Diagonals `A C` and `B D` of a quadrilateral `A B C D` intersect each at `Pdot` Show That: `a r(A P B) xx ar ( CPD)= ar (A P D) xx ar ( P B C)`

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)