ABCD is a parallelogram. E is any point on side BC when produced lines DC and AE meet each other at F. Show that $a r e a (Δ A D F) = a r e a (A B F C)$

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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: $a r e a (Δ A D F) = a r e a (A B F C)$ .
Proof
$a r e a (Δ A B C) = a r e a (Δ A B F)$ .....(1) (Triangles on the same base AB and between same parallels, $A B | | C F$ are equal in area)
$a r e a (Δ A B C) = a r e a (Δ A C D)$ ....(2) (Diagonal of a parallelogram divides it into two triangles of equal area )
Now,
$a r e a (Δ A D F) = a r e a (Δ A C D) + a r e a (Δ A C F)$
$∴ a r e a (Δ A D F) = a r e a (Δ A B C) + a r e a (Δ A C F)$ (From (2))
$\Rightarrow a r e a (Δ A D F) = a r e a (Δ A B F) + a r e a (Δ A C F)$ (From (1))
$\Rightarrow a r e a (Δ A D F) = a r e a (A B F C)$

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In a parallelogram ABCD,any point E is taken on the sides BC.AE and DC when produced meet at a point M.Prove that $a r (△ A D M) = a r (A B M C)$ .

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