Chapter 3.2, Problem 14WE

To determine

To find: the values of x , y and z .

Answer to Problem 14WE

  $x=56$

  $y=100$

  $z=25$

Explanation of Solution

Given:

The diagram:

Concept Used:

Theorem 3-2: If two parallel lines are cut by a transversal then the alternate interior angles are congruent.

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

Label the given figure as shown below:

In the above figure, lines $\overline{AB}\parallel \overline{DC}$ and $\overline{AC}$ is a transversal, so by theorem 3-2,

  $\angle BAC\cong \angle ACD$ (Alternate interior angles)

That is,

  $\begin{array}{c}m\angle BAC=m\angle ACD\\ x=56\end{array}$

Now, lines $\overline{AB}\parallel \overline{DC}$ and $\overline{AD}$ is a transversal, so by theorem 3-3,

  $m\angle BAD+m\angle ADC=180$ (Same-side interior angles)

But $\angle BAD=\angle BAC+\angle CAD$ , so,

  $m\angle BAD=m\angle BAC+m\angle CAD$

This implies that,

  $\begin{array}{c}m\angle BAD+m\angle ADC=180\\ m\angle BAC+m\angle CAD+m\angle ADC=180\\ 56+24+4z=180\\ 80+4z=180\end{array}$

  $\begin{array}{c}4z=180-80\\ 4z=100\\ z=\frac{100}{4}\\ z=25\end{array}$

Similarly, lines $\overline{BC}\parallel \overline{AD}$ and $\overline{AB}$ is a transversal, so by theorem 3-3,

  $m\angle BAD+m\angle ABC=180$ (Same-side interior angles)

But $\angle BAD=\angle BAC+\angle CAD$ , so,

  $m\angle BAD=m\angle BAC+m\angle CAD$

This implies that,

  $\begin{array}{c}m\angle BAD+m\angle ABC=180\\ m\angle BAC+m\angle CAD+m\angle ABC=180\\ 56+24+y=180\end{array}$

  $\begin{array}{c}80+y=180\\ y=180-80\\ y=100\end{array}$