91Ó°ÊÓ

The given diagram is:

As, $DE=BF$, therefore it can be noticed that:

$\begin{array}{c}DE=BF\\ DE+EF=BF+EF\\ DF=BE\end{array}$

Therefore, $DFâ‰\dots BE$.

AsABCD is a parallelogram, therefore $ABâ‰\dots CD$,$ABâˆ\yen CD$ and BD is a transversal.

From the given diagram it can be noticed that the angles $∠FDC$and $∠EBA$ are the alternate interior angles.

Therefore, $∠FDCâ‰\dots ∠EBA$.

In the triangles $â–³DCF$and $â–³BAE$, it can be noticed that $ABâ‰\dots CD$, $∠FDCâ‰\dots ∠EBA$and $DFâ‰\dots BE$.

Therefore, the triangles $â–³DCF$and $â–³BAE$are congruent triangles by using the SAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be noticed that $CFâ‰\dots AE$.

AsABCD is a parallelogram, therefore $ADâ‰\dots CB$,$ADâˆ\yen BC$ and BD is a transversal.

From the given diagram it can be noticed that the angles $∠FDA$and $∠EBC$ are the alternate interior angles.

Therefore, $∠FDAâ‰\dots ∠EBC$.

In the triangles $â–³DAF$and $â–³BCE$, it can be noticed that $ADâ‰\dots CB$, $∠FDAâ‰\dots ∠EBC$and $DFâ‰\dots BE$.

Therefore, the triangles $â–³DAF$and $â–³BCE$are congruent triangles by using the SAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be noticed that $AFâ‰\dots CE$.

Therefore, it can be noticed that $CFâ‰\dots AE$and $AFâ‰\dots CE$.

Therefore, it can be said the pairs of opposite sides of the quadrilateral AFCE are congruent.

If the pairs of opposite sides of a quadrilateral are congruent then the quadrilateral is a parallelogram.

As, $CFâ‰\dots AE$and $AFâ‰\dots CE$, therefore AFCE is a parallelogram.