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Answer :

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).

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