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Question
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar (ΔBPC)
Solution
Construction: Draw BQ ⊥ AC and DR ⊥ AC
Proof:
L.H.S
= ar (Δ APB ) × ar (ΔCP)
= `1/2[ ( AP xx BQ )] xx (1/2 xx PC xx DR)`
=`( 1/2 xx PC xx BQ )xx (1/2 xx AP xx DR)`
= ar (Δ BPC ) × ar (APR)
= RHS
∴ LHS = RHS
Hence proved.
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