Math Core

A central angle has its vertex at the center of a circle. But what if the vertex sits on the circle itself? Angles like that appear whenever you connect three points on a circle, and they follow one of the most useful rules in geometry: they are always exactly half of the arc they cut off.

What an inscribed angle is

Definition

Inscribed angle

An inscribed angle is an angle whose vertex is on a circle and whose sides are chords of the circle. The arc that lies inside the angle, with its endpoints on the sides, is the intercepted arc.

In the figure, $\angle ACB$ is an inscribed angle. Its vertex $C$ is on the circle, and it intercepts the arc $\overset{\frown}{AB}$ at the bottom of the circle. The dashed segments show the central angle $\angle AOB$ that intercepts the same arc.

Inscribed angle ACB and central angle AOB (dashed) intercept the same arc AB.

Compare the two angles. The central angle is wide and the inscribed angle is narrower. If you measured them, you'd find the central angle is exactly twice as big.

Inscribed Angle Theorem

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

$m\angle ACB = \dfrac{1}{2}\, m\overset{\frown}{AB}.$

Equivalently, the intercepted arc is twice the inscribed angle.

Why it works

Look at the special case where one side of the inscribed angle is a diameter. Say $\overline{CD}$ is a diameter through the center $O$ , and $A$ is another point on the circle. Draw radius $\overline{OA}$ .

$OA = OC$ because both are radii, so $\triangle AOC$ is isosceles and its base angles are equal. Call each one $x$ .
The central angle $\angle AOD$ is an exterior angle of $\triangle AOC$ , so it equals the sum of the two remote interior angles: $x + x = 2x$ .
The arc $\overset{\frown}{AD}$ has the same measure as $\angle AOD$ , which is $2x$ . The inscribed angle $\angle ACD$ is $x$ , exactly half.

Any other inscribed angle can be split (or extended) by a diameter into two angles of this special kind, so the half rule holds for every inscribed angle.

Worked example: Using the theorem both ways

An inscribed angle intercepts an arc of $110^\circ$ . Find the angle.
An inscribed angle measures $38^\circ$ . Find its intercepted arc.

Solutions.

The angle is half the arc: $\dfrac{1}{2}(110^\circ) = 55^\circ$ .
The arc is twice the angle: $2(38^\circ) = 76^\circ$ .

Common mistake

Don't confuse the two rules. A central angle equals its arc. An inscribed angle is half its arc. Before you write anything, find the vertex: at the center, or on the circle?

Three consequences

The Inscribed Angle Theorem leads to three facts you will use constantly.

1. Angles that intercept the same arc are congruent. If $\angle ACB$ and $\angle ADB$ are both inscribed and both intercept $\overset{\frown}{AB}$ , each equals half of the same arc, so they are equal.

2. An angle inscribed in a semicircle is a right angle. If $\overline{AB}$ is a diameter and $C$ is any other point on the circle, then $\angle ACB$ intercepts a $180^\circ$ arc, so $m\angle ACB = 90^\circ$ . The reverse is true too: if an inscribed angle is a right angle, the chord connecting its endpoints is a diameter.

AB is a diameter, so angle ACB is a right angle wherever C is on the circle.

3. Opposite angles of an inscribed quadrilateral are supplementary. A quadrilateral whose four vertices are on a circle is inscribed in the circle. Opposite angles $\angle A$ and $\angle C$ intercept arcs that together make the whole circle, so

$m\angle A + m\angle C = \dfrac{1}{2}(360^\circ) = 180^\circ.$

The same is true for $\angle B$ and $\angle D$ .

Worked example: A right triangle in a circle

Triangle $ABC$ is inscribed in a circle with $\overline{AB}$ as a diameter. If $AC = 9$ and $BC = 12$ , find the radius of the circle.

Since $\overline{AB}$ is a diameter, $\angle C = 90^\circ$ . By the Pythagorean theorem,

$AB = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15.$

The diameter is $15$ , so the radius is $7.5$ .

Worked example: An inscribed quadrilateral

Quadrilateral $PQRS$ is inscribed in a circle. $m\angle P = (2x + 12)^\circ$ and $m\angle R = (4x - 6)^\circ$ . Find $m\angle P$ .

Opposite angles are supplementary:

$(2x + 12) + (4x - 6) = 180 \quad\Longrightarrow\quad 6x + 6 = 180 \quad\Longrightarrow\quad x = 29.$

So $m\angle P = 2(29) + 12 = 70^\circ$ (and $m\angle R = 110^\circ$ ).

Tip

Not every quadrilateral can be inscribed in a circle. A parallelogram has equal opposite angles, and they can only be equal and supplementary if both are $90^\circ$ . So the only parallelograms that can be inscribed are rectangles (including squares).