Question

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

Answer 1

Given : Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.  

 

To Prove : ar(Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(Δ BPC).

Construct : BQ and DR two perpendiculars on AC.

Proof :  

Now,

L.H.S. = ar(ΔAPB) x ar(ΔCDP)

= (½ x AP × BQ) × (1/2 × PC × DR)

= (1/2 x PC × BQ) × (1/2 × AP × DR)

[Diagonals of a quadrilateral bisect each other i. e, AP = PC ]

= ar(ΔBPC) × ar(ΔAPD)  

= R.H.S.

Hence proved.

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Answer 2

Answer:-

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Given :

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P.  

 

To Prove :

ar(Δ APB) × ar(Δ CPD) = ar(Δ APD) × ar(Δ BPC).

Construct :

BQ and DR two perpendiculars on AC.

Proof :  

Now,

L.H.S. = ar(ΔAPB) x ar(ΔCDP)

= (½ x AP × BQ) × (1/2 × PC × DR)

= (1/2 x PC × BQ) × (1/2 × AP × DR)

[Diagonals of a quadrilateral bisect each other i. e, AP = PC ]

= ar(ΔBPC) × ar(ΔAPD)  

= R.H.S.

Hence proved.

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