Exploring Circle Theorems: Outside Angle Theorem and Inscribed Angle Theorem

An angle can be constructed by the lines and line segments that intersect a circle. In this lesson, the angles related to circles and their properties will be explored.

Catch-Up and Review

Here is some recommended reading before getting started with this lesson.

Angle
Circle
Radius
Chord
Secant

Explore

Investigating Inscribed Angles of a Circle

Consider a circle with a radius of $4$ units. An angle whose sides are two chords of the circle is formed as shown. Move the points $B$ and $C$ along the circle so that the angle is a right angle.

Think about the following questions.

When the measure of the angle is $90\Deg,$ what can the chord joining $B$ and $C$ be called?
Does the arc that the angle intercepts have a special name? If yes, what is it?

Explore

Inscribed and Central Angles of a Circle

On the circle above, construct an angle with a vertex at the center of the circle. The angle being constructed should also cut off the same arc. In other words, construct the corresponding central angle.

Recall that the measure of an arc is the measure of the central angle that includes the arc. Now, move the points and make conjectures that explain the relationship between the measures of angles and arcs.

Discussion

Inscribed Angle of a Circle

In the circles shown previously, the angle formed by two chords $\angle BAC$ is called an inscribed angle.

An inscribed angle is an angle whose vertex lies on the circumference of a circle and whose sides are secant to the circle.

Discussion

Inscribed Angle Theorem

Observe the measures of the angle and the arc intercepted by the angle. Start by moving $B$ so that $B,$ $O,$ and $A$ are collinear. Then, move it once again so that $B,$ $O,$ and $C$ are collinear.

As can be seen, when $B,$ $O,$ and $C$ are collinear, $\Seg{BC}$ becomes the diameter of the circle, and the angle cuts off the semicircle. Furthermore, the measure of $\Arc{BC}$ is twice the measure of an inscribed angle that intercepts it. This statement can be restated as a theorem.

The measure of an inscribed angle is half the measure of its intercepted arc.

In this figure, the measure of $\angle BAC$ is half the measure of $\overgroup{BC}.$

$m\angle BAC=\dfrac{1}{2}m\overgroup{BC}$

Proof

An inscribed angle can be created in three different ways depending on where the angle is located in relation to the center of the circle.

Consider Case II. A diameter can be drawn from $A$ through $O,$ dividing the inscribed angle into two.

Let the measures of $\angle BAD$ and $\angle DAC$ be $\alpha$ and $\beta,$ respectively.

Because the radii of a circle are all congruent, two isosceles triangles can be obtained by drawing $\Seg{OB}$ and $\Seg{OC}.$

Therefore, by the Isosceles Triangle Theorem, the measures of $\angle OBA$ and $\angle OCA$ will also be $\alpha$ and $\beta,$ respectively.

By the Triangle Exterior Angle Theorem, it is recognized that $m\angle BOD=2\alpha$ and $m\angle COD = 2\beta.$

The measure of an intercepted arc is the same as the measure of its central angle. Therefore, the measures of $\overgroup{BD}$ and $\overgroup{CD}$ are $2\alpha$ and $2\beta,$ respectively. \begin{gathered} m\overgroup{BD}=2\alpha \quad \text{and} \quad m\overgroup{CD}=2\beta \end{gathered} From here, using the Angle Addition Postulate and the Arc Addition Postulate, $m\angle BAC$ and $m\overgroup{BC}$ can be written in terms of $\alpha$ and $\beta.$ \begin{aligned} m\angle BAC &= \alpha + \beta\\[0.7em] m\overgroup{BC} &= 2\alpha + 2\beta \\[0.7em] m\overgroup{BC} &= 2\left(\alpha+\beta\right) \end{aligned} Consequentially, the measure of $\angle BAC$ is half the measure of $m\overgroup{BC}.$ The proof of Case II has been completed.

$m\angle BAC=\dfrac{1}{2}m\overgroup{BC}$

By applying similar logic as the procedure above, Case I and Case III can be proven.

