Angle Subtended by an Arc

This is the most beautiful theorem in basic circle geometry: the angle subtended by an arc at the centre of a circle is exactly twice the angle subtended by the same arc at any point on the remaining arc. One arc, two angles, with a constant factor of two. From this single fact a cascade of corollaries flows, including the right angle in a semicircle and the equal angles in the same segment.

Definitions

Given a circle and a chord $AB$ , the chord divides the circle into two arcs. The angle subtended at the centre by arc $AB$ is the angle $\angle AOB$ at the centre $O$ . The angle subtended at the circumference by arc $AB$ is the angle $\angle APB$ at any point $P$ on the other arc.

The theorem

Theorem. The angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc at any point on the remaining part of the circle.

In symbols: if $O$ is the centre and $P$ is a point on the major arc (with the chord $AB$ creating the minor arc), then $\angle AOB = 2 \angle APB.$

(The same statement applies when $P$ is on the minor arc and we look at the reflex angle at the centre , but in standard configurations we use the non-reflex version.)

Proof (sketch)

Given. Circle with centre $O$ , points $A, B, P$ on the circle.

Construction. Join $PO$ and extend it to meet the circle at $Q$ .

Proof outline. In $\triangle OAP$ , $OA = OP$ (radii), so the triangle is isosceles. Hence $\angle OAP = \angle OPA = \alpha$ , say. Then $\angle AOQ$ , the exterior angle of $\triangle OAP$ at $O$ , equals $2\alpha$ by the exterior-angle theorem.

Similarly in $\triangle OBP$ , $OB = OP$ (radii), so $\angle OBP = \angle OPB = \beta$ . Then $\angle BOQ = 2\beta$ .

Adding: $\angle AOB = \angle AOQ + \angle BOQ = 2\alpha + 2\beta = 2(\alpha + \beta) = 2 \angle APB$ . Q.E.D.

The proof has one construction (extending $OP$ ) and uses two isosceles triangles. The exterior-angle theorem closes the argument.

Five corollaries , the gold mine

Corollary 1 (Angles in the same segment are equal). Two angles subtended at the circumference by the same arc are equal.

Why. Both are half the (single) central angle. So they are equal to each other.

Corollary 2 (Angle in a semicircle is a right angle). The angle inscribed in a semicircle is $90^\circ$ .

Why. The chord is a diameter, so the "arc" subtended is half the circle, giving a central angle of $180^\circ$ . By the theorem, the angle at the circumference is $\tfrac{1}{2} \cdot 180^\circ = 90^\circ$ .

Corollary 3 (Cyclic quadrilateral , opposite angles supplementary). In any cyclic quadrilateral $ABCD$ , $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$ .

Why. $\angle A$ and $\angle C$ are inscribed angles subtending opposite arcs that together make the whole circle. So the two central angles sum to $360^\circ$ , and each inscribed angle is half , totalling $180^\circ$ .

Corollary 4 (Equal chords subtend equal angles at the centre). In a circle, two equal chords subtend equal angles at the centre.

Why. The chord determines the angle at the centre. Equal chords $\Rightarrow$ equal triangles formed by the radii $\Rightarrow$ equal central angles.

Corollary 5 (Angle in alternate segment). The angle between a chord and a tangent at the point of contact equals the inscribed angle in the alternate segment. (We will study tangents in Class X. Mentioning the result here for completeness.)

Using the theorem

The single theorem and its five corollaries cover the great majority of circle-angle problems.

Example. In a circle, the central angle subtended by a chord is $80^\circ$ . Find the angle subtended by the chord at any point on the major arc.

Half: $40^\circ$ .

Example. A triangle $ABC$ is inscribed in a semicircle with $AB$ as diameter. Find $\angle C$ .

$\angle C = 90^\circ$ (angle in a semicircle).

Example. In a cyclic quadrilateral $ABCD$ , $\angle A = 110^\circ$ . Find $\angle C$ .

$\angle A + \angle C = 180^\circ \Rightarrow \angle C = 70^\circ$ .

Worked examples

Example 1. A chord of a circle subtends $60^\circ$ at the centre. Find the angle subtended on the major arc.

Half: $30^\circ$ .

Example 2. A chord of a circle subtends $80^\circ$ on the circumference. Find the central angle.

Double: $160^\circ$ .

Example 3. $AB$ is a diameter of a circle, and $C$ is a point on the circle. Find $\angle ACB$ .

$\angle ACB = 90^\circ$ (angle in a semicircle).

Example 4. In a cyclic quadrilateral $ABCD$ , $\angle B = 95^\circ$ . Find $\angle D$ .

Opposite angles: $\angle D = 180 - 95 = 85^\circ$ .

Example 5. A chord $AB$ subtends $\angle APB = 40^\circ$ at $P$ on the major arc and $\angle AQB$ at $Q$ on the same arc. Find $\angle AQB$ .

Equal angles in the same segment: $\angle AQB = 40^\circ$ .

Try it yourself

State the angle-subtended-by-an-arc theorem.
State the five corollaries.
A chord subtends $50^\circ$ on the major arc. Find the central angle.
A chord subtends $120^\circ$ at the centre. Find the angle on the major arc.
In a cyclic quadrilateral, $\angle A = 70^\circ$ . Find $\angle C$ .
Why is the angle in a semicircle $90^\circ$ ?
Two points $P, Q$ are on the major arc of a circle with chord $AB$ . What is the relationship between $\angle APB$ and $\angle AQB$ ?
A chord subtends $130^\circ$ at the centre. Find the inscribed angle on the major arc and on the minor arc.
In cyclic quadrilateral $PQRS$ , $\angle Q + \angle S = ?$
Prove the angle-in-semicircle theorem in your own words.

Pitfalls / Insight

The point at the circumference must be on the remaining arc. A point on the same side as the chord gives a different (reflex-based) relationship.
The "twice" relation is exact. Always double / halve.
For cyclic quadrilateral opposite angles, the sum is $180^\circ$ , not just any pair.

Insight. This single "angle at centre is twice angle at circumference" theorem is the heart of all circle geometry. Every other angle property here follows from it. Memorise it carefully and the rest of the chapter falls into place.