Chapter 3.3, Problem 11CE

To determine

To state: the segments in the figure which are parallel given that $m\angle 1+m\angle 2+m\angle 5+m\angle 6=180$ .

Answer to Problem 11CE

  $\overline{PA}\parallel \overline{LR}$

Explanation of Solution

Given:

The diagram:

Concept Used:

Theorem 3-6: If two lines are cut by a transversal and same-side interior angles are supplementary, then the lines are parallel.

Consider the given figure:

From the above figure, it can be observed that:

  $m\angle 1+m\angle 2=m\angle PLR$

  $m\angle 5+m\angle 6=m\angle APL$

So, the given data can be rewritten as:

  $\begin{array}{c}m\angle 1+m\angle 2+m\angle 5+m\angle 6=180\\ m\angle APL+m\angle PLR=180\end{array}$

Now the angles $\angle APL$ and $\angle PLR$ are the same-side interior angles which are formed when the lines $\overline{PA}$ and $\overline{LR}$ are cut by the transversal $\overline{PL}$ .

Since,

  $m\angle APL+m\angle PLR=180$

It implies that, the same-side interior angles $\angle APL$ and $\angle PLR$ are supplementary.

Thus, by the theorem 3-6, it can be said that the lines $\overline{PA}$ and $\overline{LR}$ are parallel.