Proof of the Circle Theorems

What you'll learn

How to prove circle theorems using only basic geometry facts.
Why equal radii create isosceles triangles.
How the “angle at the centre is twice the angle at the circumference” theorem leads to several others.
How to write clear GCSE proof steps for Grade 8/9 marks.

1. The basic ingredients

Before proving circle theorems, you need a few reliable facts.

Definition

Key circle words

Basic circle vocabulary: centre, radius, chord, diameter, tangent, and a subtended angle.

The centre is the fixed point in the middle of the circle, usually labelled $O$ .
A radius is a line from the centre to the circumference.
A chord is a straight line joining two points on the circumference.
A diameter is a chord through the centre.
A tangent is a straight line that touches the circle at exactly one point.
An angle is subtended by a chord or arc if its arms meet the two ends of that chord or arc.

The most important non-circle-theorem fact is: all radii of the same circle are equal. So triangles made from two radii are isosceles.

Key Idea

The proof engine

Most circle theorem proofs start by drawing radii, then using isosceles triangle base angles and angles in a triangle.

Example

Using radii to create equal angles

Points $A$ and $B$ lie on a circle with centre $O$ . Prove that $∠OAB=∠ABO\angle OAB = \angle ABO$ .

Triangle AOB is isosceles because OA and OB are equal radii of the same circle.

Since $O A$ and $O B$ are both radii of the same circle, $O A = O B$ .
Therefore triangle $A O B$ is isosceles.
In an isosceles triangle, the base angles are equal, so $∠OAB=∠ABO\angle OAB = \angle ABO$ .

2. Angle at the centre theorem

The first big proof is the one that powers many others.

Definition

Angle at the centre theorem

The angle at the centre of a circle is twice the angle at the circumference when both angles stand on the same arc.

Suppose $A$ , $B$ and $C$ are on a circle with centre $O$ . To prove that $∠AOC=2∠ABC\angle AOC = 2\angle ABC$ , you draw $O B$ and use two isosceles triangles.

Example

Proving the centre angle is double

Points $P$ , $Q$ and $R$ are on a circle with centre $O$ . Prove that the angle at the centre standing on arc $P R$ is twice the angle at $Q$ .

Joining OQ splits the proof into two isosceles triangles, OPQ and ORQ.

Join $O Q$ . Let $∠PQO=x\angle PQO = x$ and $∠OQR=y\angle OQR = y$ .
Since $O P = O Q$ , triangle $O P Q$ is isosceles, so $∠OPQ=x\angle OPQ = x$ .
Since $O R = O Q$ , triangle $O R Q$ is isosceles, so $∠ORQ=y\angle ORQ = y$ .
In triangle $O P Q$ , $∠POQ=180∘−2x\angle POQ = 180^\circ - 2x$ .
In triangle $O R Q$ , $∠QOR=180∘−2y\angle QOR = 180^\circ - 2y$ .
Angles around point $O$ add to 360°, so the angle at the centre standing on arc $P R$ is:
$360∘−(180∘−2x)−(180∘−2y)=2x+2y360^\circ - (180^\circ - 2x) - (180^\circ - 2y) = 2x + 2y$
The angle at the circumference is $∠PQR=x+y\angle PQR = x + y$ .
Therefore the angle at the centre is $2 (x + y)$ , which is twice the angle at the circumference.

Common Mistake

Using the theorem to prove itself

If the question says “do not use circle theorems”, you must not write “angle at centre is twice angle at circumference”. Prove it from radii and triangle angle facts instead.

3. Angle in a semicircle

A semicircle is half a circle. If a triangle has one side as the diameter of a circle, the angle opposite the diameter is a right angle.

Example

Proving the angle in a semicircle

Points $A$ , $B$ and $C$ lie on a circle with centre $O$ , and $A C$ is a diameter. Prove that $∠ABC=90∘\angle ABC = 90^\circ$ without using circle theorems.

A triangle with AC as a diameter can be split into two isosceles triangles using OB.

Join $O B$ . Let $∠ABO=x\angle ABO = x$ and $∠OBC=y\angle OBC = y$ .
Since $O A = O B$ , triangle $A O B$ is isosceles, so $∠BAO=x\angle BAO = x$ .
Since $O B = O C$ , triangle $B O C$ is isosceles, so $∠BCO=y\angle BCO = y$ .
In triangle $A O B$ , $∠AOB=180∘−2x\angle AOB = 180^\circ - 2x$ .
In triangle $B O C$ , $∠BOC=180∘−2y\angle BOC = 180^\circ - 2y$ .
Since $A O C$ is a straight line, $∠AOB+∠BOC=180∘\angle AOB + \angle BOC = 180^\circ$ .
So $(180∘−2x)+(180∘−2y)=180∘(180^\circ - 2x) + (180^\circ - 2y) = 180^\circ$ , giving $x+y=90∘x+y=90^\circ$ .
Since $∠ABC=x+y\angle ABC = x+y$ , $∠ABC=90∘\angle ABC = 90^\circ$ .

