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# 9.2: Inscribed Angles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the measure of inscribed angles and the arcs they intercept.

## Inscribed Angle, Intercepted Arc

An inscribed angle is an angle whose vertex is on the circle and whose sides contain chords of the circle.

• The __________________________ of an inscribed angle is on the circle.
• The sides of an inscribed angle are ________________________ of the circle.

An inscribed angle is said to intercept an arc of the circle. In the diagram above, the intercepted arc is shown between the two sides of the angle.

• The sides of an inscribed angle intercept an _________________ of the circle.

We will prove shortly that the measure of an inscribed angle is half of the measure of the arc it intercepts.

Notice that the vertex of the inscribed angle can be anywhere on the circumference of the circle; it does not need to be diametrically opposite the intercepted arc.

Measure of Inscribed Angle

The measure of a central angle is twice the measure of the inscribed angle that intercepts the same arc.

Proof

COB\begin{align*}\angle COB\end{align*} and CAB\begin{align*}\angle CAB\end{align*} both intercept CBˆ\begin{align*}\widehat{CB}\end{align*}. COB\begin{align*}\angle COB\end{align*} is a central angle and CAB\begin{align*}\angle CAB\end{align*} is an inscribed angle.

We draw the diameter of the circle through points A,O,\begin{align*}A, O,\end{align*} and D,\begin{align*}D,\end{align*} and let mCAO=x\begin{align*}m \angle CAO = x^\circ\end{align*} and mBAO=y\begin{align*}m \angle BAO = y^\circ\end{align*}, as labeled in the diagram above.

We see that ΔAOC\begin{align*}\Delta AOC\end{align*} is isosceles because AO¯¯¯¯¯¯¯¯\begin{align*}\overline{AO}\end{align*} and AC¯¯¯¯¯¯¯¯\begin{align*}\overline{AC}\end{align*} are radii of the circle and are therefore congruent. You can make congruent marks on these segments in the diagram.

From this we can conclude that mACO=x\begin{align*}m \angle ACO = x^\circ\end{align*}.

Similarly, we can conclude that mABO=y\begin{align*}m \angle ABO = y^\circ\end{align*}.

We use the property that the sum of angles inside a triangle equals 180\begin{align*}180^\circ\end{align*} to find that: mAOC=1802x\begin{align*}m \angle AOC = 180^\circ - 2x^\circ\end{align*} and mAOB=1802y\begin{align*}m \angle AOB = 180^\circ - 2y^\circ\end{align*}.

Then, mCOD=180mAOC=180(1802x)=2x\begin{align*}m \angle COD = 180^\circ -m \angle AOC = 180^\circ -(180^\circ -2x^\circ) = 2x^\circ\end{align*} and mBOD=180mAOB=180(1802y)=2y\begin{align*}m \angle BOD = 180^\circ -m \angle AOB = 180^\circ-(180^\circ - 2y^\circ) = 2y^\circ\end{align*}

Therefore: mCOB=2x+2y=2(x+y)=2(mCAB)\begin{align*}m \angle COB = 2x^\circ + 2y^\circ = 2(x^\circ + y^\circ) = 2(m \angle CAB)\end{align*}

This proves that the measure of a central angle is two times the measure of the inscribed angle that intercepts the same arc.

Another way of looking at this is that the measure of the inscribed angle is half the measure of the __________________________ angle that intercepts the same arc.

Do you remember what a corollary is?

A corollary is a statement that follows a theorem.

If we know that a theorem is true, we can often make other statements that are true. These are called corollaries.

• A corollary is a statement that follows a ___________________________.

## Inscribed Angle Corollaries

The theorem we just proved has several corollaries, which you can prove on your own:

1. Inscribed angles intercepting the same arc are congruent.
2. Opposite angles of an inscribed quadrilateral are supplementary.
3. An angle inscribed in a semicircle is a right angle.
4. An inscribed right angle intercepts a semicircle.
• A _________________________ angle is inscribed in a semicircle.
• Opposite angles of an inscribed quadrilateral are __________________________.

Example 1

Find the angle marked x\begin{align*}x\end{align*} in the circle:

mEOB\begin{align*}m \angle EOB\end{align*} is twice the measure of the inscribed angle OAB\begin{align*}\angle OAB\end{align*} because EOB\begin{align*}\angle EOB\end{align*} is a central angle and both angles intercept the same arc, EBˆ\begin{align*}\widehat{EB}\end{align*}.

So mOAB=35\begin{align*}m \angle OAB = 35^\circ\end{align*}.

This means that the angle marked x=18035=145\begin{align*}x = 180^\circ - 35^\circ = 145^\circ\end{align*}.

Example 2

Find the angle marked x\begin{align*}x\end{align*} in the circle:

We know that an arc has the same measure as the central angle that intercepts it and double the measure of the inscribed angle that intercepts it.

• An arc has the _______________ measure as the central angle that intercepts it.
• An arc has ______________ the measure of the inscribed angle that intercepts it.

Therefore,

mADCˆsomABCˆandmABˆsomBCˆ=295=190=360190=170=235=70=17070=100\begin{align*}m \widehat{ADC} & = 2 \cdot 95^\circ = 190^\circ\\ \text{so} \quad m \widehat{ABC} & = 360^\circ - 190^\circ = 170^\circ\\ \text{and} \quad m \widehat{AB} & = 2 \cdot 35^\circ = 70^\circ\\ \text{so} \quad m \widehat{BC} & = 170^\circ - 70^\circ = 100^\circ\end{align*}

mBDC\begin{align*}m \angle BDC\end{align*} (which is the angle marked x\begin{align*}x\end{align*}) is the inscribed angle that intercepts the arc BCˆ\begin{align*}\widehat{BC}\end{align*} so \begin{align*}x\end{align*} must be half of the measure of the arc:

\begin{align*}m \angle BDC & = \frac{1}{2} ( m \widehat{BC})\\ m \angle BDC & = \frac{1}{2} ( \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}) = \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\end{align*}

Therefore, \begin{align*}x = 50^\circ\end{align*}.

Example 3

Find the angles marked \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in the circle:

First we use \begin{align*}\Delta ABD\end{align*} to find the measure of angle \begin{align*}x\end{align*}, since all three angles in the triangle add up to \begin{align*}180^\circ\end{align*}:

\begin{align*}x + 15^\circ + 32^\circ & = 180^\circ\\ x + 47^\circ & = 180^\circ\\ x & = 133^\circ\end{align*}

So \begin{align*}m \angle CBD = 180^\circ - 133^\circ = 47^\circ\end{align*} because it is a linear pair with \begin{align*}\angle ABD\end{align*} or \begin{align*}x\end{align*}.

\begin{align*}m \angle BCE = m \angle BDE\end{align*} because they are both inscribed angles and both intercept the same arc, \begin{align*}\widehat{EB}\end{align*}. This makes \begin{align*}m \angle BCE = 15^\circ\end{align*}.

Using \begin{align*}\Delta BFC\end{align*} to find the measure of angle \begin{align*}y\end{align*}:

\begin{align*}y + 47^\circ + 15^\circ & = 180^\circ\\ y + 62^\circ & = 180^\circ\\ y & = 118^\circ\end{align*}

Therefore, \begin{align*}x = 133^\circ\end{align*} and \begin{align*}y = 118^\circ\end{align*}.

1. What was the most important thing you learned about inscribed angles in this lesson?

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2. How does a central angle relate to the inscribed angle that shares the same intercepted arc?

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