Example

Finding the Inscribed Angle of a Circle

In the diagram, the vertex of $\angle UVT$ is on the circle $O,$ and the sides of the angle are chords of the circle. Given the measure of $\Arc{UT}$, find the measure of the angle.

Write the answer without the degree symbol.

Hint

The Inscribed Angle Theorem can be used to find the measure of the angle.

Solution

The angle shown in the diagram fits the definition of an inscribed angle. For this reason, the measure of the angle can be found using the Inscribed Angle Theorem. The theorem states that the measure of an inscribed angle is half the measure of its intercepted arc.

The arc intercepted by $\angle UVT$ measures $40\Deg.$

$m\angle UVT = \dfrac{1}{2}m\Arc{UT}$

Substitute

\Substitute{m\Arc{UT} }{\colIII{40\Deg}}

$m\angle UVT = \dfrac{1}{2}(\colIII{40\Deg})$

Multiply

\Multiply

$m\angle UVT =20\Deg$

Therefore, $\angle UVT$ measures $20\Deg.$

Pop Quiz

Practice Finding Inscribed Angles

Find the measure of the inscribed angle in the circle.

Example

Finding the Central Angle of a Circle

Similarly, given the measure of an inscribed angle, the measure of its corresponding central angle can be found using the Inscribed Angle Theorem. This can be done because the measure of the central angle is the same as the measure of the arc that the central angle cuts off.

In the circle, $\angle KLM$ measures $67\Deg.$

Find the measure of the corresponding central angle.

Hint

Start by drawing the corresponding central angle.

Solution

Recall that a central angle is an angle whose vertex lies at the center of the circle. Additionally, the inscribed angle and its corresponding central angle intercept the same arc $\Arc{KM}$ for this example. Therefore, the corresponding central angle is $\angle KOM.$

The arc $\Arc{KM}$ is included in $\angle KOM.$ Therefore, the measure of $\angle KOM$ is equal to the measure of $\Arc{KM}.$ Using the Inscribed Angle Theorem, the measure of $\Arc{KM}$ can be found.

$m\angle KLM = \dfrac{1}{2}m\Arc{KM}$

Substitute

\Substitute{m\angle KLM }{67\Deg}

$\col{67\Deg} = \dfrac{1}{2}m\Arc{KM}$

▼

\MMSolve{m\Arc{KM} }

MultEqn

\MultEqn{2}

$134 = m\Arc{KM}$

RearrangeEqn

\RearrangeEqn

$m\Arc{KM} = 134 \Deg$

Therefore, $\angle KOM$ also measures $134\Deg.$

Pop Quiz

Practice Finding Central Angles

Given the measure of an inscribed angle, find the measure of its corresponding central angle.

Discussion

Inscribed Angles of a Circle Theorem

Up to now, the relationship between inscribed angles and their corresponding central angles has been discussed. Now the relationship between two inscribed angles that intercept the same arc will be investigated.

As can be observed, the angles are congruent, so long as they intercept the same arc.

If two inscribed angles of a circle intercept the same arc, then they are congruent.

By this theorem, $\angle ADB$ and $\angle ACB$ in the above diagram are congruent angles.

$\angle ADB \cong \angle ACB$

Proof

Consider two inscribed angles $\col{\angle ADB}$ and $\colIV{\angle ACB}$ that intercept the same arc $\colII{\Arc{AB}}$ in a circle.

By the Inscribed Angle Theorem, the measures of $\angle ADB$ and $\angle ACB$ are half the measure of $\Arc{AB}.$ \begin{gathered} m\col{\angle ADB} = \dfrac{1}{2}m\colII{\Arc{AB}}\\[1em] m\colIV{\angle ACB} = \dfrac{1}{2}m\colII{\Arc{AB}} \end{gathered} Therefore, by the Transitive Property of Equality, it can be said that these two angles have the same measure. \begin{gathered} m\col{\angle ADB} = m\colIV{\angle ACB} \end{gathered} Consequently, by the definition of congruent angles, $\angle ADB$ and $\angle ACB$ are congruent.

