Chapter 9, Problem 13CT

Solution

To prove: $\overline{AC}\parallel \overline{DB}$

Given:

The figure:

  $m\stackrel{⌢}{B}C=m\stackrel{⌢}{A}B$

Explanation:

Consider the given figure:

By theorem 9-7, an inscribed angle is equal to one half the measure of it’s intercepted, so

  $m\angle DBA=\frac{1}{2}m\stackrel{⌢}{A}B$ ….. (1)

And

  $m\angle BAC=\frac{1}{2}m\stackrel{⌢}{B}C$ ….. (2)

Now, it is given that,

  $m\stackrel{⌢}{B}C=m\stackrel{⌢}{A}B$

So,

  $\frac{1}{2}m\stackrel{⌢}{B}C=\frac{1}{2}m\stackrel{⌢}{A}B$ ….. (3)

From (1), (2) and (3), it can be said that

  $m\angle BAC=m\angle DBA$

This gives that,

  $\angle BAC\cong \angle DBA$

It can be observed that the angles $\angle BAC$ and $\angle DBA$ are the alternate interior angles for the lines $\overline{AC}$ and $\overline{DB}$ and the transversal $\overline{BA}$ .

Thus, by the theorem 3-5, it can be said that the lines

$\overline{AC}$ and $\overline{DB}$ are parallel.