How do I tell which arc an inscribed angle intercepts?

Follow the two sides of the angle out to where they hit the circle; those two points are the endpoints of the intercepted arc. Then take the arc that lies inside the angle, not behind the vertex. A quick check: the vertex is never on the intercepted arc.

Why is an inscribed angle exactly half the central angle?

Draw the radius from the center to the inscribed angle's vertex. The radii create isosceles triangles, and the exterior angle of an isosceles triangle equals twice its base angle. That doubling at the center is where the factor of 2 comes from, and every case of the theorem reduces to that picture by adding or subtracting two such triangles.

Is arc measure the same as arc length?

No. Arc measure is in degrees and describes what fraction of the full turn the arc covers; it does not depend on the circle's size. Arc length is an actual distance and equals n/360 · 2 r for an arc of n degrees in a circle of radius r.

Are all quadrilaterals cyclic?

No. A quadrilateral can be inscribed in a circle only if its opposite angles are supplementary. Rectangles, squares, and isosceles trapezoids qualify; a non-rectangular parallelogram, a general kite, or most random quadrilaterals do not. Checking the opposite-angle sum is the fastest test.

GEOM-9.2

Central & Inscribed Angles

Master central and inscribed angles in Geometry 9.2: arc measures, the half-the-arc rule, congruent inscribed angles, semicircle right angles, and cyclic quadrilaterals.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Central & Inscribed Angles, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Central & Inscribed Angles free

What this lesson covers

A circle has no corners, yet it is full of angles. Some have their vertex at the center; others have their vertex on the circle itself. That single difference — center versus edge — changes everything about how the angle relates to the arc it opens onto. A central angle equals its arc. An inscribed angle is exactly half of it.

In this lesson you will learn why that halving happens, how to read arc measures off a diagram quickly, and how three powerful corollaries fall right out of the inscribed angle theorem: inscribed angles that share an arc are congruent, any angle inscribed in a semicircle is a right angle, and the opposite angles of a quadrilateral inscribed in a circle are supplementary. These tools show up constantly for the rest of the circles unit, so build the habit now of asking one question about every angle you see: where is its vertex?

Central Angles and the Measure of an Arc

A central angle has its vertex at the center of the circle, and its sides are two radii. The arc that lies in the interior of the angle is the intercepted arc, and the definition you need is short: the measure of a minor arc equals the measure of its central angle.

If

O

is the center and

m\angle AOB = 74^\circ

, then

m\widehat{AB} = 74^\circ

for the minor arc, and the major arc going the long way around measures

360^\circ - 74^\circ = 286^\circ

. A full circle is

360^\circ

, and a semicircle — the arc cut off by a diameter — measures

180^\circ

.

Two cautions here. First, arc measure in degrees is not arc length. Two circles of different sizes can both have a

60^\circ

arc; the bigger circle's arc is physically longer. Degrees describe how much of the turn around the center the arc covers, nothing about size. Second, notation matters:

\widehat{AB}

with two letters is normally read as the minor arc, so when you want the major arc you name it with a third point on it, like

\widehat{ACB}

.

Because arcs on a circle add the way segments on a line do, the Arc Addition Postulate lets you write

m\widehat{AB} + m\widehat{BC} = m\widehat{ABC}

whenever

B

lies between

A

and

C

on the arc. Most circle problems are solved by combining that one fact with the angle theorems in the next section, then setting the total equal to

360^\circ

The Inscribed Angle Theorem

An inscribed angle has its vertex on the circle and its sides are two chords. The theorem: an inscribed angle measures half its intercepted arc.

m\angle ABC = \tfrac{1}{2}\,m\widehat{AC}

Equivalently, the arc is twice the angle. Students go wrong far more often by using the relationship backwards than by forgetting it, so anchor it with a picture you trust: an angle inscribed in a semicircle intercepts a

180^\circ

arc and clearly is not

360^\circ

— it is

90^\circ

. Half, not double, for the angle on the circle.

