What if your family went to Washington DC over the summer and saw the White House? The closest you can get to the White House are the walking trails on the far right. You got as close as you could (on the trail) to the fence to take a picture (you were not allowed to walk on the grass). Where else could you have taken your picture from to get the same frame of the White House? Where do you think the best place to stand would be? Your line of sight in the camera is marked in the picture as the grey lines. The white dotted arcs do not actually exist, but were added to help with this problem. After completing this Concept, you will be able to use inscribed angles to answer this question.
Watch This
CK12 Foundation: Chapter9InscribedAnglesinCirclesA
Learn more about inscribed angles by watching the video at this link.
Guidance
An inscribed angle is an angle with its vertex is the circle and its sides contain chords. The intercepted arc is the arc that is on the interior of the inscribed angle and whose endpoints are on the angle. The vertex of an inscribed angle can be anywhere on the circle as long as its sides intersect the circle to form an intercepted arc.
Let's investigate the relationship between the inscribed angle, the central angle and the arc they intercept.
Investigation: Measuring an Inscribed Angle
Tools Needed: pencil, paper, compass, ruler, protractor
1. Draw three circles with three different inscribed angles. For \begin{align*}\bigodot A\end{align*}, make one side of the inscribed angle a diameter, for \begin{align*}\bigodot B\end{align*}, make \begin{align*}B\end{align*} inside the angle and for \begin{align*}\bigodot C\end{align*} make \begin{align*}C\end{align*} outside the angle. Try to make all the angles different sizes.
2. Using your ruler, draw in the corresponding central angle for each angle and label each set of endpoints.
3. Using your protractor measure the six angles and determine if there is a relationship between the central angle, the inscribed angle, and the intercepted arc.
\begin{align*}& m \angle LAM=\underline{\;\;\;\;\;\;\;\;\;} && m \angle NBP= \underline{\;\;\;\;\;\;\;\;\;} && m \angle QCR= \underline{\;\;\;\;\;\;\;\;\;}\\ & m \widehat{LM}= \underline{\;\;\;\;\;\;\;\;\;} && m \widehat{NP}= \underline{\;\;\;\;\;\;\;\;\;} && m \widehat{QR} =\underline{\;\;\;\;\;\;\;\;\;}\\ & m \angle LKM=\underline{\;\;\;\;\;\;\;\;\;} && m \angle NOP=\underline{\;\;\;\;\;\;\;\;\;} && m \angle QSR=\underline{\;\;\;\;\;\;\;\;\;}\end{align*}
Inscribed Angle Theorem: The measure of an inscribed angle is half the measure of its intercepted arc.
In the picture, \begin{align*}m \angle ADC=\frac{1}{2} m \widehat{AC}\end{align*}. If we had drawn in the central angle \begin{align*}\angle ABC\end{align*}, we could also say that \begin{align*}m \angle ADC=\frac{1}{2} m \angle ABC\end{align*} because the measure of the central angle is equal to the measure of the intercepted arc. To prove the Inscribed Angle Theorem, you would need to split it up into three cases, like the three different angles drawn from the Investigation.
Congruent Inscribed Angle Theorem: Inscribed angles that intercept the same arc are congruent.
Inscribed Angle Semicircle Theorem: An angle that intercepts a semicircle is a right angle.
In the Inscribed Angle Semicircle Theorem we could also say that the angle is inscribed in a semicircle. Anytime a right angle is inscribed in a circle, the endpoints of the angle are the endpoints of a diameter. Therefore, the converse of the Inscribed Angle Semicircle Theorem is also true.
Example A
Find \begin{align*}m \widehat{DC}\end{align*} and \begin{align*}m \angle ADB\end{align*}.
From the Inscribed Angle Theorem, \begin{align*}m \widehat{DC} =2 \cdot 45^\circ =90^\circ\end{align*}. \begin{align*}m \angle ADB=\frac{1}{2} \cdot 76^\circ=38^\circ\end{align*}.
Example B
Find \begin{align*}m \angle ADB\end{align*} and \begin{align*}m \angle ACB\end{align*}.
The intercepted arc for both angles is \begin{align*}\widehat{AB}\end{align*}. Therefore, \begin{align*}m \angle ADB=m \angle ACB=\frac{1}{2} \cdot 124^\circ=62^\circ\end{align*}
Example C
Find \begin{align*}m \angle DAB\end{align*} in \begin{align*}\bigodot C\end{align*}.
Because \begin{align*}C\end{align*} is the center, \begin{align*}\overline{DB}\end{align*} is a diameter. Therefore, \begin{align*}\angle DAB\end{align*} inscribes semicircle, or \begin{align*}180^\circ\end{align*}. \begin{align*}m \angle DAB=\frac{1}{2} \cdot 180^\circ=90^\circ\end{align*}.
Watch this video for help with the Examples above.