Tip

Spot the diameter

If a line passes through the centre and both ends are on the circle, it is a diameter. The angle opposite it on the circumference will be 90°.

The diameter is the side opposite the right angle on the circumference.

4. Angles in the same segment

A segment is the region between a chord and an arc. Angles in the same segment stand on the same chord and sit on the same side of it.

Example

Proving angles in the same segment are equal

Points $A$ , $B$ , $C$ and $D$ lie on a circle. Angles $∠ABD\angle ABD$ and $∠ACD\angle ACD$ both stand on chord $A D$ . Prove that they are equal.

Angles ABD and ACD are in the same segment because they stand on the same chord AD.

Draw the radii $O A$ and $O D$ .
The angle at the centre standing on chord $A D$ is $∠AOD\angle AOD$ .
By the angle at the centre theorem, $∠AOD=2∠ABD\angle AOD = 2\angle ABD$ .
By the same theorem, $∠AOD=2∠ACD\angle AOD = 2\angle ACD$ .
Since both angles are half of the same central angle, $∠ABD=∠ACD\angle ABD = \angle ACD$ .

5. Opposite angles in a cyclic quadrilateral

Definition

Cyclic quadrilateral

A cyclic quadrilateral is a four-sided shape whose four vertices all lie on the circumference of a circle.

The theorem says: opposite angles in a cyclic quadrilateral add to 180°.

Example

Proving opposite angles add to 180°

A quadrilateral $A B C D$ is drawn inside a circle, with its vertices on the circumference. Prove that $∠ABC+∠ADC=180∘\angle ABC + \angle ADC = 180^\circ$ .

In a cyclic quadrilateral, opposite angles such as ABC and ADC add to 180°.

Join $O A$ and $O C$ .
Angle $∠ABC\angle ABC$ stands on the arc from $A$ to $C$ that does not contain $B$ .
Angle $∠ADC\angle ADC$ stands on the other arc from $A$ to $C$ .
The two central angles standing on these two arcs add to 360° because they make a full turn around $O$ .
Each circumference angle is half its matching central angle.
Therefore $∠ABC+∠ADC\angle ABC + \angle ADC$ is half of 360°, so $∠ABC+∠ADC=180∘\angle ABC + \angle ADC = 180^\circ$ .

Common Mistake

Only opposite angles

In a cyclic quadrilateral, opposite angles add to 180°. Adjacent angles do not necessarily add to 180°.

6. The tangent and chord theorem

The tangent-chord theorem is often called the alternate segment theorem.

It says: the angle between a tangent and a chord equals the angle in the opposite segment.

You also need the fact that a radius meets a tangent at 90°.

Example

Proving the alternate segment theorem

A tangent touches a circle at $C$ . Points $A$ and $B$ are also on the circumference, and chord $B C$ is drawn. Prove that the angle between the tangent and $B C$ equals $∠BAC\angle BAC$ .

The radius to the point of contact is perpendicular to the tangent, and the tangent-chord angle matches the angle in the opposite segment.

Join $O C$ and $O B$ .
The radius $O C$ is perpendicular to the tangent, so the angle between $O C$ and the tangent is 90°.
Let $∠OCB=y\angle OCB = y$ . Then the angle between the tangent and chord $B C$ is $90∘−y90^\circ - y$ .
Since $O B = O C$ , triangle $O B C$ is isosceles, so $∠OBC=y\angle OBC = y$ .
In triangle $O B C$ , $∠BOC=180∘−2y\angle BOC = 180^\circ - 2y$ .
By the angle at the centre theorem, $∠BAC\angle BAC$ is half of $∠BOC\angle BOC$ .
So $∠BAC=90∘−y\angle BAC = 90^\circ - y$ .
Therefore the angle between the tangent and chord $B C$ equals $∠BAC\angle BAC$ .

Exam technique

In the exam

Draw in the centre and any missing radii before starting the proof.
Use named basic facts clearly: “radii are equal”, “base angles in an isosceles triangle are equal”, and “angles in a triangle add to 180°”.
If you use a theorem already proved, state it precisely and make sure it applies to the correct chord or arc.

Self review

Check yourself

Can you prove the angle at the centre theorem without quoting any circle theorem?
In a cyclic quadrilateral, which pair of angles must add to 180°?
Why does drawing a radius to the point of contact help in a tangent proof?

Proof of the Circle Theorems

What you'll learn

How to prove circle theorems using only basic geometry facts.
Why equal radii create isosceles triangles.
How the “angle at the centre is twice the angle at the circumference” theorem leads to several others.
How to write clear GCSE proof steps for Grade 8/9 marks.

1. The basic ingredients

Before proving circle theorems, you need a few reliable facts.

Definition

Key circle words

Basic circle vocabulary: centre, radius, chord, diameter, tangent, and a subtended angle.

