9.4: Inscribed Angles
Learning Objectives
 Find the measure of inscribed angles and the arcs they intercept.
Review Queue
We are going to use #14 from the homework in the previous section.
 What is the measure of each angle in the triangle? How do you know?
 What do you know about the three arcs?
 What is the measure of each arc?
 What is the relationship between the angles in the triangles and the measure of each arc?
Know What? 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.
Inscribed Angles
We have discussed central angles so far in this chapter. We will now introduce another type of angle, the inscribed angle.
Inscribed Angle: An angle with its vertex is the circle and its sides contain chords.
Intercepted Arc: 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.
Now, we will investigation the relationship between the inscribed angle, the central angle and the arc they intercept.
Investigation 94: 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 Investigation 94. We will touch on the algebraic proofs in the review exercises.
Example 1: Find \begin{align*}m \widehat{DC}\end{align*} and \begin{align*}m \angle ADB\end{align*}.
Solution: 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 2: Find \begin{align*}m \angle ADB\end{align*} and \begin{align*}m \angle ACB\end{align*}.
Solution: 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*}
This example leads us to our next theorem.
Theorem 98: Inscribed angles that intercept the same arc are congruent.
To prove Theorem 98, you would use the similar triangles that are formed by the chords.
Example 3: Find \begin{align*}m \angle DAB\end{align*} in \begin{align*}\bigodot C\end{align*}.
Solution: 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*}.
Theorem 99: An angle that intercepts a semicircle is a right angle.
In Theorem 99 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 Theorem 99 is also true.
When the three vertices of a triangle are on the circle, like in Example 3, we say that the triangle is inscribed in the circle. We can also say that the circle is circumscribed around (or about) the triangle. Any polygon can be inscribed in a circle.
Example 4: 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*}.
Solution:
\begin{align*}m \angle PMN=m \angle PLN=68^\circ\end{align*} by Theorem 98.
\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 Theorem 99.
\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*}.
Inscribed Quadrilaterals
The last theorem for this section involves inscribing a quadrilateral in a circle.
Inscribed Polygon: A polygon where every vertex is on a circle.
Note, that not every quadrilateral or polygon can be inscribed in a circle. Inscribed quadrilaterals are also called cyclic quadrilaterals. For these types of quadrilaterals, they must have one special property. We will investigate it here.
Investigation 95: Inscribing Quadrilaterals
Tools Needed: pencil, paper, compass, ruler, colored pencils, scissors
 Draw a circle. Mark the center point \begin{align*}A\end{align*}.
 Place four points on the circle. Connect them to form a quadrilateral. Color the 4 angles of the quadrilateral 4 different colors.
 Cut out the quadrilateral. Then cut the quadrilateral into two triangles, by cutting on a diagonal.
 Line up \begin{align*}\angle B\end{align*} and \begin{align*}\angle D\end{align*} so that they are adjacent angles. What do you notice? What does this show?
This investigation shows that the opposite angles in an inscribed quadrilateral are supplementary. By cutting the quadrilateral in half, through the diagonal, we were able to show that the other two angles (that we did not cut through) formed a linear pair when matched up.
Theorem 910: A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary.
Example 5: Find the value of the missing variables.
a)
b)
Solution:
a) \begin{align*}x+80^\circ=180^\circ\end{align*} by Theorem 910. \begin{align*}x=100^\circ\end{align*}.
\begin{align*}y+71^\circ=180^\circ\end{align*} by Theorem 910. \begin{align*}y=109^\circ\end{align*}.
b) It is easiest to figure out \begin{align*}z\end{align*} first. It is supplementary with \begin{align*}93^\circ\end{align*}, so \begin{align*}z=87^\circ\end{align*}. Second, we can find \begin{align*}x\end{align*}. \begin{align*}x\end{align*} is an inscribed angle that intercepts the arc \begin{align*}58^\circ + 106^\circ = 164^\circ\end{align*}. Therefore, by the Inscribed Angle Theorem, \begin{align*}x=82^\circ\end{align*}. \begin{align*}y\end{align*} is supplementary with \begin{align*}x\end{align*}, so \begin{align*}y=98^\circ\end{align*}.
