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Inscribed Angles in Circles

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Point  A is the center of the circle below. What can you say about \Delta CBD ?

Watch This

https://www.youtube.com/watch?v=aRGI4lafk8Y Brightstorm: Inscribed Angles

Guidance

An inscribed angle is an angle with its vertex on the circle. The sides of an inscribed angle will be chords of the circle. Below,  \angle CED is an inscribed angle.

Inscribed angles are inscribed in arcs. You can say that  \angle  CED is inscribed in \widehat{CED} . You can also say that \angle  CED   intercepts \widehat{CD} .

The measure of an inscribed angle is always half the measure of the arc it intercepts. You will prove and then use this theorem in the examples.

Example A

Consider the circle below with center at point A . Prove that m \angle AEC = \frac{1}{2}  m \angle CAD = \frac{1}{2} m \widehat{CD} .

Solution:   \overline{CA} and  \overline{EA} are both radii of the circle and are therefore congruent. This means that  \Delta CAE is isosceles. The base angles of isosceles triangles are congruent so m \angle ACE = m \angle AEC . \angle CAD is an exterior angle of \Delta CAE , so its measure must be the sum of the measures of the remote interior angles. Therefore, m \angle CAD = m \angle ACE + m \angle AEC . By substitution, m \angle CAD = m \angle AEC + m \angle AEC . This means  m \angle CAD = 2m \angle AEC and \frac{1}{2} m \angle CAD = m \angle AEC . A central angle has the same measure as its intercepted arc, so m \angle CAD = m \widehat{CD} . Therefore by substitution, \frac{1}{2} m \widehat{CD} = m \angle AEC .

This proves that when an inscribed angle passes through the center of a circle, its measure is half the measure of the arc it intercepts. In Example B, you will extend this proof.

Example B

Use the result from Example A to prove that m \angle CED = \frac{1}{2}m \widehat{CD} .

Solution: Draw a diameter through points  E and A .

From Example A, you know that m \angle FEC = \frac{1}{2} m \angle FAC .

You also know that m \angle FED = \frac{1}{2} m \angle FAD .

Because m \angle FED = m \angle FEC + m \angle CED , by substitution \frac{1}{2} m \angle FAD = \frac{1}{2} m \angle FAC + m \angle CED . This means m \angle CED = \frac{1}{2} (m \angle FAD - m \angle FAC) . m \angle FAD - m \angle FAC = m \angle CAD , so m \angle CED = \frac{1}{2} m \angle CAD .

Because m \angle CAD = m \widehat{CD}, m \angle CED = \frac{1}{2} m \widehat{CD} .

This proves in general that the measure of an inscribed angle is half the measure of its intercepted arc.

Example C

Find m \angle CED .

Solution: Notice that both  \angle CED and  \angle CBD intercept \widehat{CD} . This means that their measures are both half the measure of \widehat{CD} , so their measures must be equal. m \angle CED=38^\circ .

Concept Problem Revisited

Point  A is the center of the circle below. What can you say about \Delta CBD ?

If point  A is the center of the circle, then  \overline{CB} is a diameter and it divides the circle into two equal halves. This means that m\widehat{CB}=180^\circ\angle CDB is an inscribed angle that intercepts \widehat{CB} , so its measure must be half the measure of \widehat{CB} . Therefore,  m \angle CDB=90^\circ and   \Delta CBD is a right triangle.

In general, if a triangle is inscribed in a semicircle then it is a right triangle.

Vocabulary

A central angle for a circle is an angle with its vertex at the center of the circle. The measure of a central angle is equal to the measure of its intercepted arc.

An inscribed angle is an angle with its vertex on the circle. The measure of an inscribed angle is half the measure of its intercepted arc.

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 .

A 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 #1-#2, you will use the circle below to prove that when two chords intersect inside a circle, the products of their segments are equal.

  1. Prove that \Delta EFC \sim \Delta BFD . Hint: Look for congruent angles!
  2. Prove that EF \cdot FD = BF \cdot FC .
  3. For the circle below, find BF .

Answers:

1.  \angle CED \cong  \angle CBD because both are inscribed angles that intercept the same arc (\widehat{CD})\angle EFC \cong \angle BFD because they are vertical angles. Therefore, \Delta EFC \sim  \Delta BFD by AA \sim .

2. Because \Delta EFC \sim \Delta BFD , its corresponding sides are proportional. This means that \frac{EF}{BF}=\frac{FC}{FD} . Multiply both sides of the equation by BF \cdot FD and you have EF \cdot FD = BF \cdot FC . This proves that in general, when two chords intersect inside a circle, the products of their segments are equal.

3. Based on the result of #2, you know that 15 \cdot 6=BF \cdot 9 . This means that BF=10 .

Practice

1. How are central angles and inscribed angles related?

In the picture below, \overline{BD} \ \| \ \overline{EC} . Use the picture below for #2-#6.

2. Find m \widehat{BD} .

3. Find m \angle ABD .

4. Find m \widehat{EB} .

5. Find m \angle ECD .

6. What type of triangle is \Delta ACD ?

Solve for  x in each circle. If  x is an angle, find the measure of the angle.

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10. m \widehat{DC} = 95^\circ

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15. In the picture below,  \overline{BD}\ || \ \overline{EC} . Prove that \widehat{BC} \cong \widehat{ED} .

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