Chapter 10.7, Problem 8PSA

To determine

Prove the given statement.

Answer to Problem 8PSA

  $\overrightarrow{BG}bisect\stackrel{⌢}{DC}$ Proved.

Explanation of Solution

Given information:

The bisector of an angle of an inscribed triangle also bisects the arc cut off by the opposite side.

Calculation:

As per statement the figure is,

  

  $\Delta BCD$ is inscribed in circle $A$ ,

  $\begin{array}{l}\overrightarrow{BG}bi\sec ttheangle\angle CBD\\ find\overrightarrow{BG}bi\sec t\stackrel{⌢}{DC}\end{array}$

If a ray bisects the angle it divides it in to congruent angles,

  $\begin{array}{l}\angle DBG\cong \angle GBC\\ m\angle DBG=\angle GBC\end{array}$

The measure of an inscribed angle is one half the intercepted arc,

  $\begin{array}{l}m\angle DBG=\frac{1}{2}m\stackrel{⌢}{DG}\\ m\angle GBC=\frac{1}{2}m\stackrel{⌢}{GC}\\ \frac{1}{2}m\stackrel{⌢}{DG}=\frac{1}{2}m\stackrel{⌢}{GC}\\ m\stackrel{⌢}{DG}=m\stackrel{⌢}{GC}\\ \stackrel{⌢}{DG}\cong \stackrel{⌢}{GC}\end{array}$

Hence , $\overrightarrow{BG}bisect\stackrel{⌢}{DC}$