91Ó°ÊÓ

The given diagram is:

It is being given that ABCD is a parallelogram.

In the triangles $â–³AFB$and $â–³CED$, it can be noticed that:

$∠BAF=∠DCE\left(\text{alternateinteriorangle}\right)$

$∠CED=∠AFB=90°\left(given\right)$

$ABâ‰\dots CD\left(\text{asABCDisaparallelogram}\right)$

Therefore, the triangles $â–³AFB$and $â–³CED$are the congruent triangles by AAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be said that $DEâ‰\dots BF$.

In the triangles $â–³EBF$and $â–³FDE$, it can be noticed that:

$DEâ‰\dots BF$

$∠CED=∠AFB=90°\left(given\right)$

$EFâ‰\dots EF\left(common\right)$

Therefore, the triangles $â–³EBF$and $â–³FDE$are the congruent triangles by the SAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be said that $EBâ‰\dots FD$.

Therefore, it can be noticed that $DEâ‰\dots BF$and $EBâ‰\dots FD$.

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

As, $DEâ‰\dots BF$and $EBâ‰\dots FD$, therefore the quadrilateral DEBF is a parallelogram.