Given: $\overline{\mathbf{AD}}\cong \overline{\mathbf{BC}}$; $\overline{\mathbf{AD}}\parallel \overline{\mathbf{BC}}$.

Prove: $\overline{\mathbf{EF}}\cong \overline{\mathbf{FG}}$.

The given diagram is:

It is given that $\overline{AD}\cong \overline{BC}$ and $\overline{AD}\parallel \overline{BC}$.

If a pair of opposite sides of a quadrilateral are congruent and parallel to each other then the quadrilateral is a parallelogram.

As, $\overline{AD}\cong \overline{BC}$ and $\overline{AD}\parallel \overline{BC}$, therefore the quadrilateral $ABCD$ is a parallelogram.

In the parallelogram, the diagonals bisect each other.

Therefore, $AF=FC$ and $DF=BF$.

As $ABCD$ is a parallelogram, therefore $CD\parallel AB$ and $AC$ is a transversal.

The angles $\angle GCF$ and $\angle FAE$ are the alternate interior angles.

Therefore, $\angle GCF=\angle FAE$.

In the triangles $AFE$ and $CFG$, it can be noticed that:

$AF=FC$

$\angle GCF=\angle FAE$

$\angle CFG=\angle AFE\left(\text{vertically opposite angles}\right)$

Therefore, in the triangles $AFE$ and $CFG$, it can be seen that $AF=FC$, $\angle GCF=\angle FAE$and $\angle CFG=\angle AFE$.

Therefore, the triangles $AFE$and $CFG$are the congruent triangles by using the AAS postulate.

Therefore, by using the corresponding parts of congruent triangles it can be obtained that $\overline{EF}\cong \overline{FG}$.

Hence proved.