Angle at the Centre and Angles in a Circle | IEB Grade 11 Maths

The angle at the centre theorem is the key to most circle angle results: angles in the same segment and the angle in a semicircle both follow from it. Its proof is examinable in Grade 11 and Grade 12 (acute-angle case).

Angle at the centre is double the angle at the circumference

Theorem. The angle that an arc subtends at the centre is double the angle it subtends at any point on the circumference. (Reason: $∠$ at centre $= 2 \times ∠$ at circumference)

Proof. Join $C O$ and extend it to $D$ .

$O A = O C$ (radii), so $\hat{A} = \hat{C}_{1}$ ( $∠$ s opp equal sides)
$\hat{O}_{1} = \hat{A} + \hat{C}_{1} = 2 \hat{C}_{1}$ (ext $∠$ of $△$ )
Similarly $\hat{O}_{2} = 2 \hat{C}_{2}$
So $\hat{A O B} = \hat{O}_{1} + \hat{O}_{2} = 2 (\hat{C}_{1} + \hat{C}_{2}) = 2 \hat{A C B}$

Example. If the angle at the centre is $2 x + 10°$ and the angle at the circumference on the same arc is 50°, then $2 x + 10° = 100°$ and $x = 45°$ .

Angles in the same segment

Theorem. Angles subtended by the same arc in the same segment are equal. (Reason: $∠$ s in the same seg)

Both are half of the same angle at the centre.

Example. $A$ , $B$ , $C$ and $D$ lie on a circle. If $\hat{A D B} = 62°$ , then $\hat{A C B} = 62°$ , as long as $C$ and $D$ are on the same side of $A B$ .

The angle in a semicircle

Theorem. The angle subtended by a diameter at the circumference is 90°. (Reason: $∠$ in semi-circle)

This is the angle at the centre with a central angle of 180°.

Example. $B D$ is a diameter and $A$ is on the circle. If $\hat{A B D} = 28°$ , then $\hat{B A D} = 90°$ and $\hat{A D B} = 62°$ .

Equal chords subtend equal angles

Theorem. Equal chords (or equal arcs) subtend equal angles at the circumference. (Reason: equal chords; equal $∠$ s)

Example. $A$ , $B$ , $C$ and $D$ lie on a circle with $A B = B C$ . Then $\hat{A D B} = \hat{B D C}$ , so $D B$ bisects $\hat{A D C}$ .

The converse also holds: equal angles at the circumference stand on equal chords.

Key terms

Subtend: A chord or arc subtends an angle at a point when lines from its ends meet at that point.
Segment: The part of a circle cut off by a chord.
Semicircle: Half a circle, cut off by a diameter.

Common mistakes

Using the angle at the centre theorem for angles on different arcs.
Using 'angles in the same segment' for angles on opposite sides of the chord.
Writing reasons in your own words instead of the accepted form.
Assuming a chord is a diameter because it looks like one.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

$O$ is the centre of a circle. The angle at the centre is $2 x + 10°$ and the angle at the circumference subtended by the same arc is 50°. Calculate $x$ .

Worked solution

Hidden - try the question first

Solution

$2 x + 10° = 2 \times 50°$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
$2 x = 90°$
$x = 45°$

Question 2

5 marks total

routine

In the diagram, $O$ is the centre of the circle and $A$ , $B$ and $C$ lie on the circle. Prove the theorem that the angle subtended by an arc at the centre of a circle is double the angle it subtends at the circumference, that is, $\hat{A O B} = 2 \hat{A C B}$ .

Worked solution

Hidden - try the question first

Solution

Construction: join $C O$ and extend to $D$
$O A = O C$ (radii), so $\hat{A} = \hat{C}_{1}$
$\hat{O}_{1} = \hat{A} + \hat{C}_{1} = 2 \hat{C}_{1}$ (ext $∠$ of $△$ )
Similarly $\hat{O}_{2} = 2 \hat{C}_{2}$
$\hat{A O B} = 2 (\hat{C}_{1} + \hat{C}_{2}) = 2 \hat{A C B}$

Question 3

6 marks total

complex

$A$ , $B$ , $C$ and $D$ lie on a circle with centre $O$ . $B D$ is a diameter, and $A$ and $C$ lie on opposite sides of $B D$ . $\hat{A B D} = 28°$ .

Calculate, with reasons:

(a) $\hat{B A D}$ (1)

(b) $\hat{A D B}$ (1)

(d) $\hat{A O B}$ (2)

Worked solution

Hidden - try the question first

Solution

(a) $C O$ ( $∠$ in semi-circle)
(b) $O A = O C$ (sum of $∠$ s in $△$ )
(c) $\hat{A C B} = \hat{A D B}$ ( $∠$ s in the same seg)
$= 62°$
(d) $\hat{A O B} = 2 \hat{A D B}$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
$= 124°$

Question 4

8 marks total

problem-solving

In the diagram, $A$ , $B$ , $C$ and $D$ lie on a circle with $A B = B C$ . $B$ and $D$ lie on opposite sides of $A C$ . Chords $D B$ and $A C$ are drawn.

(a) Prove that $D B$ bisects $\hat{A D C}$ . (3)

(b) If $\hat{A D C} = 70°$ , calculate $\hat{A C B}$ , $\hat{B A C}$ and $\hat{A B C}$ . (5)

Worked solution

Hidden - try the question first

Solution

(a) $C O$ (given)
$\hat{A D B} = \hat{B D C}$ (equal chords; equal $∠$ s)
So $D B$ bisects $\hat{A D C}$
(b) $\hat{A D B} = 35°$
$\hat{A C B} = \hat{A D B} = 35°$ ( $∠$ s in the same seg)
$∠$ ( $∠$ s opp equal sides)
$\hat{A B C} = 180° - 70°$ (sum of $∠$ s in $△$ )
$= 110°$

FAQs

Is the angle at the centre proof examinable?

Yes, in Grade 11 and Grade 12, for the case where the centre lies inside the angle. The converse is not examinable.

Why is the angle in a semicircle 90°?

The diameter makes an angle of 180° at the centre, and the angle at the circumference is half of that.

How do I spot angles in the same segment?

Look for two angles whose arms end at the same two points on the circle, with both vertices on the same side of the chord.

Angle at the centre, angles in the same segment, angle in a semicircle

Angle at the centre is double the angle at the circumference

Angles in the same segment

The angle in a semicircle

Equal chords subtend equal angles

Key terms

Common mistakes

Practice questions

FAQs

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