Question

Diagonals AC and BD of ABCD intersect at P. Show that ar(APB) *ar(CPD) = ar(APD) *ar(BPC)

Answer 1

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)