Chapter 5.3, Problem 2PSA

To determine

To Prove: $\overline{EJ}\cong \overline{JH}$ .

Explanation of Solution

Given:

  

  $\overline{EF}\parallel \overline{GH},\overline{EF}\cong \overline{GH}$

Concept Used:

When two parallel lines are intersected by a transversal alternate interior angles are equal

Proof:

Here, we have

  

Since, when two parallel lines are intersected by a transversal alternate interior angles are equal.

  $\begin{array}{l}\overline{EF}\parallel \overline{GH},\overline{EF}\cong \overline{GH}\\ \text{In }△JEF\text{ and }△JHG\\ \Rightarrow \angle JEF=\angle JHG\left(\because \text{ Alternate interior angles}\right)\\ \Rightarrow \overline{EF}\cong \overline{GH}\left(\because \text{ Given}\right)\\ \Rightarrow \angle EFJ=\angle HGJ\left(\because \text{ Alternate interior angles}\right)\\ \therefore △JEF\cong △JHG\\ \Rightarrow \overline{EJ}\cong \overline{JH}\end{array}$