Inscribed Angles

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Learning goals

Identify the intercepted arc that excludes an inscribed angle's vertex
Prove the inscribed-angle theorem for all three positions of the center
Compare inscribed angles with vertices on the same or opposite arcs
Apply the semicircle right-angle theorem and its converse
Choose the correct arc for an angle between a tangent ray and a chord

Choose the arc away from the vertex

An inscribed angle has its vertex on a circle and its sides along chords to two other, distinct points on that circle. For $\angle AVB$ , the intercepted arc is the arc from $A$ to $B$ that does not contain $V$ .

Recall that an arc’s degree measure comes from its central angle; this is a measure of turning, not a length. Circles, Chords, and Arcs gives the arc notation and addition rules.

An inscribed angle measures half its intercepted arc. An arc of $82^\circ$ gives an angle of $41^\circ$ . Equal radii create an isosceles triangle whose two equal angles explain the factor of two, as we will prove for every position of the center.

Worked example 1 An intercepted arc can be major

Problem. Points $A,V,B,R$ lie on a circle in that order. The minor arc $AVB$ measures $94^\circ$ . Find $m\angle AVB$ and name its intercepted arc.

Points A, V, B and R occur in that circle order; the angle has vertex V.

Solution. The intercepted arc is the major arc $ARB$ , because the other arc contains $V$ .

\begin{aligned} m(\widehat{ARB})&=360^\circ-94^\circ,\\ m(\widehat{ARB})&=266^\circ. \end{aligned}

Half of this arc gives the angle.

\begin{aligned} m\angle AVB&=266^\circ/2,\\ m\angle AVB&=133^\circ. \end{aligned}

Check your understanding

Points $A,V,B,C$ occur in that order on a circle. Which arc does $\angle AVB$ intercept?

A The arc

AVB

B The arc

ACB

C The arc

VAC

D The whole circle

Why half the arc works in every case

We use equal-radius isosceles triangles and the triangle angle sum. The earlier Isosceles and Equilateral Triangles explains why equal sides give equal opposite angles.

Case 1: the center is on one side

Let $VA$ be a diameter, with center $O$ between $V$ and $A$ . Join $O$ to $B$ .

Case 1. The dashed diameter VA lies on one angle side. Highlighted arc AB excludes V.

First use $34^\circ$ for the angle at $V$ . Radii $OV$ and $OB$ are equal, so the angle at $B$ in triangle $VOB$ is also $34^\circ$ . The triangle angle sum gives the center angle on the $V$ side.

\begin{aligned} m\angle VOB&=180^\circ-34^\circ-34^\circ,\\ m\angle VOB&=112^\circ. \end{aligned}

Rays $OV$ and $OA$ are opposite, so the remaining central angle is $68^\circ$ .

\begin{aligned} m\angle BOA&=180^\circ-112^\circ,\\ m\angle BOA&=68^\circ. \end{aligned}

That angle measures arc $BA$ , twice the original $34^\circ$ angle.

Inscribed-angle theorem when the center is on a side#

Now let $m\angle AVB=x$ . Since $O$ lies on ray $VA$ , the angle $OVB$ also measures $x$ . Equal radii make triangle $VOB$ isosceles, so its angle $VBO$ measures $x$ .

The triangle angle sum gives

m\angle VOB=180^\circ-2x.

The straight angle at $O$ then gives

\begin{aligned} m\angle BOA&=180^\circ-(180^\circ-2x),\\ m\angle BOA&=2x. \end{aligned}

Arc $BA$ excluding $V$ has this measure. Nothing depended on choosing $34^\circ$ , so the inscribed angle is half that arc in every such configuration.

Case 2: the center is inside the angle

Draw diameter $VD$ . Here ray $VD$ splits $\angle AVB$ , and $D$ splits its intercepted arc $AB$ .

Case 2. Solid blue arc AD and dashed red arc DB together make the intercepted arc AB.

