Question

$DE$ is parallel to $FB$

Answer 1

Consider $△DEA$ and $△BFC$,

$\Rightarrow AD=BC$ and $AE=FC$

$\therefore \angle DAE=\angle BCF$ ...... (Diagonal bisect)

So it is congruent by SAS

$\therefore DE\parallel FB$ and $DE=FB$ .... [By CPCT]