A point E is taken as the mid point of the side BC OF the parallelogram ABCD. AE produced to DC to meet at F.prove that ar(ADF) =ar(ABFC) .

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Solution

Given : ABCD is a parallelogram. E is a point on BC. AE and DC are produced to meet at F.
To prove : Area $(Δ A D F)$ = area (ABFC).
Area $(Δ A B C)$ = area $(Δ A B F) . . . (1)$ (Triangles on the same base AB and between same parallelels, AB||CF are equal in area)
Area $(Δ A B C) = a r e a (Δ A C D) . . . (2)$ Diagonal of a parallelogram divides it into two triangles of equal area)
Now,
Area $(Δ A D F)$ = area $(Δ A C D) + a r e a (Δ A C F)$
$∴ A r e a (Δ A D F) = a r e a (Δ A B C) + a r e a (Δ A C F) (F r o m (2))$
$\Rightarrow a r e a (Δ A D F) = a r e a (Δ A B F) + a r e a (Δ A C F) (F r o m (1))$
$\Rightarrow a r e a (Δ A D F) = a r e a (Δ A B F C)$

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