Given: parallelogram ABCD; $\overline{\mathrm{DE}}\perp \overline{\mathrm{AC}}$; $\overline{\mathrm{BF}}\perp \overline{\mathrm{AC}}$.

Prove: DEBF is a parallelogram.

The given diagram is:

It is being given that ABCD is a parallelogram.

In the triangles $△AFB$and $△CED$, it can be noticed that:

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

$\angle CED=\angle AFB=90°\left(given\right)$

$AB\cong CD\left(\text{asABCDisaparallelogram}\right)$

Therefore, the triangles $△AFB$and $△CED$are the congruent triangles by AAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be said that $DE\cong BF$.

In the triangles $△EBF$and $△FDE$, it can be noticed that:

$DE\cong BF$

$\angle CED=\angle AFB=90°\left(given\right)$

$EF\cong EF\left(common\right)$

Therefore, the triangles $△EBF$and $△FDE$are the congruent triangles by the SAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be said that $EB\cong FD$.

Therefore, it can be noticed that $DE\cong BF$and $EB\cong FD$.

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

As, $DE\cong BF$and $EB\cong FD$, therefore the quadrilateral DEBF is a parallelogram.