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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)

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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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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 14 Areas of Parallelograms and Triangles
Exercise 14.3 | Q 16 | Page 46
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