Example 6: Algebra Connection Find \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the picture below.
Solution: The opposite angles are supplementary. Set up an equation for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.
\begin{align*}(7x+1)^\circ+105^\circ &= 180^\circ && (4y+14)^\circ+(7y+1)^\circ = 180^\circ\\ 7x+106^\circ &= 180^\circ && \qquad \qquad \quad \ 11y+15^\circ = 180^\circ\\ 7x &= 84^\circ && \qquad \qquad \qquad \quad \quad 11y = 165^\circ\\ x &= 12^\circ && \qquad \qquad \qquad \qquad \quad y = 15^\circ\end{align*}
Example 7: Find \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the picture below.
Solution: To find \begin{align*}x\end{align*}, use \begin{align*}\triangle ACE\end{align*}. \begin{align*}m \angle ACE=14^\circ\end{align*} because it is half of \begin{align*}m \widehat{BE}\end{align*} by the Inscribed Angle Theorem.
\begin{align*}32^\circ+14^\circ+x^\circ &= 180^\circ\\ x &= 134^\circ\end{align*}
To find \begin{align*}y\end{align*}, we will use \begin{align*}\triangle EFD\end{align*}.
\begin{align*}m \angle FED &= 180^\circ  x^\circ\\ m \angle FED &= 180^\circ  134^\circ =46^\circ\end{align*}
\begin{align*}m \angle BDE=m \angle ACE=14^\circ\end{align*} because they intercept the same arc, Theorem 98. Let’s solve for \begin{align*}y\end{align*} in \begin{align*}\triangle EFD\end{align*}, using the Triangle Sum Theorem.
\begin{align*}46^\circ+14^\circ+y^\circ &= 180^\circ\\ y &= 120^\circ\end{align*}
Know What? 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.
Review Questions
Quadrilateral \begin{align*}ABCD\end{align*} is inscribed in \begin{align*}\bigodot E\end{align*}. Find:
 \begin{align*}m \angle DBC\end{align*}
 \begin{align*}m \widehat{BC}\end{align*}
 \begin{align*}m \widehat{AB}\end{align*}
 \begin{align*}m \angle ACD\end{align*}
 \begin{align*}m \angle ADC\end{align*}
 \begin{align*}m \angle ACB\end{align*}
Quadrilateral \begin{align*}ABCD\end{align*} is inscribed in \begin{align*}\bigodot E\end{align*}. Find:
 \begin{align*}m \angle A\end{align*}
 \begin{align*}m \angle B\end{align*}
 \begin{align*}m \angle C\end{align*}
 \begin{align*}m \angle D\end{align*}
Find the value of \begin{align*}x\end{align*} and/or \begin{align*}y\end{align*} in \begin{align*}\bigodot A\end{align*}.
Algebra Connection Solve for the variables.
Use the diagram below to find the measures of the indicated angles and arcs in problems 2428.
 \begin{align*}m \angle EBO\end{align*}
 \begin{align*}m \angle EOB\end{align*}
 \begin{align*}m \widehat{BC}\end{align*}
 \begin{align*}m \angle ABO\end{align*}
 \begin{align*}m \angle A\end{align*}
 \begin{align*}m \angle EDC\end{align*}
 Fill in the blanks of one proof of the Inscribed Angle Theorem. 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*} 
 Use the diagram below to write a proof of Theorem 98.
 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*}. Prove that \begin{align*}D\end{align*} is the midpoint of \begin{align*}\widehat{BC}\end{align*}.
Review Queue Answers
 \begin{align*}60^\circ\end{align*}, it is an equilateral triangle.
 They are congruent because the chords are congruent.
 \begin{align*}\frac{360^\circ}{3} = 120^\circ\end{align*}
 The arcs are double each angle.
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