Chapter 5, Problem 12CR

To determine

To prove that $SR\cong FD$ .

Explanation of Solution

Given:

A parallelogram $CDEF$ .

  $S$ and $T$ are the midpoints of $EF$ and $ED$ .

  

Proof:

  $CDEF$ is a parallelogram.

  $S$ and $T$ are midpoints of $EF$ and $ED$ respectively.

  $\Rightarrow ES\cong SF$ and $ET\cong TD$ .

Now, $\angle STE\cong \angle RTD$ (vertically opposite angle)

Vertical opposite angle is congruent

Now, $FS$ is parallel to $DR$ (opposite sides of a parallelogram are parallel)

  $\Rightarrow \angle SET\cong \angle RDT$ (alternate interior angles are congruent)

In $\Delta STE$ and $\Delta RTD,$

  $\angle STE\cong \angle RTD$

  $ET\cong TD$

  $\angle SET\cong \angle RDT$

  $\therefore \Delta STE\cong \Delta RTD$ by Angle-Side-Angle (ASA) congruency criteria.

  $\Rightarrow SE\cong RD$ (corresponding parts of congruent triangles are congruent)

  $\Rightarrow SF\cong RD$ (transitive property of congruency)

  $\therefore FSRD$ is a parallelogram.

  $\Rightarrow SR\cong FD$

Hence Proved.

Conclusion:

Hence, $SR\cong FD$ .