For example, suppose the angles $AVD$ and $DVB$ measure $24^\circ$ and $39^\circ$ . Case 1 makes their arcs $48^\circ$ and $78^\circ$ . Add the angles, and separately add the arcs.

\begin{aligned} m\angle AVB&=24^\circ+39^\circ,\\ m\angle AVB&=63^\circ,\\ m(\widehat{ADB})&=48^\circ+78^\circ,\\ m(\widehat{ADB})&=126^\circ. \end{aligned}

Inscribed-angle theorem when the center is inside#

Let the parts $AVD$ and $DVB$ measure $u$ and $v$ . Case 1 applies to each because $VD$ is a diameter. Arc $AD$ measures $2u$ , and arc $DB$ measures $2v$ .

Angle addition gives

m\angle AVB=u+v.

Arc addition gives

\begin{aligned} m(\widehat{ADB})&=2u+2v,\\ m(\widehat{ADB})&=2(u+v). \end{aligned}

The whole arc is twice the whole angle, just as in the numerical instance.

Worked example 2 Add the two pieces

Problem. Diameter $VD$ lies inside $\angle AVB$ . Point $D$ lies on the intercepted arc $AB$ . Arc $AD$ measures $58^\circ$ , and arc $DB$ measures $86^\circ$ . Find the angle parts $AVD$ and $DVB$ , and the whole angle $AVB$ .

The given arc pieces are AD and DB. The angle parts are unknown.

Solution. Each part has the diameter along one side, so Case 1 gives

\begin{aligned} m\angle AVD&=58^\circ/2,\\ m\angle AVD&=29^\circ,\\ m\angle DVB&=86^\circ/2,\\ m\angle DVB&=43^\circ. \end{aligned}

Add the angle parts.

\begin{aligned} m\angle AVB&=29^\circ+43^\circ,\\ m\angle AVB&=72^\circ. \end{aligned}

Case 3: the center is outside the angle

Draw diameter $VD$ again. This time rays $VA$ and $VB$ lie on the same side of $VD$ . Take the names so that $VA$ lies between $VD$ and $VB$ . Along the arc away from $V$ , the order is $D,A,B$ .

Case 3. Remove dashed arc DA from the full arc DB to keep solid arc AB.

Suppose $DVB$ measures $47^\circ$ and $DVA$ measures $19^\circ$ . Case 1 gives arcs of $94^\circ$ and $38^\circ$ . Subtract on both the angle side and the arc side.

\begin{aligned} m\angle AVB&=47^\circ-19^\circ,\\ m\angle AVB&=28^\circ,\\ m(\widehat{AB})&=94^\circ-38^\circ,\\ m(\widehat{AB})&=56^\circ. \end{aligned}

Inscribed-angle theorem when the center is outside#

Let the angles $DVB$ and $DVA$ measure $p$ and $q$ . The ray order gives $p>q$ . Case 1 makes arc $DB$ measure $2p$ and arc $DA$ measure $2q$ .

Subtract the smaller angle from the larger.

m\angle AVB=p-q.

Subtract the smaller arc from the larger.

\begin{aligned} m(\widehat{AB})&=2p-2q,\\ m(\widehat{AB})&=2(p-q). \end{aligned}

The remaining arc is twice the remaining angle. Reflecting the picture, or swapping $A$ and $B$ , covers the other orientation.

The center is on a side, inside the angle, or outside it, so these cases cover every inscribed angle. The inscribed-angle theorem is therefore

m\angle AVB=\frac12\,m(\text{arc }AB\text{ excluding }V).

Check your understanding

Diameter $VD$ lies outside $\angle AVB$ , with arc order $D,A,B$ away from $V$ . Arc $DB$ measures $170^\circ$ and arc $DA$ measures $74^\circ$ . Find $m\angle AVB$ .

96^\circ

122^\circ

48^\circ

85^\circ

Move the vertex, keep track of its arc

If a vertex moves along one arc between fixed chord endpoints, its intercepted arc stays the same. An intercepted arc of $138^\circ$ gives $69^\circ$ at every valid vertex on the other arc.

