Question

diagonals AC and BD of a quadrilateral ABCD intersect at P. show that ar(APB)×ar(CPD) = ar(BPC)×ar(APD)

Answer 1

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)

Answer 2

Let ABCD is a quadrilateral having diagonal AC and BD. These diagonals intersect at point P.

Now area of a triangle  = 1/2 * base * height

=> Area(ΔAPB) * Area(ΔCPD)  = {1/2 * BP*AM}*{1/2 * PD*CN}

=> Area(ΔAPB) * Area(ΔCPD)  = 1/4 * BP*AM* PD*CN .........1

Again

     Area(ΔAPD) * Area(ΔBPC) = {1/2 * PD*AM}*{1/2 *CN*BP}

=> Area(ΔAPD) * Area(ΔBPC)  = 1/4 * BP*AM* PD*CN ..........2

From equation 1 and 2, we get

Area(ΔAPB) * Area(ΔCPD)  = Area(ΔAPD) * Area(ΔBPC)