Angles Outside a Circle
An angle is outside a circle if its vertex is outside the circle and its sides are tangents or secants. The possibilities are: an angle formed by two tangents, an angle formed by a tangent and a secant, and an angle formed by two secants.
Outside Angle Theorem: The measure of an angle formed by two secants, two tangents, or a secant and a tangent from a point outside the circle is half the difference of the measures of the intercepted arcs.
\begin{align*}m\angle D = \frac{m\widehat{EF}m\widehat{GH}}{2}\end{align*}
What if you were given a circle with either two secants, two tangents, or one of each that share a common point outside the circle? How could you use the measure of the arcs formed by those circle parts to find the measure of the angle they make outside the circle?
Examples
Example 1
Find the measure of \begin{align*}x\end{align*}
Use the Outside Angle Theorem.
\begin{align*}x=\frac{125^\circ27^\circ}{2}=\frac{98^\circ}{2}=49^\circ\end{align*}
Example 2
Find the measure of \begin{align*}x\end{align*}
Use the Outside Angle Theorem.
\begin{align*}40^\circ\end{align*}
Example 3
Find the value of \begin{align*}x\end{align*}
\begin{align*}x=\frac{72^\circ  22^\circ}{2}=\frac{50^\circ}{2}=25^\circ\end{align*}
Example 4
Find the value of \begin{align*}x\end{align*}
\begin{align*}x=\frac{120^\circ  32^\circ}{2}=\frac{88^\circ}{2}=44^\circ\end{align*}
Example 5
Find the value of \begin{align*}x\end{align*}
First note that the missing arc by angle \begin{align*}x\end{align*}
Review
Find the value of the missing variable(s).
Solve for \begin{align*}x\end{align*}
 Fill in the blanks of the proof for the Outside Angle Theorem
Given: Secant rays \begin{align*}\overrightarrow{AB}\end{align*}
Prove: \begin{align*}m\angle a = \frac{1}{2} \left (m\widehat{BC}m\widehat{DE} \right )\end{align*}
Statement  Reason 

1. Intersecting secants \begin{align*}\overrightarrow{AB}\end{align*} and \begin{align*}\overrightarrow{AC}\end{align*}.  1. 
2. Draw \begin{align*}\overline{BE}\end{align*}.

2. Construction 
3. \begin{align*}m\angle BEC &= \frac{1}{2} m\widehat{BC}\\ m\angle DBE &= \frac{1}{2} m\widehat{DE}\end{align*}  3. 
4. \begin{align*}m\angle a+m\angle DBE=m\angle BEC\end{align*}  4. 
5.  5. Subtraction PoE 
6.  6. Substitution 
7. \begin{align*}m\angle a=\frac{1}{2} \left (m\widehat{BC}m\widehat{DE} \right )\end{align*}  7. 
Review (Answers)
To see the Review answers, open this PDF file and look for section 9.8.