In the circle below,
Watch This
http://www.youtube.com/watch?v=b1E3eFF43fY Brightstorm: Geometry Secants
Guidance
Recall that a line that intersects a circle in exactly one point is called a tangent line. A line that intersects a circle in two points is called a secant line. Below,
When two secants or a tangent and a secant are drawn, they can interact in four ways. In each case, arcs, angles and line segments have special relationships. These ideas are summarized below, and will be explored further and proved in the examples and practice.
Case #1: Two secants intersect outside the circle.
Relevant Theorems:

BF⋅CF=EF⋅DF (This will be explored in Example A) 
mCDˆ−mBEˆ2=m∠BFE (This will be explored in Example B)
Case #2: Two secants intersect inside the circle.
Relevant Theorems:

CF⋅FB=DF⋅FE (This was previously proved as a property of intersecting chords) 
mCDˆ−mBEˆ2=m∠BFE=m∠CFD (This will be explored in Example C)
Case #3: A secant and a tangent intersect on the circle.
Relevant Theorem:

m∠BFG=mBFˆ2 (This will be explored in Guided Practice #1)
Case #4: A secant and a tangent intersect outside the circle.
Relevant Theorems:

FB⋅FH=FE2 (This will be explored in the practice problems) 
mHGEˆ−mBEˆ2=m∠BFE (This will be explored in the practice problems)
Example A
Prove that
Solution: Draw chords
Two triangles are created,
Example B
Prove that
Solution: You are trying to prove that the measure of the angle is equal to half the difference between the measures of the red arc and the blue arc. As in Example A, draw chords
Consider how the angles are arcs are related.

m∠CED=mCDˆ2 (inscribed angle) 
m∠ECF=mBEˆ2 (inscribed angle) 
m∠CED=m∠ECF+m∠BFE (exterior angle equals the sum of the remote interior angles)
Make two substitutions and you have:
Therefore,
Example C
Prove that
Solution: This logic of this proof is similar to the logic used in Example B. Start by drawing chord
Consider how the angles and arcs are related.

m∠CBD=mCDˆ2 (inscribed angle) 
m∠BDE=mBEˆ2 (inscribed angle) 
m∠CFD=m∠CBD+m∠∠BDE (exterior angle equals the sum of the remote interior angles)
Make two substitutions and you have:
Therefore,
Concept Problem Revisited
In the circle below,
This is an example of two secants intersecting outside the circle. The intersection angle of the two secants is equal to half the difference between their intercepted arcs. In other words,
\begin{align*}m\widehat{BE} &=360^\circ  m\widehat{CD}  m\widehat{BC}  m\widehat{DE}\\ m\widehat{BE} &=360^\circ  100^\circ  120^\circ  100^\circ \\ m\widehat{BE} &= 40^\circ\end{align*}
Now, solve for the measure of \begin{align*}\angle BFE\end{align*}.
\begin{align*}m\angle BFE &=\frac{m\widehat{CD}m\widehat{BE}}{2}\\ m\angle BFE &=\frac{100^\circ  40^\circ}{2} \\ m\angle BFE &=30^\circ \end{align*}
Vocabulary
When a line intersects a circle in exactly one point the line is said to be tangent to the circle or a tangent of the circle.
A line that intersects a circle in two points is a secant line.
A chord is a segment that connects two points on a circle. If a chord passes through the center of the circle then it is a diameter.
Guided Practice
1. \begin{align*}\overleftrightarrow{FG}\end{align*} is tangent to circle \begin{align*}A\end{align*} at point \begin{align*}F\end{align*}. Prove that \begin{align*}m\angle BFG=\frac{m\widehat{BF}}{2}\end{align*}.
2. \begin{align*}m\widehat{FCB}=280^\circ\end{align*}. Find \begin{align*}m\angle BFG\end{align*}.
3. \begin{align*}m\widehat{CD}=70^\circ\end{align*} and \begin{align*}m\widehat{BE}=40^\circ\end{align*}. Find \begin{align*}m\angle CFE\end{align*}.
Answers:
1. Draw a diameter through points \begin{align*}F\end{align*} and \begin{align*}A\end{align*}. This segment will be perpendicular to \begin{align*}\overleftrightarrow{FG}\end{align*}.
First note that \begin{align*}m\widehat{CB}+m\widehat{BF}=180^\circ\end{align*} because the two arcs make a semicircle. This means that \begin{align*}\frac{m\widehat{CB}}{2}+\frac{m\widehat{BF}}{2}=90^\circ\end{align*} and thus \begin{align*}\frac{m\widehat{BF}}{2}=90^\circ  \frac{m\widehat{CB}}{2}\end{align*}.
Now consider other angle and arc relationships:
 \begin{align*}m\angle CFB=\frac{m\widehat{CB}}{2}\end{align*} (inscribed angle)
 \begin{align*}m\angle CFB+m\angle BFG=90^\circ\end{align*} (two angles make a right angle)
By substitution, \begin{align*}\frac{m\widehat{CB}}{2}+m\angle BFG=90^\circ\end{align*}. Therefore, \begin{align*}m\angle BFG=90^\circ  \frac{m\widehat{CB}}{2}\end{align*}.
Consider the two highlighted statements. Both \begin{align*}\frac{m\widehat{BF}}{2}\end{align*} and \begin{align*}m\angle BFG\end{align*} are equal to \begin{align*}90^\circ  \frac{m\widehat{CB}}{2}\end{align*}. Therefore, \begin{align*}m\angle BFG=\frac{m\widehat{BF}}{2}\end{align*}.
2. If \begin{align*}m\widehat{FCB}=280^\circ\end{align*}, then \begin{align*}m\widehat{FB}=360^\circ  280 ^\circ =80^\circ\end{align*}. Therefore, \begin{align*}m\angle {BFG}=\frac{80^\circ}{2}=40^\circ\end{align*}.
3. \begin{align*}m\widehat{CD}=70^\circ\end{align*} and \begin{align*}m\widehat{BE}=40^\circ\end{align*}. \begin{align*}m\angle CFD\end{align*} is the average of the measure of the intercepted arcs.
\begin{align*}m\angle CFD=\frac{70^\circ + 40^\circ}{2}=55^\circ\end{align*}
Therefore, \begin{align*}m\angle CFE=180^\circ55^\circ=125^\circ\end{align*}.
Practice
1. What's the difference between a secant and a tangent?
Use the relationships explored in this concept to solve for \begin{align*}x\end{align*} or \begin{align*}\theta\end{align*} in each circle.
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In #8#12 you will explore Case #4: A secant and a tangent intersect outside the circle.
8. Draw chord \begin{align*}\overline{BE}\end{align*}. Explain why \begin{align*}\angle FEB \cong \angle EHB\end{align*}.
9. Prove that \begin{align*}\Delta EHF \sim \Delta BEF\end{align*}.
10. Prove that \begin{align*}FB \cdot FH=FE^2\end{align*}.
11. Prove that \begin{align*}\frac{m\widehat{HGE}}{2}=\frac{m\widehat{BE}}{2}+m\angle BFE\end{align*} (Use Example B to help).
12. Prove that \begin{align*}\frac{m\widehat{HGE}m\widehat{BE}}{2}=m\angle BFE\end{align*}.
13. How is the theorem proved in #11#12 related to the theorem proved in Examples B?
Solve for \begin{align*}x\end{align*} or \begin{align*}\theta\end{align*} in each circle.
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