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Central Angles and Chords

Radius  $\overline{AD}$ bisects  $\angle BAC$ in the circle below. How does  $\overline{AD}$ relate to chord $\overline{BC}$ ? Prove your ideas.

Watch This

https://www.youtube.com/watch?v=o5cq9BA_hlE Brightstorm: Central Angles and Intercepted Arcs

Guidance

A central angle for a circle is an angle with its vertex at the center of the circle.

In the circle above,  $A$ is the center and  $\angle BAC$ is a central angle. Notice that the central angle meets the circle at two points ( $B$ and $C$ ), dividing the circle into two sections. Each of circle portions is called an arc . The smaller arc (blue below) is called the minor arc , and is considered the arc that is intercepted by the central angle. The larger arc (red below) is called the major arc .

The minor arc above is named $\widehat{B C}$ . Notice that  $B$ and  $C$ are the endpoints of the arc and there is an arc symbol above the letters indicating that you are referring to an arc.

The major arc above is named $\widehat{B D C}$ . When naming a major arc you should use three letters so that it is clear you are referring to the larger portion of the circle.

Arcs can be measured in degrees just like angles. In general, the degree measure of a minor arc is equal to the measure of the central angle that intercepts it . Because there are  $360^\circ$ in a circle, the sum of the measures of a minor arc and its corresponding major arc will be $360^\circ$ .

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 . In the circle below, both  $\overline{BD}$ and  $\overline{CE}$ are chords.

Notice that each chord has a corresponding arc.  $\overline{CE}$ is a chord and  $\widehat{CE}$ is an arc. In Example B you will prove that two chords are congruent if and only if their corresponding arcs are congruent.

Example A

Find  $m \angle CAE$ and $m \widehat{C D E}$ .

Solution: The degree measure of a minor arc is equal to the measure of the central angle that intercepts it. Therefore, $m \angle CAE=140^\circ$ . A full circle is $360^\circ$ , so $m \widehat{C D E}=360^\circ - 140^\circ=220^\circ$ .

Example B

Prove that two chords are congruent if and only if their corresponding arcs are congruent.

Solution: This statement has two parts that you must prove.

1. If two arcs are congruent then their corresponding chords are congruent.
2. If two chords are congruent then their corresponding arcs are congruent.

Both statements can be proved by finding congruent triangles. Consider the circle below with center $A$ . Note that $\overline{CA}$ , $\overline{DA}$ , $\overline{EA}$ , and  $\overline{FA}$ are all radii of the circle and therefore are all congruent.

Start by proving statement #1. Assume that $\widehat{C D} \cong \widehat{F E}$ . This would imply that $m \widehat{C D}=m \widehat{F E}$ . Because the measure of an arc is the same as the measure of its corresponding central angle, it must be true that  $m \angle CAD=m \angle FAE$ and thus $\angle CAD \cong \angle FAE$$\overline{CA} \cong \overline{DA} \cong \overline{EA} \cong \overline{FA}$ because they are all radii of the circle. Therefore,  $\Delta CAD \cong \Delta FAE$ by $SAS \cong$ . This means  $\overline{CD} \cong \overline{FE}$ because they are corresponding parts of congruent triangles.

Now prove the converse (statement #2). Assume that  $\overline{CD} \cong \overline{FE}$$\overline{CA} \cong \overline{DA} \cong \overline{EA} \cong \overline{FA}$ because they are all radii of the circle. Therefore,  $\Delta CAD \cong \Delta FAE$ by $SSS \cong$ . This means  $\angle CAD \cong \angle FAE$ because they are corresponding parts of congruent triangles. This implies that the arcs intercepted by these angles are congruent and therefore $\widehat{CD} \cong \widehat{FE}$ .

Example C

$A$  is the center of the circle below. Find the shortest distance from  $A$ to $\overline{EF}$ .

Solution: The shortest distance from  $A$ to  $\overline{EF}$ will be the length of a segment from  $A$ to  $\overline{EF}$ that is perpendicular to $\overline{EF}$ . Because  $A$ is the center of the circle,  $\overline{AC}$ is a radius and thus the length of any radius will be 3 units. Draw two radii from  $A$ to  $E$ and  $A$ to $F$ . Also draw a segment from  $A$ to  $\overline{EF}$ that is perpendicular to $\overline{EF}$ .

