Question

A point E is taken on the side BC of a paralellogram ABCD . AE and DC are produced to meet at F . prove that ar(ADF)=ar(ABFC).

Answer 1

HEY MATE HERE IS YOUR ANSWER

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)---  eq 1

⇒ area (ΔADF) = area (ΔABF) + area (ΔACF)---  eq 2

⇒ area (ΔADF) = area (ΔABFC)

HOPE THIS HELPS YOU......✌✌

STAY COOL MATE!!