at P.

To prove: ar (APB) × ar (CPD) = ar (APD) × ar (BPC)

Construction: Draw AM ⊥ BD and CN ⊥ BD.

Proof:

LHS = ar (APB) × ar (CPD)

= ar (BPC) × ar (APD) 

= ar (APD) × ar (BPC) 

= RHS [Hence proved] 
Regards