Central and Inscribed Angles: The Inscribed Angle Theorem

The parts of a circle

Term	Meaning
radius	a segment from the center to the circle
chord	a segment with both ends on the circle
diameter	a chord through the center; twice the radius
arc	a piece of the circle between two points
central angle	vertex at the center, sides along two radii
inscribed angle	vertex on the circle, sides along two chords
tangent	a line touching the circle at exactly one point

An arc is measured in degrees, and its measure is defined as the measure of the central angle that stands on it. A central angle of $140°$ cuts off an arc of $140°$.

The minor arc between two points is the shorter way round, less than $180°$; the major arc is the longer way.

An inscribed angle is half the central angle

Both angles stand on the lower arc from $A$ to $B$. The central angle $\angle AOB$ measures $140°$; the inscribed angle $\angle ACB$ measures $70°$.

$$m\angle ACB = \frac{1}{2} \, m\angle AOB$$

That is the inscribed angle theorem, and it holds wherever $C$ sits on the major arc.

Why the inscribed angle is half

Start with the case where one side of the inscribed angle passes through the center — so $C$, $O$ and $B$ are in a line.

Statement	Reason
$OC = OA$	radii of the same circle
$\angle OCA \cong \angle OAC$	base angles of isosceles $\triangle OCA$
$m\angle AOB = m\angle OCA + m\angle OAC$	exterior angle theorem, at $O$
$m\angle AOB = 2 \, m\angle OCA$	substitution
$m\angle ACB = \tfrac{1}{2} \, m\angle AOB$	$\angle ACB$ is $\angle OCA$

One isosceles triangle does the work, and the two equal radii are what make it isosceles.

The other cases reduce to this one. If the center lies inside the inscribed angle, draw the diameter from $C$ through $O$. It splits the inscribed angle and the central angle into two pieces each, and each inscribed piece is half its central piece by the case already proved — so the sums are in the same ratio. If the center lies outside, the same diameter gives a difference instead of a sum.

The exterior angle theorem is the step to watch. $\angle AOB$ sits outside $\triangle OCA$, next to its angle at $O$, so it equals the sum of the two far angles.

Two consequences

Inscribed angles on the same arc are equal. Every one is half of the same central angle. Move $C$ anywhere on the major arc and $\angle ACB$ stays $70°$.

An angle in a semicircle is a right angle. If $\overline{AB}$ is a diameter, the central angle on it is a straight angle of $180°$. Any inscribed angle standing on that diameter is half of it:

$$\frac{1}{2} \times 180° = 90°$$

This is often called Thales’s theorem, and it is how a carpenter’s square finds the center of a circle: two right angles set in the circle give two diameters, and they cross at the center.

A tangent meets its radius at a right angle

A tangent touches the circle at one point, $T$. Every other point of the tangent line lies outside the circle, so it is farther from the center than $T$ is.

So $\overline{OT}$ is the shortest segment from $O$ to the tangent line. The shortest segment from a point to a line is the perpendicular one. Therefore

$$\overline{OT} \perp \text{tangent}$$

Two tangents drawn from an outside point $P$, touching at $A$ and $B$, form a circumscribed angle $\angle APB$. Quadrilateral $OAPB$ has right angles at $A$ and $B$, and its four angles total $360°$, so

$$m\angle APB + m\angle AOB = 180°$$

The tangent segments from $P$ are also equal: $\triangle OAP$ and $\triangle OBP$ are right triangles sharing the hypotenuse $\overline{OP}$ with equal legs $OA = OB$, so they are congruent.

Chords and the center

A line from the center perpendicular to a chord bisects the chord.

The two radii to the chord’s ends are equal, so the triangle they make with the chord is isosceles. In an isosceles triangle the perpendicular from the apex lands on the midpoint of the base.

This gives a quick way to find distances. A chord $16$ cm long lies $6$ cm from the center. Find the radius. Half the chord is $8$, and the radius is the hypotenuse of a right triangle with legs $6$ and $8$:

$$r = \sqrt{6^2 + 8^2} = 10 \text{ cm}$$

Worked examples

Common mistakes

Practice problems