Moving the vertex to the opposite arc changes the intercepted arc to $222^\circ$ , so the new angle is $111^\circ$ . At an endpoint itself there is no inscribed angle with three distinct points.

In general, let one arc measure $a$ . Vertices on the other arc see half of $a$ . Vertices on the first arc see half of the remaining $360^\circ-a$ .

\begin{aligned} \text{first angle}&=a/2,\\ \text{other angle}&=180^\circ-a/2. \end{aligned}

Thus angles intercepting the same arc are equal. Angles whose vertices lie on opposite arcs between the same endpoints sum to $180^\circ$ . If the endpoints form a diameter, both arcs are semicircles and both angles are $90^\circ$ .

Worked example 3 Same arc and opposite arc

Problem. Points $P,Q$ lie on the same open arc between $A,B$ . Point $R$ lies on the other open arc. Given $m\angle APB=68^\circ$ , find $m\angle AQB$ and $m\angle ARB$ .

P and Q exclude the same arc AB. R excludes the other arc.

Solution. Vertices $P,Q$ intercept the same arc, so

m\angle AQB=68^\circ.

That intercepted arc measures $136^\circ$ . The other arc measures $224^\circ$ , giving

\begin{aligned} m\angle ARB&=224^\circ/2,\\ m\angle ARB&=112^\circ. \end{aligned}

Explore the intercepted arc choice

Before moving a vertex, predict which arc it will exclude. Positions on the same open arc give equal angles; positions on opposite arcs give angles that sum to $180^\circ$ . Measurements illustrate the proved theorem.

Choose the arc that excludes the vertex

Vertex V

V is on the major arc AB (position 3). The intercepted minor arc AB measures 120°. Angle AVB = 60°.

A and B stay fixed. The highlighted intercepted arc excludes V; moving V to the other arc changes the intercepted arc.

Check your understanding

$P,Q$ lie on one open arc between $A,B$ , and $R$ on the other. Given $m\angle APB=52^\circ$ , which pair is correct?

$m\angle AQB=52^\circ$

$m\angle ARB=128^\circ$

$m\angle AQB=52^\circ$

$m\angle ARB=52^\circ$

$m\angle AQB=128^\circ$

$m\angle ARB=128^\circ$

$m\angle AQB=128^\circ$

$m\angle ARB=52^\circ$

A diameter and a right angle

A diameter gives a right angle at every other point on the circle. Its intercepted semicircle measures $180^\circ$ , and half of $180^\circ$ is $90^\circ$ .

The angle at C intercepts the semicircle AB away from C.

The semicircle right-angle theorem and its converse#

If $AB$ is a diameter and $C$ is any other point on the circle, the arc $AB$ excluding $C$ is a semicircle. The inscribed-angle theorem gives $m\angle ACB=90^\circ$ .

For the converse, suppose $A,B,C$ are three distinct points on a circle and $m\angle ACB=90^\circ$ . Doubling the angle makes the intercepted arc $180^\circ$ . The radii to $A,B$ form a straight angle, so $AB$ passes through the center and is a diameter.

All three points must lie on the given circle for the converse.

Worked example 4 Use the converse, then the triangle sum

Problem. Triangle $ACB$ has all three vertices on a circle with center $O$ . Given $m\angle ACB=90^\circ$ and $m\angle CAB=37^\circ$ , identify a diameter and find $m\angle ABC$ .

Solution. The right angle at $C$ intercepts an arc of $180^\circ$ . By the converse, $AB$ is a diameter.

The statement supplies the right angle. The solution identifies the diameter.

The triangle’s angles sum to $180^\circ$ .

\begin{aligned} m\angle ABC&=180^\circ-90^\circ-37^\circ,\\ m\angle ABC&=53^\circ. \end{aligned}

Check your understanding

Distinct points $A,B,C$ lie on a circle, and $m\angle ACB=90^\circ$ . Which segment must be a diameter?

