Chapter 3.2, Problem 12WE

To determine

To find: the values of x and y .

Answer to Problem 12WE

  $x=55$

  $y=75$

Explanation of Solution

Given:

The diagram:

Concept Used:

Postulate 10: If two parallel lines are cut by a transversal then the corresponding angles are congruent

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

Sum of measures of all angles of a triangle is 180.

Label the given figure as shown below:

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

  $\angle ABC\cong \angle BCD$ (Alternate interior angles)

That is,

  $m\angle ABC=m\angle BCD$ ….. (1)

Similarly, lines $\overline{BC}\parallel \overline{DE}$ and $\overline{CD}$ is a transversal, so by theorem 3-2,

  $\angle BCD\cong \angle CDE$ (Alternate interior angles)

That is,

  $m\angle BCD=m\angle CDE$ ….. (2)

From (1) and (2),

  $\begin{array}{c}m\angle CDE=m\angle ABC\\ x=55\end{array}$

Now, in triangle $△CDE$ and using that sum of measures of all angles of a triangle is 180,

  $\begin{array}{c}m\angle CDE+m\angle DEC+m\angle ECD=180\\ 55+50+m\angle ECD=180\\ 105+m\angle ECD=180\end{array}$

  $\begin{array}{c}m\angle ECD=180-105\\ m\angle ECD=75\end{array}$

In the above figure, lines $\overline{AB}\parallel \overline{CD}$ and $\overline{AC}$ is a transversal, so by postulate 10,

  $\angle CAB\cong \angle ECD$ (Corresponding angles)

That is,

  $\begin{array}{c}m\angle CAB=m\angle ECD\\ y=75\end{array}$