Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar(APB) x ar(CPD) = ar(APD) x ar(BPC). [Hint From A and C, draw perpendiculars to BD.]

We draw AM \(\perp \) BD and CN \(\perp \) BD

Area of triangle = \(\frac{1}{2} \) × b × h

ar(APB) × ar(CPD )
= [\(\frac{1}{2} \) × BP × AM ]×[ \(\frac{1}{2} \) × PD × CN ]
= \(\frac{1}{4} \) × BP × AM × PD × CN

And ar(APD) × ar(BPC )
=[ \(\frac{1}{2} \) × PD × AM] × [\(\frac{1}{2} \) × BP × CN ]
= \(\frac{1}{4} \) × BP × AM × PD × CN


\(\therefore \) ar(APB) x ar(CPD) = ar(APD) x ar(BPC).