AC

BC

C None is forced D

AB

A tangent ray and a chord

The angle between a tangent and a chord also comes from half an arc, but choosing the tangent ray matters. A tangent is perpendicular to the radius at its contact point, as proved in Tangents and Secants.

Take a chord $TA$ that is not a diameter. Let the minor central angle $TOA$ be $104^\circ$ . Equal radii make the base angles of triangle $OTA$ equal.

\begin{aligned} m\angle OTA&=(180^\circ-104^\circ)/2,\\ m\angle OTA&=38^\circ. \end{aligned}

Choose tangent ray $TR$ on the same side of line $OT$ as $A$ . It makes a right angle with $TO$ . Subtract the triangle angle.

\begin{aligned} m\angle RTA&=90^\circ-38^\circ,\\ m\angle RTA&=52^\circ. \end{aligned}

The opposite tangent ray $TS$ gives the supplementary angle, $128^\circ$ . It corresponds to the major arc, whose measure is $256^\circ$ .

R and A are on the same side of line OT. S is on the opposite side.

The tangent-chord theorem with both ray choices#

Let the minor arc $TA$ measure $\theta$ , where $0^\circ<\theta<180^\circ$ . Triangle $OTA$ is isosceles because $OT$ and $OA$ are radii. Its angle sum gives

m\angle OTA=\frac{180^\circ-\theta}{2}.

For tangent ray $TR$ on the same side of line $OT$ as $A$ , chord ray $TA$ lies inside the right angle $RTO$ . Subtracting gives

\begin{aligned} m\angle RTA&=90^\circ-\frac{180^\circ-\theta}{2},\\ m\angle RTA&=\theta/2. \end{aligned}

For opposite tangent ray $TS$ , use a straight angle.

\begin{aligned} m\angle STA&=180^\circ-\theta/2,\\ m\angle STA&=(360^\circ-\theta)/2. \end{aligned}

Thus $RTA$ is half the minor arc $TA$ , and $STA$ is half the major arc $TA$ . The numerical argument followed these same steps. Reflecting the configuration covers a chord on the other side of line $OT$ .

If $TA$ is a diameter, it lies on the radius line. Both tangent rays are perpendicular to it, giving $90^\circ$ , half of either semicircle.

This is the tangent-chord theorem. Equivalently, a tangent-chord angle equals an inscribed angle intercepting the same chosen arc: put the vertex on the other arc, so it is excluded from the chosen one.

Worked example 5 Keep the tangent rays separate

Problem. A circle has center $O$ and minor arc $TA$ of $138^\circ$ . Line $RTS$ is tangent at $T$ . Ray $TR$ lies on the same side of line $OT$ as $A$ , and $TS$ is its opposite ray. Find $m\angle RTA$ and $m\angle STA$ , naming each intercepted arc.

Opposite tangent rays form two angles with the same chord TA.

Solution. Ray $TR$ selects the minor arc $TA$ .

\begin{aligned} m\angle RTA&=138^\circ/2,\\ m\angle RTA&=69^\circ. \end{aligned}

Ray $TS$ selects the major arc, whose measure is $222^\circ$ .

\begin{aligned} m\angle STA&=222^\circ/2,\\ m\angle STA&=111^\circ. \end{aligned}

Check your understanding

Minor arc $TA$ measures $156^\circ$ . Tangent ray $TR$ lies on the same side of line $OT$ as $A$ ; $TS$ is its opposite ray. Which pair is correct?

$m\angle RTA=102^\circ$

$m\angle STA=78^\circ$

$m\angle RTA=78^\circ$

$m\angle STA=102^\circ$

$m\angle RTA=78^\circ$

$m\angle STA=78^\circ$

$m\angle RTA=102^\circ$

$m\angle STA=102^\circ$

Inscribed Angles

Learning goals

Choose the arc away from the vertex

Why half the arc works in every case

Case 1: the center is on one side

Case 2: the center is inside the angle

Case 3: the center is outside the angle

Move the vertex, keep track of its arc

Explore the intercepted arc choice

A diameter and a right angle

A tangent ray and a chord

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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