Chapter 5.5, Problem 15PSB

To determine

To prove:The segments $DF\text{and}EB$ are parallel to each other.

Explanation of Solution

Given Information:

  $ABCD$ is a parallelogram

  $AF=CE$

Formula used:

Opposite angles of a parallelogram are equal

Opposite sides of a parallelogram are parallel and equal

Corresponding parts of congruent triangles are congruent

Proof:

  $ABCD$ is a parallelogram

We know that, the opposite sides of a parallelogram are equal

  $\Rightarrow AB\parallel CD\text{and}AB=CD\mathrm{...}\left(\text{i}\right)$

In $\Delta ABE\text{and}\Delta CDF,$

  $AB=CD$ … From (i)

  $\begin{array}{l}AE=AF+FE\mathrm{...}\left(\text{Given}\right)\\ \Rightarrow AE=CE+FE=CF\end{array}$

  $\angle BAE=\angle DCF$ … (Alternate interior angles)

  $\therefore \Delta ABE\cong \Delta CDF$ … (By SAS congruence rule)

As the corresponding parts of congruent triangles are congruent, so we have

  $\angle AEB=\angle CFD$

Now, $AC$ is the line through $BE\text{and}DF$ and we have $\angle AEB=\angle CFD$

Thus, $AC$ becomes the transversal through $BE\text{and}DF$

Hence, $DF\parallel EB.$