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

Answer:

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(Δ

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