$\angle ADB \cong \angle ACB$

Example

Using the Inscribed Angles of a Circle Theorem

Mark and Jordan have been asked to find the measure of $\angle G.$

Mark claims that the measure of $\angle G$ is the same as the measure of $\angle E.$ Jordan, however, thinks that its measure is half the measure of $\angle E.$ Who is correct?

Find the measure of $ \angle G.$

Hint

Determine which angles intercept the same arc. Use the Inscribed Angles of a Circle Theorem to find $m\angle G.$

Solution

The inscribed angles $E$ and $H$ intercept $\Arc{FG}.$

By the Inscribed Angles of a Circle Theorem, $\angle E$ is congruent to $\angle H.$ \begin{gathered} \angle E \cong \angle H \\ \Updownarrow \\ m\angle E = m\angle H = 48\Deg \end{gathered} Accordingly, half the measure of $\angle E$ can be found as $\frac{48\Deg}{2}=24\Deg.$ Additionally, $\angle F$ and $\angle G$ intercept the same arc, $\Arc{EH}.$

Again by the Inscribed Angles of a Circle Theorem, $\angle F$ is congruent to $\angle G.$ Therefore, $m\angle G = 25\Deg.$ \begin{gathered} \angle F \cong \angle G \\ \Updownarrow \\ m\angle F =25\Deg = m\angle G \end{gathered} Since $m\angle G = 25\Deg,$ neither Mark's claim nor Jordan's claim is correct.

Discussion

Constructing Angles Outside of a Circle

Inscribed angles, or the central angles, are not the only angles related to circles. In the next part, the angles constructed outside the circles will be examined. To construct an angle outside a circle, tangents can be used.

Concept

Tangent

A tangent is a line, segment, or ray that intersects a circle at exactly one point. The point is called point of tangency, and the line, segment, or ray is said to be tangent to the circle.

Rule

Tangent to a Circle Theorem

A line is a tangent to a circle if and only if the line is perpendicular to the endpoint of a radius on the circle's circumference.

Based on the diagram, the following relation holds true.

Line $m$ is tangent to $\odot Q$ $ \quad \Leftrightarrow\quad $ $m \perp \Seg{QP}$

Proof

The theorem will be proven in two parts as it is a biconditional statement. Each will be proven by using an indirect proof.

Part I: If a Line Is a Tangent to a Circle, Then It Is Perpendicular to the Endpoint of a Radius on the Circle’s Circumference

Assume that line $m$ is tangent to the circle centered at $Q$ and not perpendicular to $\Seg{QP}.$ By the Perpendicular Postulate, there is another segment from $Q$ that is perpendicular to $m.$ Let that segment be $\colV{\overline{QT}}.$ The goal is to prove that $\overline{QP}$ must be that segment. The following diagram shows the mentioned characteristics.

Since $\overline{QT}$ is said to be perpendicular to $m,$ $\triangle QTP$ is a right triangle. In this case, $\Seg{QP},$ whose length is $r$ units, is the hypotenuse and, therefore, the longest side. As a result, $QT$ is less than $QP.$ \begin{aligned} \colV{QT} < \colII{QP} \end{aligned} There is a part of $\Seg{QT}$ that lies outside of the circle. Let $b$ be the length of that part. The other part of $\Seg{QT}$ is the radius of $\odot Q,$ because it is a segment from the center of the circle to a point on the circle.

By the Segment Addition Postulate, $\colV{QT}$ is equal to the sum of $\colIII{r}$ and $\col{b}.$ This expression, together with $QP=r,$ can be substituted into the aforementioned inequality.

$QT < QP$

SubstituteExpressions

\SubstituteExpressions

$\colV{r+b} < \colIII{r}$

SubIneq

\SubIneq{< }{r}

$b< 0$

Since a length cannot be less than $0,$ the assumption that $\Seg{QP}$ is not perpendicular to the line $m$ must be false. Therefore, $\Seg{QP}$ is perpendicular to the tangent $m$ at its endpoint on the circle. This concludes the first part of the proof.

