Question

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

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

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)