$\Delta AFE$  is an equilateral triangle because it has three sides of length 3. This means that  $m \angle AFD=60^\circ$ and  $\Delta AFD$ is a 30-60-90 triangle. According to the 30-60-90 pattern, $AD=\frac{3 \sqrt{3}}{2}$ .

Concept Problem Revisited

Radius  $\overline{AD}$ bisects  $\angle BAC$ in the circle below. How does  $\overline{AD}$ relate to chord $\overline{BC}$ ? Prove your ideas.

Two possible conjectures are that  $\overline{AD}$ bisects  $\overline{BC}$ and  $\overline{AD}$ is perpendicular to $\overline{BC}$ . Both of these conjectures can be proved by first proving that $\Delta ABE \cong \Delta ACE$ .

To prove that $\Delta ABE \cong \Delta ACE$ , first note that  $\overline{AB}$ and  $\overline{AC}$ are both radii of the circle, and therefore $\overline{AB} \cong \overline{AC}$ . Also note that  $\overline{AE} \cong \overline{AE}$ by the reflexive property. Since it was assumed that  $\overline{AD}$ bisects $\angle BAC$ , $\angle BAE \cong \angle CAE$ . $\Delta ABE \cong \Delta ACE$ by $SAS \cong$ .

Because  $\Delta ABE \cong \Delta ACE$$\overline{BE} \cong \overline{EC}$ because they are corresponding parts. Therefore,  $\overline{AD}$ bisects $\overline{BC}$ . Also because  $\Delta ABE \cong \Delta ACE$$\angle CEA \cong \angle BEA$ because they are corresponding parts.  $\angle CEA$ and  $\angle BEA$ are supplementary because they form a line. Therefore,  $m \angle CEA=m \angle BEA=90^\circ$ and  $\overline{AD}$ is perpendicular to $\overline{BC}$ .

This means that if a radius bisects a central angle, then it is the perpendicular bisector of the related chord.

Vocabulary

central angle for a circle is an angle with its vertex at the center of the circle.

An  arc is a portion of a circle. If an arc is less than half a circle it is called a minor arc . If an arc is more than half a circle it is called a major arc .

chord is a segment that connects two points on the circle. If a chord passes through the center of the circle then it is a diameter .

Guided Practice

In the circle below, diameters  $\overline{EB}$ and  $\overline{CF}$ are perpendicular and $m \angle EAD=30^\circ$ .

1. Find $m \widehat{B C}$ .

2. Find $m \widehat{D F}$ .

3. Find $m \angle BAD$ .

1.  $\overline{EB}$ and  $\overline{CF}$ are perpendicular. This means that  $m \angle CAB=90^\circ$ and therefore $m \widehat{B C}=90^\circ$ .

2.  $m \angle EAF=90^\circ$ because  $\angle EAF$ and  $\angle CAB$ are vertical angles and are therefore congruent. This means that  $m \angle DAF=60^\circ$ and therefore $m \widehat{D F}=60^\circ$ .

3.  $m \angle BAF=90^\circ$ because it is supplementary with $\angle BAC$$m \angle BAD=m \angle BAF+m \angle DAF$ . Therefore, $m \angle BAD=90^\circ+60^\circ=150^\circ$ .

Practice

1. Draw an example of a central angle and its intercepted arc .

2. What's the relationship between a central angle and its intercepted arc?

3. Draw an example of a chord.

4. A chord that passes through the center of the circle is called a _______________.

In the circle below,  $\overline{CB}$ and  $\overline{ED}$ are diameters,  $\overline{AG}$ bisects $\angle EAB$$m \angle DAB=50^\circ$ and $m \angle CAF=20^\circ$ . Use this circle for #5-#9.

5. Find $m \widehat{F E}$ .

6. Find $m \widehat{C D}$ .

7. Find $m \angle EAG$ .

8. Find $m \widehat{G B}$ .

9. How is  $\widehat{E B}$ related to $\overline{AG}$ ?

10. Prove that when a radius bisects a chord, it is perpendicular to the chord. Use the picture below and prove that $m \angle AFD=90^\circ$ .

11. Prove that when a radius is perpendicular to a chord it bisects the chord. Use the picture below and prove that $\overline{EF} \cong \overline{FD}$ .

In the circle below with center $A$$AB=12$ and $DE=16$ .

12. Find $DF$ .

13. Find $AC$ .

14. Find $AF$ .

15. Find $CF$ .

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