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Question

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:

ar (APB)×ar(CPD)=ar(APD)×ar(BPC)

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Solution

IN quadrilateral ABCD, diagonal AC and BD intersect each other as P
To prove : ar (APB)×ar(CPD)=ar(APD)×ar(BPC)
Construction : Draw Al and CN perpendiculars on BD

Proof: ar(APD)×ar(BPC)=(12×AL×DP)×(12×CN×BP)=(12×BP×AL)×(12×DP×CM)=ar(APB)×ar(CPD)


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