Question

ABCD is a parallelogram and P is any point on the side CD. Prove that area of triangle APD + area of triangle BCP =area of triangle ABP.

Answer 1

Given,
ABCD is a ||gm and P is a pt on CD 
R.T.P : Area of triangle APD+Area of triangle BCP = Area of triangle ABP
Now,
triangle APD + triangle BCP +triangle ABP = ||gm ABCD   ...(i)
but,
by the theorem on area,13.4,
which states that the area of a triangle lying on the same base and between the same parallel lines as of the ||gm has area equal to half of that of the parallelogram.
=>area of triangle ABP=1/2 area of ||gm ABCD   ...(ii)
from (i) and (ii),we have,
area of trian. APD +area of trian.BCP + 1/2 area of ||gm ABCD=area of ||gm ABCD
area of trian. APD + area of trian. BCP=(1-1/2)area of ||gm ABCD
                                                             =1/2 area of ||gm ABCD
                                                             =Area of triangle APB  (from (ii)).
Hence, proved.