Question

Given: CDEF is a parallelogram; S and T are the midpoints of $\overline{\mathit{E}\mathit{F}}$and $\overline{\mathit{E}\mathit{D}}$.

Prove: $\overline{\mathit{S}\mathit{R}}\cong \overline{\mathit{F}\mathit{D}}$.

Short answer

In the parallelogram FSRD, it can be said that $FS\mathit{\cong }RD$and $FD\mathit{\cong }SR$.

Therefore, role="math" localid="1638348936438" $SR\mathit{\cong }FD$.

Step-by-step solution

Step 1. Observe the given diagram.

The given diagram is:

Step 2. Description of step.

It is being given that CDEF is a parallelogram, therefore, $FE\parallel CD$.

Therefore it can be said that $FS\parallel DR$or $SE\parallel DR$.

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

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

Therefore,$FS\cong SE$

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

Therefore, $ET\cong TD$.

As $SE\parallel DR$and DE is a transversal and the angles $\angle SET$and $\angle RDT$are the alternate interior angles.

Therefore, $\angle SET\cong \angle RDT$.

The angles $\angle STE$and $\angle RTD$are the vertically opposite angles.

Therefore, $\angle STE\cong \angle RTD$.

Step 3. Description of step.

In the triangles $\Delta SET$and $\Delta RDT$, it can be noticed that:

$ET\cong TD$

$\angle SET\cong \angle RDT$

$\angle STE\cong \angle RTD$

Therefore, the triangles $\Delta SET$and $\Delta RDT$are congruent triangles by using the ASA postulate.

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

As, $FS\cong SE$and $SE\cong RD$, therefore $FS\cong RD$.

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 $FS\parallel DR$and $FS\cong RD$.

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 $FS\cong RD$and $FD\cong SR$.

Therefore, $SR\cong FD$.

Hence proved.