Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar (△APB)×ar(△CPD)=ar(△APD)×ar(△BPC)
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)