What are cyclic quadrilaterals?

A cyclic quadrilateral has all four vertices on a circle. Its opposite angles are supplementary, adding to 180^. An exterior angle of a cyclic quadrilateral equals the interior opposite angle.

What are tangent properties?

A tangent touches a circle at exactly one point. The radius drawn to the point of contact is perpendicular to the tangent, giving a 90^ angle. Two tangents drawn from the same external point are equal in length, and they make equal angles with the line to the centre.

What are not giving reasons?

A correct value without the named theorem loses the reasoning marks.

The angle at the centre of a circle on arc PQ is 84^. Find the angle at the circumference on the same arc. [1 mark]

In a cyclic quadrilateral one angle is 108^. Find the angle opposite it. [1 mark]

SEAB Original-style practice: ABCD is a cyclic quadrilateral. Angle ABC = 95^ and angle BCD = 70^. Find angle ADC and angle BAD.

Opposite angles of a cyclic quadrilateral are supplementary, adding to 180^. Angle ADC is opposite angle ABC: 180^ - 95^ = 85^. Angle BAD is opposite angle BCD: 180^ - 70^ = 110^. Markers reward using the supplementary opposite-angle property for both pairs, giving angle ADC = 85^ and angle BAD = 110^.

SingaporeMathsSyllabus dot point

What angle relationships hold inside a circle, and how do tangents and chords behave?

Apply the circle theorems relating angles at the centre and circumference, angles in a semicircle and the same segment, cyclic quadrilaterals, and tangent properties

A focused answer to the O-Level E-Maths outcome on circle theorems. The angle at the centre, angle in a semicircle, angles in the same segment, cyclic quadrilaterals, and the tangent-radius and tangent properties.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to apply the circle theorems to find unknown angles, quoting the correct theorem as your reason. The circle theorems are a signature topic of O-Level mathematics, and questions reward both the correct answer and a clear justification at each step.

The answer

Angle at the centre and circumference

The angle subtended by an arc at the centre is twice the angle subtended by the same arc at the circumference:

\text{angle at centre} = 2 \times \text{angle at circumference}

So an angle at the circumference is half the corresponding angle at the centre, standing on the same arc.

Angle in a semicircle

An angle subtended at the circumference by a diameter is a right angle. This is the special case of the centre-circumference theorem where the central angle is the straight angle $180^\circ$ , half of which is $90^\circ$ . Spotting a diameter often instantly gives a $90^\circ$ angle.

Angles in the same segment

Angles subtended by the same arc, on the same side, at the circumference are equal. So two angles standing on the same chord, both above it, must match. This lets you transfer a known angle to another point on the circle.

Cyclic quadrilaterals

A cyclic quadrilateral has all four vertices on a circle. Its opposite angles are supplementary, adding to $180^\circ$ . An exterior angle of a cyclic quadrilateral equals the interior opposite angle.

Tangent properties

A tangent touches a circle at exactly one point. The radius drawn to the point of contact is perpendicular to the tangent, giving a $90^\circ$ angle. Two tangents drawn from the same external point are equal in length, and they make equal angles with the line to the centre.

Worked example

$A$ , $B$ and $C$ lie on a circle. $AB$ is a diameter. Angle $BAC = 35^\circ$ . Find angle $ABC$ , giving reasons.

Step 1: Use the angle in a semicircle

Since $AB$ is a diameter, the angle $ACB$ at the circumference stands on the diameter, so angle $ACB = 90^\circ$ (angle in a semicircle).

Step 2: Identify the triangle's known angles

In triangle $ABC$ , angle $BAC = 35^\circ$ and angle $ACB = 90^\circ$ .

Step 3: Apply the triangle angle sum

\text{angle } ABC = 180^\circ - 90^\circ - 35^\circ

Step 4: Evaluate

\text{angle } ABC = 55^\circ

So angle $ABC = 55^\circ$ , using the angle in a semicircle and the triangle angle sum.

Examples in context

Example 1. Finding a centre. Because the angle in a semicircle is a right angle, drawing a right angle in a circle places the hypotenuse as a diameter through the centre. Builders and designers use this to locate the centre of a circular arc.

Example 2. Tangent lengths in design. Two tangents from an external point being equal lets engineers design symmetric fittings, such as a belt touching two pulleys, where equal tangent lengths keep the geometry balanced.

Try this

Q1. The angle at the centre of a circle on arc $PQ$ is $84^\circ$ . Find the angle at the circumference on the same arc. [1 mark]

Cue. Half the centre angle: $\dfrac{84^\circ}{2} = 42^\circ$ .

Q2. $PQ$ is a diameter and $R$ lies on the circle. State the size of angle $PRQ$ . [1 mark]

Cue. Angle in a semicircle is $90^\circ$ .

Q3. In a cyclic quadrilateral one angle is $108^\circ$ . Find the angle opposite it. [1 mark]

Cue. Opposite angles are supplementary: $180^\circ - 108^\circ = 72^\circ$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marks

A

B

and

C

lie on a circle with centre

O

. The angle

AOC

at the centre is

130^\circ

. (a) Find the angle

ABC

at the circumference. (b) State the theorem used.

Show worked answer →

(a) The angle at the centre is twice the angle at the circumference standing on the same arc $AC$ .

So angle $ABC = \dfrac{1}{2} \times 130^\circ = 65^\circ$ .

(b) The theorem used is that the angle at the centre is twice the angle at the circumference subtended by the same arc.

Markers reward halving the centre angle, the value $65^\circ$ , and naming the centre-circumference theorem.

Original4 marks

ABCD

is a cyclic quadrilateral. Angle

ABC = 95^\circ

and angle

BCD = 70^\circ

. Find angle

ADC