Why is it true? Take the easy case where one side of the inscribed angle

\angle ABC

passes through the center

O

, so

BC

is a diameter. Draw radius

OA

. Then

\triangle OAB

is isosceles because

OA = OB

, so its base angles both equal the inscribed angle

x

. The exterior angle

\angle AOC

of that triangle equals the sum of the two remote interior angles, giving

m\angle AOC = 2x

. But

\angle AOC

is a central angle, so

m\widehat{AC} = 2x

— the arc is twice the inscribed angle. The general case is handled by splitting the angle into two such pieces (or subtracting them when the center falls outside the angle).

A useful comparison:

Feature	Central angle	Inscribed angle
Vertex	Center of circle	On the circle
Sides	Two radii	Two chords
Relation to arc	Equal to arc	Half the arc
Example with a $100^\circ$ arc	$100^\circ$	$50^\circ$

Before you compute anything, locate the vertex and name the intercepted arc — the arc whose endpoints are on the angle's sides and which lies inside the angle.

Three Corollaries You Will Use Constantly

Everything below follows from the halving rule.

Congruent inscribed angles. Two inscribed angles that intercept the same arc — or congruent arcs — are congruent. If

\angle ADB

and

\angle ACB

both open onto

\widehat{AB}

, then both equal

\tfrac{1}{2}m\widehat{AB}

, so they are equal to each other. Point

D

can slide anywhere along the major arc and the angle never changes. This is the fact that creates similar triangles in circle diagrams, so watch for it whenever two chords cross.

Angle in a semicircle. If a side of an inscribed angle is a diameter, the intercepted arc is

180^\circ

and the angle is

\tfrac{1}{2}(180^\circ) = 90^\circ

. The converse also holds: if an inscribed angle is right, its intercepted arc is a semicircle, so the chord joining its sides' far endpoints is a diameter. That converse is how you prove a segment is a diameter.

Inscribed quadrilateral. If all four vertices of quadrilateral

ABCD

lie on a circle (a cyclic quadrilateral), then opposite angles are supplementary:

m\angle A + m\angle C = 180^\circ

and

m\angle B + m\angle D = 180^\circ

. Reason:

\angle A

and

\angle C

intercept the two arcs that together make the whole circle, so their measures add to

\tfrac{1}{2}(360^\circ) = 180^\circ

.

A frequent mistake is assuming adjacent angles of an inscribed quadrilateral are supplementary. They are not, in general. Only opposite pairs. Another is applying the supplementary rule to a quadrilateral whose vertices are not all on the circle — check that every vertex actually touches before using it.

Solving Strategy and Common Pitfalls

A reliable routine keeps these problems short.

First, label the center if there is one, and mark every radius as congruent — isosceles triangles hide in almost every circle diagram. Second, for each angle in question, classify it: vertex at center means angle equals arc; vertex on circle means angle equals half the arc. (Vertices inside or outside the circle belong to the next lesson, on secants and tangents, and use different rules.) Third, name the intercepted arc out loud. Fourth, write an equation, using the fact that all arcs around the circle sum to

360^\circ

if you need one more relationship.

Here are the errors that show up most on homework and quizzes:

Mistake	Fix
Doubling an inscribed angle's arc instead of halving the arc	Arc is the big number; the inscribed angle is the small one
Using the arc the angle "points away from"	The intercepted arc lies between the sides, inside the angle
Assuming a chord through the middle of the picture is a diameter	Only use the $90^\circ$ corollary if the chord is marked as passing through the center
Adding adjacent angles of a cyclic quadrilateral to $180^\circ$	Only opposite angles are supplementary
Confusing arc measure with arc length	Degrees measure turn; length needs the radius

When a problem gives you an algebraic expression, such as inscribed angles of

(3x+5)^\circ

and

(5x-11)^\circ

intercepting the same arc, set them equal rather than summing them:

3x+5 = 5x-11

gives

x = 8

, so each angle is

29^\circ

and the arc is

58^\circ

. Deciding equal versus supplementary versus half is the real skill; the algebra afterward is easy.

