Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar (ΔBPC)
Advertisement Remove all ads
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.
Concept: Concept of Area
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads
