a point E is taken on side BC of a parallelogram AbCD. AE and DC are produced to meet at F.
prove that area of triangle ADE= area of triangle ABFC

Dear Student!

Given : ABCD is a parallelogram. E is a point on BC. AE and DC are produced to meet at F.

To prove: area (ΔADF) = area (ABFC).

Proof :

area (ΔABC) = area (ΔABF)   ...(1) (Triangles on the same base AB and between same parallels, AB || CF are equal in area)

area (ΔABC) = area (ΔACD)    ...(2)    (Diagonal of a Parallelogram divides it into two triangles of equal area)

Now,

area (ΔADF) = area (ΔACD) + area (ΔACF)

∴ area (ΔADF) = area (ΔABC) + area (ΔACF)    (From (2))

⇒ area (ΔADF) = area (ΔABF) + area (ΔACF)    (From (1))

⇒ area (ΔADF) = area (ΔABFC)

Cheers!