In the following figure, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).


It is given that $A B C D$ is a parallelogram. We know that opposite sides of a parallelogram are equal.

$\therefore \mathrm{AD}=\mathrm{BC} \ldots(1)$

Similarly, for parallelograms DCEF and ABFE, it can be proved that

$\mathrm{DE}=\mathrm{CF} \ldots(2)$

And, $E A=F B \ldots(3)$

In $\triangle \mathrm{ADE}$ and $\triangle \mathrm{BCF}$,

AD = BC [Using equation (1)]

DE = CF [Using equation (2)]

EA = FB [Using equation (3)]

$\therefore \triangle \mathrm{ADE} \cong \mathrm{BCF}(\mathrm{SSS}$ congruence rule $)$

$\Rightarrow$ Area $(\triangle \mathrm{ADE})=$ Area $(\triangle \mathrm{BCF})$

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