Point \begin{align*}A\end{align*} is the center of the circle below. What can you say about \begin{align*}\Delta CBD\end{align*} ?

#### 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, \begin{align*}\angle CED\end{align*} is an inscribed angle.

Inscribed angles are ** inscribed** in arcs. You can say that \begin{align*}\angle CED\end{align*} is

**\begin{align*}\widehat{CED}\end{align*}. You can also say that \begin{align*}\angle CED\end{align*}**

*inscribed in*

*intercepts**.*

**\begin{align*}\widehat{CD}\end{align*}**

**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 \begin{align*}A\end{align*}. Prove that \begin{align*}m \angle AEC = \frac{1}{2} m \angle CAD = \frac{1}{2} m \widehat{CD}\end{align*}.

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

**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 \begin{align*}m \angle CED = \frac{1}{2}m \widehat{CD}\end{align*}.

**Solution:** Draw a diameter through points \begin{align*}E\end{align*} and \begin{align*}A\end{align*}.

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

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

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

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

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

**Example C**

Find \begin{align*}m \angle CED\end{align*}.

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

**Concept Problem Revisited**

Point \begin{align*}A\end{align*} is the center of the circle below. What can you say about \begin{align*}\Delta CBD\end{align*} ?

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

**. If an arc is more than half a circle it is called a**

*minor arc***.**

*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.

- Prove that \begin{align*}\Delta EFC \sim \Delta BFD\end{align*}.
*Hint: Look for congruent angles!* - Prove that \begin{align*}EF \cdot FD = BF \cdot FC\end{align*}.
- For the circle below, find \begin{align*}BF\end{align*}.

**Answers:**

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

2. Because \begin{align*}\Delta EFC \sim \Delta BFD\end{align*}, its corresponding sides are proportional. This means that \begin{align*}\frac{EF}{BF}=\frac{FC}{FD}\end{align*}. Multiply both sides of the equation by \begin{align*}BF \cdot FD\end{align*} and you have \begin{align*}EF \cdot FD = BF \cdot FC\end{align*}. **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 \begin{align*}15 \cdot 6=BF \cdot 9\end{align*}. This means that \begin{align*}BF=10\end{align*}.

#### Practice

1. How are central angles and inscribed angles related?

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

2. Find \begin{align*}m \widehat{BD}\end{align*}.

3. Find \begin{align*}m \angle ABD\end{align*}.

4. Find \begin{align*}m \widehat{EB}\end{align*}.

5. Find \begin{align*}m \angle ECD\end{align*}.

6. What type of triangle is \begin{align*}\Delta ACD\end{align*} ?

Solve for \begin{align*}x\end{align*} in each circle. If \begin{align*}x\end{align*} is an angle, find the measure of the angle.

7.

8.

9.

10. \begin{align*}m \widehat{DC} = 95^\circ\end{align*}

11.

12.

13.

14.

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