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# Segments from Secants

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Practice Segments from Secants
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Secant Lines to Circles

In the circle below, $m\widehat{CD}=100^\circ$ , $m\widehat{BC}=120^\circ$ , and $m\widehat{DE}=100^\circ$ . Find $m\angle BFE$ .

#### 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, $\overleftrightarrow{AB}$ is a secant.

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:

• $BF \cdot CF=EF \cdot DF$   (This will be explored in Example A)
• $\frac{m\widehat{CD}-m\widehat{BE}}{2}=m\angle BFE$ (This will be explored in Example B)

Case #2 : Two secants intersect inside the circle.

Relevant Theorems:

• $CF \cdot FB=DF \cdot FE$ (This was previously proved as a property of intersecting chords)
• $\frac{m\widehat{CD}-m\widehat{BE}}{2}=m\angle BFE =m\angle CFD$ (This will be explored in Example C)

Case #3 : A secant and a tangent intersect on the circle.

Relevant Theorem:

• $m\angle BFG=\frac{m\widehat{BF}}{2}$   (This will be explored in Guided Practice #1)

Case #4 : A secant and a tangent intersect outside the circle.

Relevant Theorems:

• $FB\cdot FH=FE^2$   (This will be explored in the practice problems)
• $\frac{m\widehat{HGE}-m\widehat{BE}}{2}=m\angle BFE$   (This will be explored in the practice problems)

Example A

Prove that $BF\cdot CF=EF\cdot DF$ .

Solution: Draw chords $\overline{CE}$ and $\overline{DB}$ .

Two triangles are created, $\Delta CEF$ and $\Delta DBF$ . Note that both triangles share $\angle F$ . Also note that both $\angle BCE$ and $\angle EDB$ are inscribed angles of $\widehat{BE}$ . Therefore, $\angle BCE \cong \angle EDB$ . Because $\Delta CEF$ and $\Delta DBF$ have two pairs of congruent angles, they are similar by $AA\sim$ . This means that corresponding sides of the triangles are proportional. In particular, $\frac{BF}{EF}=\frac{DF}{CF}$ . This means that $BF\cdot CF=EF\cdot DF$ .

Example B

Prove that $\frac{m\widehat{CD}-m\widehat{BE}}{2}=m\angle BFE$ .

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 $\overline{CE}$ and $\overline{DB}$ .

Consider how the angles are arcs are related.

• $m\angle CED=\frac{m\widehat{CD}}{2}$ (inscribed angle)
• $m\angle ECF=\frac{m\widehat{BE}}{2}$  (inscribed angle)
• $m\angle CED=m\angle ECF+m\angle BFE$  (exterior angle equals the sum of the remote interior angles)

Make two substitutions and you have:

$\frac{m\widehat{CD}}{2}=\frac{m\widehat{BE}}{2}+m\angle BFE$

Therefore, $m\angle BFE=\frac{m\widehat{CD}}{2}-\frac{m\widehat{BE}}{2}=\frac{m\widehat{CD}-m\widehat{BE}}{2}$

Example C

Prove that $\frac{m\widehat{CD}+ m\widehat{BE}}{2}=m\angle BFE=m\angle CFD$ .

Solution: This logic of this proof is similar to the logic used in Example B. Start by drawing chord $\overline{DB}$ .

Consider how the angles and arcs are related.

• $m\angle CBD=\frac{m\widehat{CD}}{2}$  (inscribed angle)
• $m\angle BDE=\frac{m\widehat{BE}}{2}$  (inscribed angle)
• $m\angle CFD=m\angle CBD+m\angle BDE$  (exterior angle equals the sum of the remote interior angles)

Make two substitutions and you have:

$m\angle CFD=\frac{m\widehat{CD}}{2}+\frac{m\widehat{BE}}{2}$

Therefore, $\frac{m\widehat{CD}+m\widehat{BE}}{2}=m\angle CFD$ . Because $\angle{CFD}$ and $\angle BFE$ are vertical angles, they are congruent and have equal measures. This means $\frac{m\widehat{CD}+m\widehat{BE}}{2}=m\angle BFE=m\angle CFD$ .