Where These Ideas Lead

The inscribed angle theorem is the seed for the rest of circle geometry. In the next lesson, angles formed by two chords meeting inside the circle turn out to be the average of two arcs, and angles formed outside the circle by secants or tangents turn out to be half the difference of two arcs. Both proofs work by drawing an extra chord and applying the inscribed angle theorem plus the exterior angle theorem — exactly the argument used above.

The corollaries have practical reach too. The semicircle corollary explains a classic shop-class trick: to find the center of a circular disk, place a carpenter's square so its right-angle corner touches the edge; the two points where its arms cross the edge are the endpoints of a diameter. Do this twice and the diameters intersect at the center.

Cyclic quadrilaterals matter because most quadrilaterals are not cyclic. A rectangle always is (its opposite angles are

90^\circ

each, summing to

180^\circ

), and so is any isosceles trapezoid, but a general parallelogram is not unless it is a rectangle. Testing whether opposite angles sum to

180^\circ

is a fast way to decide whether four given points could lie on a common circle — a question you will revisit in the coordinate-geometry lesson at the end of this unit, where you can confirm it by checking that all four points satisfy the same circle equation.

Key terms

Central angle.: An angle whose vertex is the center of the circle and whose sides are radii; its measure equals the measure of its intercepted arc.
Inscribed angle.: An angle whose vertex lies on the circle and whose sides are chords; its measure is half the measure of its intercepted arc.
Intercepted arc.: The arc lying in the interior of an angle, with endpoints on the angle's two sides.
Minor arc / major arc.: A minor arc measures less than $180^\circ$ and is named with two letters; a major arc measures more than $180^\circ$ and is named with three letters.
Semicircle.: An arc cut off by a diameter, with measure exactly $180^\circ$ ; an angle inscribed in a semicircle is a right angle.
Cyclic (inscribed) quadrilateral.: A quadrilateral whose four vertices all lie on one circle; its opposite angles are supplementary.
Arc Addition Postulate.: The measure of an arc formed by two adjacent arcs is the sum of their measures.
Arc measure.: The number of degrees of turn an arc covers around the center, independent of the circle's radius (unlike arc length).

Worked example

Quadrilateral

QRST

is inscribed in circle

O

. Diagonal

\overline{QS}

is a diameter. You are given

m\widehat{QR} = 84^\circ

and

m\angle QTS

is unknown. Also,

m\angle RQS = (2x + 6)^\circ

. Find

m\angle QRS

m\angle QTS

m\widehat{RS}

, and the value of

x

Start with the semicircle corollary. Because

\overline{QS}

is a diameter, arc

\widehat{QRS}

is a semicircle, so

m\widehat{QRS} = 180^\circ

. Angle

QRS

is inscribed and intercepts

\widehat{QTS}

, the other semicircle, which also measures

180^\circ

. Therefore

m\angle QRS = \tfrac{1}{2}(180^\circ) = 90^\circ

.

The same reasoning applies to

\angle QTS

: its vertex

T

is on the circle and it intercepts

\widehat{QRS} = 180^\circ

, so

m\angle QTS = 90^\circ

. (Check with the cyclic quadrilateral corollary:

\angle QRS

and

\angle QTS

are opposite angles, and

90^\circ + 90^\circ = 180^\circ

. Consistent.)

Now find

m\widehat{RS}

. By the Arc Addition Postulate,

m\widehat{QR} + m\widehat{RS} = m\widehat{QRS} = 180^\circ

. Substituting,

84^\circ + m\widehat{RS} = 180^\circ

, so

m\widehat{RS} = 96^\circ

.

Finally,

\angle RQS

has its vertex

Q

on the circle and its sides pass through

R

and

S

, so it intercepts

\widehat{RS}

. By the inscribed angle theorem,

m\angle RQS = \tfrac{1}{2}(96^\circ) = 48^\circ

. Set the expression equal to that value:

2x + 6 = 48

, so

2x = 42

and

x = 21

.

Quick sanity check on triangle

QRS

: its angles are

90^\circ

R

48^\circ

Q

, and the remaining angle at

S

must be

42^\circ

. Angle

QSR

intercepts

\widehat{QR} = 84^\circ

, and

\tfrac{1}{2}(84^\circ) = 42^\circ

. Everything agrees.

