Chapter 5.3, Problem 15PSB

To determine

To Prove: $\overline{GJ}\cong \overline{HK}$ .

Explanation of Solution

Given:

  

  $\begin{array}{l}\overleftrightarrow{FH}\parallel \overleftrightarrow{JM}\\ \angle 1=\angle 2\\ \overline{FH}\cong \overline{JM}\end{array}$

Concept Used:

When two parallel lines are intersected by a transversal:

• Alternate exterior angles are equal

Proof:

Here, we have

  

  $\begin{array}{l}\overleftrightarrow{FH}\parallel \overleftrightarrow{JM}\\ \angle 1=\angle 2\\ \overline{FH}\cong \overline{JM}\end{array}$

Since exterior angle is equal to the sum of interior opposite angles.

  $\begin{array}{l}\angle 1=\angle MJK+\angle JMK\\ \angle 2=\angle FHG+\angle FGH\\ \because \angle 1=\angle 2\\ \Rightarrow \angle MJK+\angle JKM=\angle FHG+\angle FGH\end{array}$

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

  $\begin{array}{l}\Rightarrow \angle FHG=\angle KJM\\ \because \angle MJK+\angle JKM=\angle FHG+\angle FGH\\ \Rightarrow \angle FGH=\angle JKM\end{array}$

  $\begin{array}{l}\text{In }△FGH\text{and }△JKM\\ \angle FGH=\angle JKM\left(\text{Proved}\right)\\ \overline{FH}\cong \overline{JM}\left(\text{Given}\right)\\ \angle FHG=\angle KJM\left(\text{Alternate exterior angles}\right)\\ \Rightarrow △FGH\cong △JKM\\ \Rightarrow \overline{GH}\cong \overline{JK}\\ \Rightarrow \overline{GH}+\overline{HJ}\cong \overline{HJ}+\overline{JK}\\ \Rightarrow \overline{GJ}\cong \overline{HK}\end{array}$