Line $m$ is tangent to $\odot Q$ $ \quad \Rightarrow\quad $ $m \perp \Seg{QP}$

Part II: If a Line Is Perpendicular to the Endpoint of a Radius on a Circle’s Circumference, Then It Is a Tangent to the Circle

For the second part, it will be assumed that $m$ is perpendicular to the radius $\Seg{QP}$ at $P,$ and that line $m$ is not tangent to $\odot Q.$ In this case, line $m$ intersects $\odot Q$ at a second point $R.$

Since $m$ is said to be perpendicular to $\Seg{QP}$ at $P,$ $\triangle QPR$ is a right triangle and $\Seg{QR}$ is the hypotenuse. Therefore, $QP$ is less than $QR.$ \begin{gathered} \colIII{QP} < \col{QR} \end{gathered} However, it can also be seen that both $\Seg{QP}$ and $\Seg{QR}$ are radii of $\odot Q.$ Therefore, they have the same length. \begin{gathered}

\colIII{QP} = \col{QR}

\end{gathered} These two statements are contradictory to each other. Therefore, both cannot be true. The contradiction came from supposing that line $m$ was not tangent to $\odot Q.$ Consequently, $m$ must be a tangent line to the circle. This completes the second part of the proof.

$m \perp \Seg{QP}$ $ \quad \Rightarrow\quad $ line $m$ is tangent to $\odot Q$

Having proven both parts, the proof of the biconditional statement of the theorem is now complete.

Line $m$ is tangent to $\odot Q$ $ \quad \Leftrightarrow\quad $ $m \perp \Seg{QP}$

The method used to prove the theorem is called indirect proof.

Example

Using the Tangent to Circle Theorem

In the diagram, $\ell$ is tangent to the circle at the point $A,$ and $\Seg{DA}$ is a diameter.

If the measure of $\angle ABC$ is $60\Deg.$ Find the measure of $\Arc{DC}.$

Hint

A tangent is perpendicular to the radius of a circle through the point of tangency.

Solution

It has been given that $m\angle ABC = 60\Deg.$ By the Tangent to Circle Theorem, $\ell$ is perpendicular to $\Seg{DA}.$ In other words, $m \angle BAD = 90\Deg.$

Since the sum of the measures of the interior angles of a triangle is $180\Deg,$ $m\angle D=30 \Deg.$ \begin{aligned} 90\Deg + 60\Deg + m\angle ADB & = 180\Deg \\ m\angle ADB & = 30 \Deg \end{aligned} Notice that $\angle ACD$ is an inscribed angle on the diameter $\Seg{AD}$ of $\odot E.$ Therefore, $\angle ACD$ measures $90\Deg.$

Once again, using the Triangle Sum Theorem, the measure of $\angle DAC$ can be found to be $60\Deg.$ \begin{aligned} 90\Deg + 30\Deg + m\angle DAC & = 180\Deg \\ m\angle DAC & = 60 \Deg \end{aligned} Since $\Arc{DC}$ is associated with $\angle DAC$ and $m\angle DAC=60\Deg,$ the measure of the arc can be found using the Inscribed Angle Theorem.

$m\angle DAC = \dfrac{1}{2} m\Arc{DC}$

Substitute

\Substitute{ m\angle DAC }{60\Deg}

$\col{60\Deg }= \dfrac{1}{2} m\Arc{DC}$

▼

\MMSolve{m\Arc{DC}}

MultEqn

\MultEqn{2}

$120\Deg = m\Arc{DC}$

RearrangeEqn

\RearrangeEqn

$m\Arc{DC} = 120\Deg$

Discussion

Circumscribed Angle Theorem

When two tangents of a circle intersect at an exterior point, the angle formed is called a circumscribed angle.

In the diagram above, $\overleftrightarrow{AB}$ and $\overleftrightarrow{AC}$ are tangents to the circle and $\angle BAC$ is a circumscribed angle.