Concept Problem Revisited

In the circle below, $m\widehat{CD}=100^\circ$ , $m\widehat{BC}=120^\circ$ , and $m\widehat{DE}=100^\circ$ . Find $m\angle BFE$ .

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, $m\angle BFE=\frac{m\widehat{CD}-m\widehat{BE}}{2}$ . You are given $m\widehat{CD}=100^\circ$ , but you don't know $m\widehat{BE}$ . Use the fact that a full circle is $360^\circ$ to find $m\widehat{BE}$ .

$m\widehat{BE} &=360^\circ - m\widehat{CD} - m\widehat{BC} - m\widehat{DE}\\m\widehat{BE} &=360^\circ - 100^\circ - 120^\circ - 100^\circ \\m\widehat{BE} &= 40^\circ$

Now, solve for the measure of $\angle BFE$ .

$m\angle BFE &=\frac{m\widehat{CD}-m\widehat{BE}}{2}\\m\angle BFE &=\frac{100^\circ - 40^\circ}{2} \\m\angle BFE &=30^\circ$

#### 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. $\overleftrightarrow{FG}$ is tangent to circle  $A$ at point $F$ . Prove that $m\angle BFG=\frac{m\widehat{BF}}{2}$ .

2. $m\widehat{FCB}=280^\circ$ . Find $m\angle BFG$ .

3. $m\widehat{CD}=70^\circ$ and $m\widehat{BE}=40^\circ$ . Find $m\angle CFE$ .

1. Draw a diameter through points  $F$ and $A$ . This segment will be perpendicular to  $\overleftrightarrow{FG}$ .

First note that $m\widehat{CB}+m\widehat{BF}=180^\circ$ because the two arcs make a semicircle. This means that $\frac{m\widehat{CB}}{2}+\frac{m\widehat{BF}}{2}=90^\circ$ and thus $\frac{m\widehat{BF}}{2}=90^\circ - \frac{m\widehat{CB}}{2}$ .

Now consider other angle and arc relationships:

• $m\angle CFB=\frac{m\widehat{CB}}{2}$  (inscribed angle)
• $m\angle CFB+m\angle BFG=90^\circ$  (two angles make a right angle)

By substitution, $\frac{m\widehat{CB}}{2}+m\angle BFG=90^\circ$ . Therefore, $m\angle BFG=90^\circ - \frac{m\widehat{CB}}{2}$ .

Consider the two highlighted statements. Both $\frac{m\widehat{BF}}{2}$ and $m\angle BFG$ are equal to $90^\circ - \frac{m\widehat{CB}}{2}$ . Therefore, $m\angle BFG=\frac{m\widehat{BF}}{2}$ .

2. If $m\widehat{FCB}=280^\circ$ , then $m\widehat{FB}=360^\circ - 280 ^\circ =80^\circ$ . Therefore, $m\angle {BFG}=\frac{80^\circ}{2}=40^\circ$ .

3. $m\widehat{CD}=70^\circ$ and $m\widehat{BE}=40^\circ$ . $m\angle CFD$ is the average of the measure of the intercepted arcs.

$m\angle CFD=\frac{70^\circ + 40^\circ}{2}=55^\circ$

Therefore, $m\angle CFE=180^\circ-55^\circ=125^\circ$ .

#### Practice

1. What's the difference between a secant and a tangent?

Use the relationships explored in this concept to solve for $x$ or $\theta$ 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 $\overline{BE}$ . Explain why $\angle FEB \cong \angle EHB$ .

9. Prove that $\Delta EHF \sim \Delta BEF$ .

10. Prove that $FB \cdot FH=FE^2$ .

11. Prove that $\frac{m\widehat{HGE}}{2}=\frac{m\widehat{BE}}{2}+m\angle BFE$ (Use Example B to help).

12. Prove that $\frac{m\widehat{HGE}-m\widehat{BE}}{2}=m\angle BFE$ .

13. How is the theorem proved in #11-#12 related to the theorem proved in Examples B?

Solve for $x$ or $\theta$ in each circle.

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