From chapter areas of parallelogram and triangles
Q. Diagonals AC and BD of a quadrilateral intersect at P. show that-
ar(APB) ar(CPD)=ar(APD) ar(BPC).
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.
LHS = ar (APB) × ar (CPD)
= ar (BPC) × ar (APD)
= ar (APD) × ar (BPC)
= RHS [Hence proved]