91Ó°ÊÓ

The given diagram is:

It is given that $\stackrel{¯}{AD}â‰\dots \stackrel{¯}{BC}$ and $\stackrel{¯}{AD}âˆ\yen \stackrel{¯}{BC}$.

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

As, $\stackrel{¯}{AD}â‰\dots \stackrel{¯}{BC}$ and $\stackrel{¯}{AD}âˆ\yen \stackrel{¯}{BC}$, 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 $CDâˆ\yen AB$ and $AC$ is a transversal.

The angles $∠GCF$ and $∠FAE$ are the alternate interior angles.

Therefore, $∠GCF=∠FAE$.

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

$AF=FC$

$∠GCF=∠FAE$

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

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

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 $\stackrel{¯}{EF}â‰\dots \stackrel{¯}{FG}$.

Hence proved.