Question

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

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Answer 1

Answer: given below

Step-by-step explanation: given below

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Answer 2

Answer:

Step-by-step explanation

We know that,

            Area of a Δ is (1/2)*base*height.

So with reference to the figure,

     Ar(ΔAPD)=(1/2)*x*DP...................(1)

    Ar(ΔBPC)=(1/2)*y*BP..................(2)

Since height for ΔAPB is x and for ΔCPD is y

So   Ar(ΔAPB)=(1/2)*x*BP...................(3)

     Ar(ΔAPD)=(1/2)*y*DP..................(4)

Thus from (1),(2),(3) and (4),

Ar(ΔAPD)*Ar(ΔBPC)=Ar(ΔAPB)*Ar(ΔAPD)