CK12 Foundation: Chapter9InscribedAnglesinCirclesB
Concept Problem Revisited
You can take the picture from anywhere on the semicircular walking path. The best place to take the picture is subjective, but most would think the pale green frame, straighton, would be the best view.
Vocabulary
A circle is the set of all points that are the same distance away from a specific point, called the center. A radius is the distance from the center to the circle. A chord is a line segment whose endpoints are on a circle. A diameter is a chord that passes through the center of the circle. The length of a diameter is two times the length of a radius. A central angle is an angle formed by two radii and whose vertex is at the center of the circle. An inscribed angle is an angle with its vertex on the circle and whose sides are chords. The intercepted arc is the arc that is inside the inscribed angle and whose endpoints are on the angle.
Guided Practice
Find \begin{align*}m \angle PMN, m \widehat{PN}, m \angle MNP, m \angle LNP\end{align*}, and \begin{align*}m \widehat{LN}\end{align*}.
Answers:
\begin{align*}m \angle PMN=m \angle PLN=68^\circ\end{align*} by the Congruent Inscribed Angle Theorem.
\begin{align*}m \widehat{PN}=2 \cdot 68^\circ=136^\circ\end{align*} from the Inscribed Angle Theorem.
\begin{align*}m \angle MNP=90^\circ\end{align*} by the Inscribed Angle Semicircle Theorem.
\begin{align*}m \angle LNP=\frac{1}{2} \cdot 92^\circ=46^\circ\end{align*} from the Inscribed Angle Theorem.
To find \begin{align*}m \widehat{LN}\end{align*}, we need to find \begin{align*}m \angle LPN\end{align*}. \begin{align*}\angle LPN\end{align*} is the third angle in \begin{align*}\triangle LPN\end{align*}, so \begin{align*}68^\circ+46^\circ+m \angle LPN=180^\circ\end{align*}. \begin{align*}m \angle LPN=66^\circ\end{align*}, which means that \begin{align*}m \widehat{LN}=2 \cdot 66^\circ=132^\circ\end{align*}.
Interactive Practice
Practice
Fill in the blanks.
 An angle inscribed in a ________________ is \begin{align*}90^\circ\end{align*}.
 Two inscribed angles that intercept the same arc are _______________.
 The sides of an inscribed angle are ___________________.
 Draw inscribed angle \begin{align*}\angle JKL\end{align*} in \begin{align*}\bigodot M\end{align*}. Then draw central angle \begin{align*}\angle JML\end{align*}. How do the two angles relate?
Find the value of \begin{align*}x\end{align*} and/or \begin{align*}y\end{align*} in \begin{align*}\bigodot A\end{align*}.
Solve for \begin{align*}x\end{align*}.
 Suppose that \begin{align*}\overline{AB}\end{align*} is a diameter of a circle centered at \begin{align*}O\end{align*}, and \begin{align*}C\end{align*} is any other point on the circle. Draw the line through \begin{align*}O\end{align*} that is parallel to \begin{align*}\overline{AC}\end{align*}, and let \begin{align*}D\end{align*} be the point where it meets \begin{align*}\widehat{BC}\end{align*}. Explain why \begin{align*}D\end{align*} is the midpoint of \begin{align*}\widehat{BC}\end{align*}.
 Fill in the blanks of the Inscribed Angle Theorem proof.
Given: Inscribed \begin{align*}\angle ABC\end{align*} and diameter \begin{align*}\overline{BD}\end{align*}
Prove: \begin{align*}m\angle ABC = \frac{1}{2} m \widehat{AC}\end{align*}
Statement  Reason 

1. Inscribed \begin{align*}\angle ABC\end{align*} and diameter \begin{align*}\overline{BD}\end{align*} \begin{align*}m\angle ABE = x^\circ\end{align*} and \begin{align*}m\angle CBE = y^\circ\end{align*} 

2. \begin{align*}x^\circ + y^\circ = m\angle ABC\end{align*}  
3.  All radii are congruent 
4.  Definition of an isosceles triangle 
5. \begin{align*}m\angle EAB = x^\circ\end{align*} and \begin{align*}m\angle ECB = y^\circ\end{align*}  
6. \begin{align*}m\angle AED = 2x^\circ\end{align*} and \begin{align*}m\angle CED = 2y^\circ\end{align*}  
7. \begin{align*}m\widehat{AD}= 2x^\circ\end{align*} and \begin{align*}m \widehat{DC} = 2y^\circ\end{align*}  
8.  Arc Addition Postulate 
9. \begin{align*}m\widehat{AC} = 2x^\circ + 2y^\circ\end{align*}  
10.  Distributive PoE 
11. \begin{align*}m\widehat{AC} = 2m\angle ABC\end{align*}  
12. \begin{align*}m\angle ABC=\frac{1}{2} m \widehat{AC}\end{align*} 