Answer
$\angle EDF \cong \angle EBF$
Work Step by Step
The opposite pairs of angles of a parallelogram are congruent. So, we can write $\angle ADF \cong \angle CBE$
Use the angle addition postulate.
$m\angle ADF =m\angle ADE+m \angle EDF$ and $m\angle CBE =m\angle CBF+m \angle EBF$
or, $m \angle EDF=m\angle ADF -m\angle ADE+$ and $m \angle EBF=m\angle CBE -m\angle CBF$
We are given that $\angle ADE \cong \angle CBF$
Thus, we have: $m \angle EDF=m \angle EBF$
or, $\angle EDF \cong \angle EBF$
