DE = FB

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Solution

From $△ D E A$ and $△ B F C$
$A D = B C; A E = F C$
$∴ ∠ D A E = ∠ B C F$ ......... [Diagonal biesct]
So, congruent by SAS
$\Rightarrow D E = F B$ ..... [CPCT]

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