Chapter 9.6, Problem 10CE

To determine

ToProve: $m\angle 1=\frac{1}{2}(m\stackrel{⌢}{AC}-m\stackrel{⌢}{BD})$

Explanation of Solution

Giveninformation:

Given: Secants $\overline{PA}$ and $\overline{PC}$

Given figure:

  

Considerthegivenfigure,

Inabovefiguresecantsare $\overline{PA}$ and $\overline{PC}$ .

Considerthefollowing to prove $m\angle 1=\frac{1}{2}(m\stackrel{⌢}{AC}-m\stackrel{⌢}{BD})$

1. Sincethroughanytwopointsthereisexactlyoneline.

Drawchord $\overline{BC}$

2. The sum of two remote interior angles of the triangle is the measure of an exterior angle.

  $\begin{array}{l}m\angle 1+m\angle 2=m\angle 3\hfill \\ m\angle 1=m\angle 3-m\angle 2\hfill \end{array}$

3. An inscribed arc is equal to half of the measure of an intercepted angle,

  $\begin{array}{l}m\angle 2=\frac{1}{2}m\stackrel{⌢}{BD}\hfill \\ m\angle 3=\frac{1}{2}m\stackrel{⌢}{AC}\hfill \end{array}$

Putthevaluesfromabovetoget,

  $\begin{array}{cc}m\angle 1& =\frac{1}{2}m\stackrel{⌢}{AC}-\frac{1}{2}m\stackrel{⌢}{BD}\\ & =\frac{1}{2}(m\stackrel{⌢}{AC}-m\stackrel{⌢}{BD})\end{array}$

Sincelefthandsideequaltorighthandsidevalue.

Therefore, $m\angle 1=\frac{1}{2}(m\stackrel{⌢}{AC}-m\stackrel{⌢}{BD})$ isproved.