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In Figure, A B C D ,\ A B F E\ a n d\ C ...

In Figure, `A B C D ,\ A B F E\ a n d\ C D E F` are parallelograms. Prove that `a r( /_\A D E)=a r( /_\B C F)`

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In parallelogram, `ABCD`, `AD = BC`
In parallelogram, `DCFE`, `DE = CF`
In parallelogram, `ABFE`, `AE = AF`
Now, in `Delta ADE` and `Delta BCF`
`AD = BC`
`DE = CF`
`AE = AF`
So, from SSS criterion, `Delta ADE ~= Delta BCF`
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