A circumscribed angle is supplementary to the central angle it cuts off.

The measure of a circumscribed angle is equal to $180\Deg$ minus the measure of the central angle that intercepts the same arc.

Considering the above diagram, the following relation holds true.

$m\angle ADB = 180\Deg -m\angle ACB$

Proof

By definition, a circumscribed angle is an angle whose sides are tangents to a circle. Since $\angle ADB$ is a circumscribed angle, $\Ray{DA}$ and $\Ray{DB}$ are tangents to $\odot C$ at points $A$ and $B,$ respectively. By the Tangent to Circle Theorem, $\Seg{CA}$ is perpendicular to $\Ray{DA}$ and $\Seg{CB}$ is perpendicular to $\Ray{DB}.$

Notice that $ADBC$ is a quadrilateral and two of its angles are right angles. Recall that the sum of all of the angles in a quadrilateral is $360\Deg.$ Substituting the known angle measures and solving for $m\angle ADB$ will give the desired equation.

$m\angle ADB + m\angle DAC + m\angle ACB + m\angle CBD = 360\Deg$

SubstituteII

\SubstituteII{m\angle DAC}{90\Deg}{m\angle CBD}{90\Deg}

$m\angle ADB + \col{90\Deg} + m\angle ACB + \colII{90\Deg} = 360\Deg$

▼

\MMSolve{m\angle ADB}

AddTerms

\AddTerms

$m\angle ADB + m\angle ACB + 180 \Deg = 360\Deg$

SubEqn

\SubEqn{180\Deg}

$m\angle ADB+ m\angle ACB = 180 \Deg$

SubEqn

\SubEqn{m\angle ACB}

$m\angle ADB = 180 \Deg - m\angle ACB$

This completes the proof.

Pop Quiz

Practice Using the Circumscribed Angle Theorem

Find the measure of the central angle.

Example

Using Circle Theorems to Solve Problems

The following example involving circumscribed angles and inscribed angles could require the use of the previously learned theorems.

Two tangents from $P$ to $\odot M$ are drawn. The measure of $\angle KPL$ is $40\Deg.$

Find the measure of the inscribed angle that intercepts the same arc as $\angle KPL?$

Hint

Use the Circumscribed Angle Theorem and Inscribed Angle Theorem.

Solution

The inscribed angle that intercepts $\Arc{KL}$ is $\angle KNL.$ By the Inscribed Angle Theorem, $m\angle KNL$ is half $m\Arc{KL}.$ \begin{aligned} m\angle KNL = \dfrac{1}{2} m \Arc{KL} \end{aligned} Recall that the measure of an arc is the same as the measure of its corresponding central angle. Therefore, $m\angle KML$ is equal to $m\Arc{KL}.$

Substitute $m \angle KML$ into the above equation. \begin{aligned} m\angle KNL & = \dfrac{1}{2} \colIII{m \Arc{KL}} \\[0.8em] m\angle KNL & = \dfrac{1}{2} \colIII{m \angle KML} \end{aligned} Now, by the Circumscribed Angle Theorem, $m \angle KML$ is $180 \Deg$ minus $m \angle KPL.$ \begin{aligned} m\angle KNL = \dfrac{1}{2} (180\Deg -m \angle KPL) \end{aligned} It is given that $m \angle KPL = 40 \Deg.$ Substituting that measure into the equation will give the measure of the inscribed angle.

$m\angle KNL = \dfrac{1}{2} (180\Deg -m \angle KPL)$

Substitute

\Substitute{m \angle KPL}{40 \Deg}

$m\angle KNL = \dfrac{1}{2} (180\Deg - \col{40\Deg})$

▼

\MMEvalRHS

SubTerm

\SubTerm

$m\angle KNL = \dfrac{1}{2} (140\Deg)$

Multiply

\Multiply

$m\angle KNL = 70\Deg$

Closure

Summarizing Angle Theorems Related to Circles

This lesson defined three angles related to circles as well as the relationships between these angles. The diagram below shows the definitions and the main theorems of this lesson.