Central and Inscribed Angles

One arc, two angles

Let $A$ and $B$ be points on a circle with centre $O$. An arc subtends an angle when the two arms of the angle meet the endpoints of that arc.

The radii $OA$ and $OB$ form the angle at the centre, $\angle AOB$. If $C$ is another point on the circumference, the chords $CA$ and $CB$ form the angle at the circumference, $\angle ACB$. Both angles are subtended by the same arc $AB$ because their arms end at the same two points, $A$ and $B$.

The theorem states:

$$\angle AOB=2\angle ACB$$

The relevant central angle and circumference angle face the same arc. Equivalently, they lie in opposite sectors of the circle.

Applying the theorem

Once you have matched the endpoints, the calculation is a doubling or halving relationship. For example, if an angle at the centre is $106^\circ$, the angle at the circumference subtended by the same arc is

$$106^\circ\div2=53^\circ.$$

In the other direction, an angle of $41^\circ$ at the circumference corresponds to a central angle of

$$2\times41^\circ=82^\circ.$$

The appearance of the diagram does not determine whether the theorem applies. Lines may cross and the points may occupy unfamiliar positions. Trace the arms of each angle instead: if both pairs of arms finish at the same two points on the circumference, the angles are connected by the theorem.

Choosing the correct central angle

There are two possible angles between $OA$ and $OB$: the smaller angle and the reflex angle. Use the one lying in the sector opposite $C$.

Suppose the smaller central angle is $126^\circ$, but $C$ lies on the minor arc $AB$. The angle at the circumference faces the major arc, so the relevant central angle is reflex:

$$360^\circ-126^\circ=234^\circ.$$

Therefore, the angle at the circumference is

$$234^\circ\div2=117^\circ.$$

This is why an obtuse angle at the circumference can correspond to a reflex angle at the centre. Automatically halving the smaller central angle would use the wrong arc.

Proving the theorem

Join the centre $O$ to $A$, $B$ and $C$. All three lines are radii, so $OA=OC$ and $OB=OC$. This creates two isosceles triangles.

Let

$$\angle ACO=x \quad\text{and}\quad \angle OCB=y.$$

The whole angle at the circumference is therefore

$$\angle ACB=x+y.$$

In isosceles triangle $AOC$, the base angles are equal, so $\angle OAC=x$. The angle at $O$, bounded by $OA$ and $OC$, is

$$\angle AOC=180^\circ-2x.$$

Similarly, in isosceles triangle $BOC$, the angle at $O$ bounded by $OC$ and $OB$ is

$$\angle COB=180^\circ-2y.$$

Angles around $O$ total $360^\circ$. The central angle $\angle AOB$ opposite $C$ is what remains after subtracting those two adjacent central angles:

$$\begin{aligned} \angle AOB &=360^\circ-(180^\circ-2x)-(180^\circ-2y)\\ &=2x+2y\\ &=2(x+y). \end{aligned}$$

Since $x+y=\angle ACB$,

$$\angle AOB=2\angle ACB.$$

The proof uses three established facts: radii of the same circle are equal, base angles in an isosceles triangle are equal, and angles around a point total $360^\circ$. The relationship remains true when the diagram is rearranged; the relevant central angle may then be reflex.

The theorem

For points $A$, $B$ and $C$ on a circle with centre $O$,

$$\angle AOB=2\angle ACB$$

provided both angles are subtended by the same arc $AB$.

• Centre to circumference: divide by $2$.
• Circumference to centre: multiply by $2$.

Match the correct angles

Check that both angles have arms ending at the same two points. Use the central angle in the sector opposite the circumference angle; it may be reflex. If only the smaller central angle is given,

$$\text{reflex angle}=360^\circ-\text{smaller angle}.$$

Proof skeleton

Join the points to the centre. Equal radii form two isosceles triangles. If the circumference angle is split into $x$ and $y$, the two adjacent central angles are $180^\circ-2x$ and $180^\circ-2y$. Hence

$$360^\circ-(180^\circ-2x)-(180^\circ-2y)=2(x+y),$$

so the angle at the centre is twice the angle at the circumference.