Question

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

Answer 1

Step-by-step explanation:

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

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

Construction: Draw AM⊥DB,CN⊥DB

Proof: ar.(△APB)×ar.(△CPD)=

=(

2

1

×PB×AM)×(

2

1

×PD×CN)

=

4

1

×PB×AM×PD×CN....(i)

ar.(△APD)×ar.(△BPC)=

=(

2

1

×PD×AM)×(

2

1

×PB×CN)

=

4

1

×PD×AM×PB×CN....(ii)

From (i) and (ii)

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

Answer 2

Answer:

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

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

Construction: Draw AM⊥DB,CN⊥DB

Proof: ar.(△APB)×ar.(△CPD)=

=(

2

1

×PB×AM)×(

2

1

×PD×CN)

=

4

1

×PB×AM×PD×CN....(i)

ar.(△APD)×ar.(△BPC)=

=(

2

1

×PD×AM)×(

2

1

×PB×CN)

=

4

1

×PD×AM×PB×CN....(ii)

From (i) and (ii)

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