Chapter 3.3, Problem 19WE

To determine

To find: the values of x and y that makes the red lines parallel and the blue lines parallel.

Answer to Problem 19WE

  $x=35$

  $y=20$

Explanation of Solution

Given:

The diagram:

Concept Used:

Theorem 3-5: If two lines are cut by a transversal and alternate interior angles are congruent, then the lines are parallel.

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

Labeling the given figure:

From the above figure, it can be observed that the angles $\angle ABC$ and $\angle BCD$ are the alternate interior angles which are formed when the lines $\overline{AB}$ and $\overline{CD}$ are cut by the transversal $\overline{BC}$ .

Now by theorem 3-5, for the red lines $\overline{AB}$ and $\overline{CD}$ to be parallel, we must have that the angles $\angle ABC$ and $\angle BCD$ are congruent, that is,

  $\begin{array}{c}m\angle ABC=m\angle BCD\\ 3x=105\\ x=\frac{105}{3}\\ x=35\end{array}$

Now, it can also be observed that the angles $\angle BCE$ and $\angle CED$ are the same-side interior angles which are formed when the lines $\overline{BC}$ and $\overline{DE}$ are cut by the transversal $\overline{CE}$ .

Now by theorem 3-6, for the blue lines $\overline{BC}$ and $\overline{DE}$ to be parallel, we must have that the angles $\angle BCE$ and $\angle CED$ are supplementary, that is,

  $m\angle BCE+m\angle CED=180$

Now,

  $\begin{array}{c}m\angle BCE=m\angle BCD+m\angle DCE\\ =105+2y\end{array}$

And,

  $\begin{array}{c}m\angle CED=x\\ =35\end{array}$

So,

  $\begin{array}{c}m\angle BCE+m\angle CED=180\\ 105+2y+35=180\\ 105+35+2y=180\\ 140+2y=180\end{array}$

  $\begin{array}{c}2y=180-140\\ 2y=40\\ y=\frac{40}{2}\\ y=20\end{array}$