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Inscribed Angles in Circles

Angle with its vertex on a circle and sides that contain chords.

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Inscribed Angles

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

Inscribed Angles

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

Proving the Inscribed Angle Theorem  

1. 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*}.

\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.

2. Use the result from the previous problem to prove that \begin{align*}m \angle CED = \frac{1}{2}m \widehat{CD}\end{align*}.

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

From #1, 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.

Now let's find the measure of an angle. 

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

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*}.

Examples

Example 1

Earlier, you were asked what can you say about \begin{align*}\Delta CBD\end{align*}

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.

 

In #2-#3, you will use the circle below to prove that when two chords intersect inside a circle, the products of their segments are equal.

Example 2

Prove that \begin{align*}\Delta EFC \sim \Delta BFD\end{align*}. Hint: Look for congruent angles!

 \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*}.

Example 3

Prove that \begin{align*}EF \cdot FD = BF \cdot FC\end{align*}.

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.

 

Example 4

For the circle below, find \begin{align*}BF\end{align*}.

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*}.

Review

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*}.

 

Review (Answers)

To see the Review answers, open this  PDF file and look for section 8.4. 

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Vocabulary

Arc

An arc is a section of the circumference of a circle.

Intercepts

The intercepts of a curve are the locations where the curve intersects the x and y axes. An x intercept is a point at which the curve intersects the x-axis. A y intercept is a point at which the curve intersects the y-axis.

Inscribed Angle Theorem

The Inscribed Angle Theorem states that the measure of an inscribed angle is half the measure of its intercepted arc.

Semicircle Theorem

The Semicircle Theorem states that any time a right angle is inscribed in a circle, the endpoints of the angle are the endpoints of a diameter and the diameter is the hypotenuse.

Inscribed Angle

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.

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