Diagonals Ac and Bd of a Quadrilateral Abcd Intersect Each Other at P. Show That: Ar(δApb) × Ar(δCpd) = Ar(δApd) × Ar (δBpc) - Mathematics
Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar (ΔBPC)
Construction: Draw BQ ⊥ AC and DR ⊥ AC
= 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)
∴ LHS = RHS
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