ABCD is a parallelogram and E and F are two points on $¯ ¯¯¯¯¯¯ ¯ A B$ and $¯ ¯¯¯¯¯¯ ¯ B C$ respectively. show that
$a r (Δ A D F) = a r (C D E)$

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Solution

Given $A B C D$ is a parallelogram $E$ and $F$ are two points on $A B$ and $B C$ .
We have to show $a r (△ A B C) = a r (△ A B D)$
Since $△ A D F$ and parallelogram $A B C D$ are on the same base $A D$ and between the same parallels $A D$ and $B C$ .
$∴ a r (△ A D F) = \frac{1}{2} a r (A B C D) - - - - (1)$
Also $△ E C D$ and parallelogram $(A B C D)$ on same base and between the same parallels $C D$ and $A B$ .
$∴ a r (△ E C D) = \frac{1}{2} a r (A B C D) - - - - - (2)$
From $(1)$ & $(2)$
$a r (△ A D F) = a r (△ C D E)$

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ABCD is a parallelogram and E and F are two points on ¯¯¯¯¯¯¯¯AB and ¯¯¯¯¯¯¯¯BC respectively. show that ar(ΔADF)=ar(CDE)

ABCD is a parallelogram and E and F are two points on $¯ ¯¯¯¯¯¯ ¯ A B$ and $¯ ¯¯¯¯¯¯ ¯ B C$ respectively. show that
$a r (Δ A D F) = a r (C D E)$