Question

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

Answer 1

According to the question,

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:

We know that,

Triangles on the same base and between the same parallels are equal in area.

Here,

We have,

ΔABC and ΔABF are on same base AB and between same parallels, AB || CF.

Area (ΔABC) = area (ΔABF) … (1)

We also know that,

Diagonal of a Parallelogram divides it into two triangles of equal area

So,

Area (ΔABC) = area (ΔACD) … (2)

Now,

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

∴ Area (ΔADF) = area (ΔABC) + area (ΔACF) … (From equation (2))

⇒ Area (ΔADF) = area (ΔABF) + area (ΔACF) … (From equation (1))

⇒ Area (ΔADF) = area (ABFC)