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^(∘), 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 ∠ 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, BC becomes the diameter of the circle, and the angle cuts off the semicircle. Furthermore, the measure of 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 ∠ BAC is half the measure of BC.

m∠ BAC=1/2mBC

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 ∠ BAD and ∠ DAC be α and β, respectively.

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

Therefore, by the Isosceles Triangle Theorem, the measures of ∠ OBA and ∠ OCA will also be α and β, respectively.

By the Triangle Exterior Angle Theorem, it is recognized that m∠ BOD=2α and m∠ COD = 2β.

The measure of an intercepted arc is the same as the measure of its central angle. Therefore, the measures of BD and CD are 2α and 2β, respectively. mBD=2α and mCD=2β From here, using the Angle Addition Postulate and the Arc Addition Postulate, m∠ BAC and mBC can be written in terms of α and β. m∠ BAC &= α + β [0.7em] mBC &= 2α + 2β [0.7em] mBC &= 2(α+β) Consequentially, the measure of ∠ BAC is half the measure of mBC. The proof of Case II has been completed.

m∠ BAC=1/2mBC

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 ∠ UVT is on the circle O, and the sides of the angle are chords of the circle. Given the measure of 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 ∠ UVT measures 40^(∘).

m∠ UVT = 1/2mUT

Substitute

mUT = 40^(∘)

m∠ UVT = 1/2( 40^(∘))

Multiply

m∠ UVT =20^(∘)

Therefore, ∠ UVT measures 20^(∘).

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, ∠ KLM measures 67^(∘).

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 KM for this example. Therefore, the corresponding central angle is ∠ KOM.

The arc KM is included in ∠ KOM. Therefore, the measure of ∠ KOM is equal to the measure of KM. Using the Inscribed Angle Theorem, the measure of KM can be found.

m∠ KLM = 1/2mKM

Substitute

m∠ KLM = 67^(∘)

67^(∘) = 1/2mKM

▼

Solve for mKM

MultEqn

LHS * 2=RHS* 2

134 = mKM

RearrangeEqn

Rearrange equation

mKM = 134 ^(∘)

Therefore, ∠ KOM also measures 134^(∘).

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, ∠ ADB and ∠ ACB in the above diagram are congruent angles.

∠ ADB ≅ ∠ ACB

Proof

Consider two inscribed angles ∠ ADB and ∠ ACB that intercept the same arc AB in a circle.

By the Inscribed Angle Theorem, the measures of ∠ ADB and ∠ ACB are half the measure of AB. m ∠ ADB = 1/2m AB [1em] m ∠ ACB = 1/2m AB Therefore, by the Transitive Property of Equality, it can be said that these two angles have the same measure. m ∠ ADB = m ∠ ACB Consequently, by the definition of congruent angles, ∠ ADB and ∠ ACB are congruent.

∠ ADB ≅ ∠ ACB

Example

Using the Inscribed Angles of a Circle Theorem

Mark and Jordan have been asked to find the measure of ∠ G.

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

Find the measure of ∠ G.

Hint

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

Solution

The inscribed angles E and H intercept FG.

By the Inscribed Angles of a Circle Theorem, ∠ E is congruent to ∠ H. ∠ E ≅ ∠ H ⇕ m∠ E = m∠ H = 48^(∘) Accordingly, half the measure of ∠ E can be found as 48^(∘)2=24^(∘). Additionally, ∠ F and ∠ G intercept the same arc, EH.

Again by the Inscribed Angles of a Circle Theorem, ∠ F is congruent to ∠ G. Therefore, m∠ G = 25^(∘). ∠ F ≅ ∠ G ⇕ m∠ F =25^(∘) = m∠ G Since m∠ G = 25^(∘), 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 ⊙ Q ⇔ m ⊥ 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 QP. By the Perpendicular Postulate, there is another segment from Q that is perpendicular to m. Let that segment be QT. The goal is to prove that QP must be that segment. The following diagram shows the mentioned characteristics.

Since QT is said to be perpendicular to m, △ QTP is a right triangle. In this case, QP, whose length is r units, is the hypotenuse and, therefore, the longest side. As a result, QT is less than QP. QT < QP There is a part of QT that lies outside of the circle. Let b be the length of that part. The other part of QT is the radius of ⊙ Q, because it is a segment from the center of the circle to a point on the circle.

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

QT < QP

SubstituteExpressions

Substitute expressions

r+b < r

SubIneq

LHS-r< RHS-r

b< 0

Since a length cannot be less than 0, the assumption that QP is not perpendicular to the line m must be false. Therefore, 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 ⊙ Q ⇒ m ⊥ 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 QP at P, and that line m is not tangent to ⊙ Q. In this case, line m intersects ⊙ Q at a second point R.