Practice questions

In circle

P

, points

A

B

, and

C

lie on the circle and

m\angle APB = 118^\circ

, where

P

is the center. What is the measure of inscribed angle

\angle ACB

, where

C

lies on the major arc?

$59^\circ$
$118^\circ$
$236^\circ$
$62^\circ$

Answer: $59^\circ$

The central angle

\angle APB

equals its intercepted minor arc, so

m\widehat{AB} = 118^\circ

. Point

C

is on the major arc, so inscribed angle

\angle ACB

intercepts the minor arc

\widehat{AB}

and measures half of it:

\tfrac{1}{2}(118^\circ) = 59^\circ

. Choosing

118^\circ

means treating the inscribed angle as if its vertex were at the center; choosing

236^\circ

means doubling instead of halving;

62^\circ

comes from subtracting the arc from

180^\circ

instead of taking half of it.

Quadrilateral

WXYZ

is inscribed in a circle. If

m\angle W = (3x + 10)^\circ

and

m\angle Y = (2x + 20)^\circ

, find

x

and the measure of each of these two angles. Then explain why you cannot determine

m\angle X

from this information alone.

Answer: $x = 30$ , $m\angle W = 100^\circ$ , $m\angle Y = 80^\circ$ ; $m\angle X$ cannot be found because only its supplement relationship with $\angle Z$ is known, and neither is given.

\angle W

and

\angle Y

are opposite angles of a cyclic quadrilateral, so they are supplementary:

(3x+10) + (2x+20) = 180

, giving

5x + 30 = 180

5x = 150

, and

x = 30

. Then

m\angle W = 3(30)+10 = 100^\circ

and

m\angle Y = 2(30)+20 = 80^\circ

, which do sum to

180^\circ

. For

\angle X

, the only rule available is

m\angle X + m\angle Z = 180^\circ

, one equation with two unknowns, so the vertices

X

and

Z

could slide along their arcs in many ways. This is a good reminder that adjacent angles of an inscribed quadrilateral have no fixed relationship.

In circle

O

, chord

\overline{AB}

and chord

\overline{CB}

meet at

B

on the circle, and

m\angle ABC = 90^\circ

. What can you conclude about

\overline{AC}

, and why?

Answer: $\overline{AC}$ is a diameter of circle $O$ .

Angle

ABC

is inscribed and intercepts arc

\widehat{AC}

not containing

B

. By the inscribed angle theorem,

m\widehat{AC} = 2 \cdot 90^\circ = 180^\circ

. An arc of

180^\circ

is a semicircle, and the chord joining the endpoints of a semicircle passes through the center, so

\overline{AC}

is a diameter. This converse of the semicircle corollary is the standard way to prove a chord is a diameter, and it also means the center of the circle is the midpoint of

\overline{AC}

FAQ

How do I tell which arc an inscribed angle intercepts?: Follow the two sides of the angle out to where they hit the circle; those two points are the endpoints of the intercepted arc. Then take the arc that lies inside the angle, not behind the vertex. A quick check: the vertex is never on the intercepted arc.
Why is an inscribed angle exactly half the central angle?: Draw the radius from the center to the inscribed angle's vertex. The radii create isosceles triangles, and the exterior angle of an isosceles triangle equals twice its base angle. That doubling at the center is where the factor of $2$ comes from, and every case of the theorem reduces to that picture by adding or subtracting two such triangles.
Is arc measure the same as arc length?: No. Arc measure is in degrees and describes what fraction of the full turn the arc covers; it does not depend on the circle's size. Arc length is an actual distance and equals $\frac{n}{360} \cdot 2\pi r$ for an arc of $n$ degrees in a circle of radius $r$ .
Are all quadrilaterals cyclic?: No. A quadrilateral can be inscribed in a circle only if its opposite angles are supplementary. Rectangles, squares, and isosceles trapezoids qualify; a non-rectangular parallelogram, a general kite, or most random quadrilaterals do not. Checking the opposite-angle sum is the fastest test.

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