Chapter 5.3, Problem 21PSB

To determine

To Prove: The bisectors of a pair of alternate angles formed by a transversal cutting parallel lines are parallel.

Explanation of Solution

Given:

  

  $\begin{array}{l}\overline{a}\parallel \overline{b}\\ \overleftrightarrow{RL}\text{bisects }\angle \text{PLN}\\ \overleftrightarrow{MS}\text{bisects }\angle \text{UMO}\end{array}$

Concept Used:

When parallel lines is intersected by a transversal alternate exterior angles are equal.

Proof:

Here, we have

  $\begin{array}{l}\overline{a}\parallel \overline{b}\\ \text{Since, when parallel lines is intersected by a transversal alternate exterior angles are equal}\text{. }\\ \Rightarrow \angle PLN=\angle UMO\\ \because \overleftrightarrow{RL}\text{bisects }\angle \text{PLN and }\overleftrightarrow{MS}\text{bisects }\angle \text{UMO}\\ \Rightarrow \frac{1}{2}\left(m\angle PLN\right)=\frac{1}{2}\left(m\angle UMO\right)\\ \Rightarrow \angle 2=\angle 3\\ \Rightarrow \overleftrightarrow{RL}\parallel \overleftrightarrow{MS}\end{array}$