Question

Diagonal ACand BD of a quadrilateral ABCD intersect each other other at P show that ar(APB)multiply ar (CPD) =ar(APD)multiply ar (BPC)

Answer 1

✔✔ Let ABCD is the quadrilateral having diagonal AC and BD .These diagonals intersect at point P.
• Now area of Triangle = 1/2* base * height

--> area (ΔAPB) * area (ΔCPD) =
(1/2*BP *AM )* (1/2*PD *CN )
--/> area (ΔAPB) *(ΔCPD) =1/4*BP*AM *PD * CN
............(1)
again
area (ΔAPD) * area (ΔBPC) =1/2 *PD *AM *1/2*CN*BP
---> area (ΔAPD) *area (Δ BPC) =1/4*PD * AM * CN * BP
............(2)
from equation 1 n 2 ...we get
area (ΔAPB) *area (ΔCPD) =area (ΔAPD) *area (ΔBPC)

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