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  • 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)

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)

 

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Solution.

According to the question, ABCD is a parallelogram.
E is a point on BC.
AE and DC are produced to meet at F.
Using the above information, we can make the figure as follows:

We know that, triangles on the same base and between the same parallels are equal in area.
Here, we have,
\triangle ABC\triangle ABC and \triangle ABF are on same base AB and between parallels, AB and DF
Area (\triangle ABC) = area (\triangle ABF) …(1)
We also know that,
Diagonal of a parallelogram divides it into two triangles of equal area.
So, Area (\triangle ABC) = area (\triangle ACD) …(2)
Now,
Area (\triangle ADF) = area (\triangle ACD) + area (\triangle ACF)
\ Area (\triangle ADF) = area (\triangle ABC) + area(\triangle ACF) [Q From equation (2)]
\Rightarrow Area (\triangle ADF) = area (\triangle ABF) + area (\triangle ACF)[Q From equation (1)]
\Rightarrow Area (\triangle ADF) = area (ABFC)

Hence proved.

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