91Ó°ÊÓ

The given diagram is:

It is being given that CDEF is a parallelogram, therefore, $FEâˆ\yen CD$.

Therefore it can be said that $FSâˆ\yen DR$or $SEâˆ\yen DR$.

It is also being given that S and T are the midpoints of $\stackrel{¯}{EF}$and $\stackrel{¯}{ED}$.

As, S is the midpoint of $\stackrel{¯}{EF}$.

Therefore,$FSâ‰\dots SE$

As, T is the midpoint of $\stackrel{¯}{ED}$.

Therefore, $ETâ‰\dots TD$.

As $SEâˆ\yen DR$and DE is a transversal and the angles $∠SET$and $∠RDT$are the alternate interior angles.

Therefore, $∠SETâ‰\dots ∠RDT$.

The angles $∠STE$and $∠RTD$are the vertically opposite angles.

Therefore, $∠STEâ‰\dots ∠RTD$.

In the triangles $\mathrm{Δ}SET$and $\mathrm{Δ}RDT$, it can be noticed that:

$ETâ‰\dots TD$

$∠SETâ‰\dots ∠RDT$

$∠STEâ‰\dots ∠RTD$

Therefore, the triangles $\mathrm{Δ}SET$and $\mathrm{Δ}RDT$are congruent triangles by using the ASA postulate.

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

As, $FSâ‰\dots SE$and $SEâ‰\dots RD$, therefore $FSâ‰\dots RD$.

If one pair of opposite sides of a quadrilateral are both congruent and parallel then the quadrilateral is a parallelogram.

As, in the quadrilateral FSRD, it can be seen that $FSâˆ\yen DR$and $FSâ‰\dots RD$.

Therefore, as in the quadrilateral FSRD, one pair of opposite sides FS and RD are both congruent and parallel, therefore the quadrilateral FSRD is a parallelogram.

In a parallelogram both pairs of opposite sides are congruent.

Therefore, in the parallelogram FSRD, it can be said that $FSâ‰\dots RD$and $FDâ‰\dots SR$.

Therefore, $SRâ‰\dots FD$.

Hence proved.