Question

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

Answer 1

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)

Answer 2

Answer:

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Step-by-step explanation:

=>ar (∆ADF) =ar (ABFC)