Since m is said to be perpendicular to QP at P, △ QPR is a right triangle and QR is the hypotenuse. Therefore, QP is less than QR. QP < QR However, it can also be seen that both QP and QR are radii of ⊙ Q. Therefore, they have the same length. QP = QR 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 ⊙ Q. Consequently, m must be a tangent line to the circle. This completes the second part of the proof.

m ⊥ QP ⇒ line m is tangent to ⊙ Q

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

Line m is tangent to ⊙ Q ⇔ m ⊥ QP

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

Example

Using the Tangent to Circle Theorem

In the diagram, l is tangent to the circle at the point A, and DA is a diameter.

If the measure of ∠ ABC is 60^(∘). Find the measure of DC.

Hint

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

Solution

It has been given that m∠ ABC = 60^(∘). By the Tangent to Circle Theorem, l is perpendicular to DA. In other words, m ∠ BAD = 90^(∘).

Since the sum of the measures of the interior angles of a triangle is 180^(∘), m∠ D=30 ^(∘). 90^(∘) + 60^(∘) + m∠ ADB & = 180^(∘) m∠ ADB & = 30 ^(∘) Notice that ∠ ACD is an inscribed angle on the diameter AD of ⊙ E. Therefore, ∠ ACD measures 90^(∘).

Once again, using the Triangle Sum Theorem, the measure of ∠ DAC can be found to be 60^(∘). 90^(∘) + 30^(∘) + m∠ DAC & = 180^(∘) m∠ DAC & = 60 ^(∘) Since DC is associated with ∠ DAC and m∠ DAC=60^(∘), the measure of the arc can be found using the Inscribed Angle Theorem.

m∠ DAC = 1/2 mDC

Substitute

m∠ DAC = 60^(∘)

60^(∘) = 1/2 mDC

▼

Solve for mDC

MultEqn

LHS * 2=RHS* 2

120^(∘) = mDC

RearrangeEqn

Rearrange equation

mDC = 120^(∘)

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, AB and AC are tangents to the circle and ∠ 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^(∘) minus the measure of the central angle that intercepts the same arc.

Considering the above diagram, the following relation holds true.

m∠ ADB = 180^(∘) -m∠ ACB

Proof

By definition, a circumscribed angle is an angle whose sides are tangents to a circle. Since ∠ ADB is a circumscribed angle, DA and DB are tangents to ⊙ C at points A and B, respectively. By the Tangent to Circle Theorem, CA is perpendicular to DA and CB is perpendicular to 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^(∘). Substituting the known angle measures and solving for m∠ ADB will give the desired equation.

m∠ ADB + m∠ DAC + m∠ ACB + m∠ CBD = 360^(∘)

SubstituteII

m∠ DAC= 90^(∘), m∠ CBD= 90^(∘)

m∠ ADB + 90^(∘) + m∠ ACB + 90^(∘) = 360^(∘)

▼

Solve for m∠ ADB

AddTerms

Add terms

m∠ ADB + m∠ ACB + 180 ^(∘) = 360^(∘)

SubEqn

LHS-180^(∘)=RHS-180^(∘)

m∠ ADB+ m∠ ACB = 180 ^(∘)

SubEqn

LHS-m∠ ACB=RHS-m∠ ACB

m∠ ADB = 180 ^(∘) - m∠ 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 ⊙ M are drawn. The measure of ∠ KPL is 40^(∘).

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

Hint

Use the Circumscribed Angle Theorem and Inscribed Angle Theorem.

Solution

The inscribed angle that intercepts KL is ∠ KNL. By the Inscribed Angle Theorem, m∠ KNL is half mKL. m∠ KNL = 1/2 m KL Recall that the measure of an arc is the same as the measure of its corresponding central angle. Therefore, m∠ KML is equal to mKL.

Substitute m ∠ KML into the above equation. m∠ KNL & = 1/2 m KL [0.8em] m∠ KNL & = 1/2 m ∠ KML Now, by the Circumscribed Angle Theorem, m ∠ KML is 180 ^(∘) minus m ∠ KPL. m∠ KNL = 1/2 (180^(∘) -m ∠ KPL) It is given that m ∠ KPL = 40 ^(∘). Substituting that measure into the equation will give the measure of the inscribed angle.

m∠ KNL = 1/2 (180^(∘) -m ∠ KPL)

Substitute

m ∠ KPL= 40 ^(∘)

m∠ KNL = 1/2 (180^(∘) - 40^(∘))

▼

Evaluate right-hand side

SubTerm

Subtract term

m∠ KNL = 1/2 (140^(∘))

Multiply

m∠ KNL = 70^(∘)

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.

	17 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Circle Theorems

Catch-Up and Review

Proof

Hint

Solution

Hint

Solution

Proof

Hint

Solution

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

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

Hint

Solution

Proof

Hint

Solution

Circle Theorems

Recommended exercises