The centre is the fixed point in the middle of the circle, usually labelled $O$ .
A radius is a line from the centre to the circumference.
A chord is a straight line joining two points on the circumference.
A diameter is a chord through the centre.
A tangent is a straight line that touches the circle at exactly one point.
An angle is subtended by a chord or arc if its arms meet the two ends of that chord or arc.

The most important non-circle-theorem fact is: all radii of the same circle are equal. So triangles made from two radii are isosceles.

Key Idea

The proof engine

Most circle theorem proofs start by drawing radii, then using isosceles triangle base angles and angles in a triangle.

Example

Using radii to create equal angles

Points $A$ and $B$ lie on a circle with centre $O$ . Prove that $∠OAB=∠ABO\angle OAB = \angle ABO$ .

Triangle AOB is isosceles because OA and OB are equal radii of the same circle.

Since $O A$ and $O B$ are both radii of the same circle, $O A = O B$ .
Therefore triangle $A O B$ is isosceles.
In an isosceles triangle, the base angles are equal, so $∠OAB=∠ABO\angle OAB = \angle ABO$ .

2. Angle at the centre theorem

The first big proof is the one that powers many others.

Definition

Angle at the centre theorem

The angle at the centre of a circle is twice the angle at the circumference when both angles stand on the same arc.

Suppose $A$ , $B$ and $C$ are on a circle with centre $O$ . To prove that $∠AOC=2∠ABC\angle AOC = 2\angle ABC$ , you draw $O B$ and use two isosceles triangles.

Example

Proving the centre angle is double

Points $P$ , $Q$ and $R$ are on a circle with centre $O$ . Prove that the angle at the centre standing on arc $P R$ is twice the angle at $Q$ .

Joining OQ splits the proof into two isosceles triangles, OPQ and ORQ.

Join $O Q$ . Let $∠PQO=x\angle PQO = x$ and $∠OQR=y\angle OQR = y$ .
Since $O P = O Q$ , triangle $O P Q$ is isosceles, so $∠OPQ=x\angle OPQ = x$ .
Since $O R = O Q$ , triangle $O R Q$ is isosceles, so $∠ORQ=y\angle ORQ = y$ .
In triangle $O P Q$ , $∠POQ=180∘−2x\angle POQ = 180^\circ - 2x$ .
In triangle $O R Q$ , $∠QOR=180∘−2y\angle QOR = 180^\circ - 2y$ .
Angles around point $O$ add to 360°, so the angle at the centre standing on arc $P R$ is:
$360∘−(180∘−2x)−(180∘−2y)=2x+2y360^\circ - (180^\circ - 2x) - (180^\circ - 2y) = 2x + 2y$
The angle at the circumference is $∠PQR=x+y\angle PQR = x + y$ .
Therefore the angle at the centre is $2 (x + y)$ , which is twice the angle at the circumference.

Common Mistake

Using the theorem to prove itself

If the question says “do not use circle theorems”, you must not write “angle at centre is twice angle at circumference”. Prove it from radii and triangle angle facts instead.

3. Angle in a semicircle

A semicircle is half a circle. If a triangle has one side as the diameter of a circle, the angle opposite the diameter is a right angle.

Example

Proving the angle in a semicircle

Points $A$ , $B$ and $C$ lie on a circle with centre $O$ , and $A C$ is a diameter. Prove that $∠ABC=90∘\angle ABC = 90^\circ$ without using circle theorems.

A triangle with AC as a diameter can be split into two isosceles triangles using OB.

Join $O B$ . Let $∠ABO=x\angle ABO = x$ and $∠OBC=y\angle OBC = y$ .
Since $O A = O B$ , triangle $A O B$ is isosceles, so $∠BAO=x\angle BAO = x$ .
Since $O B = O C$ , triangle $B O C$ is isosceles, so $∠BCO=y\angle BCO = y$ .
In triangle $A O B$ , $∠AOB=180∘−2x\angle AOB = 180^\circ - 2x$ .
In triangle $B O C$ , $∠BOC=180∘−2y\angle BOC = 180^\circ - 2y$ .
Since $A O C$ is a straight line, $∠AOB+∠BOC=180∘\angle AOB + \angle BOC = 180^\circ$ .
So $(180∘−2x)+(180∘−2y)=180∘(180^\circ - 2x) + (180^\circ - 2y) = 180^\circ$ , giving $x+y=90∘x+y=90^\circ$ .
Since $∠ABC=x+y\angle ABC = x+y$ , $∠ABC=90∘\angle ABC = 90^\circ$ .

Tip

Spot the diameter

If a line passes through the centre and both ends are on the circle, it is a diameter. The angle opposite it on the circumference will be 90°.

The diameter is the side opposite the right angle on the circumference.

4. Angles in the same segment

A segment is the region between a chord and an arc. Angles in the same segment stand on the same chord and sit on the same side of it.

Example

Proving angles in the same segment are equal

Points $A$ , $B$ , $C$ and $D$ lie on a circle. Angles $∠ABD\angle ABD$ and $∠ACD\angle ACD$ both stand on chord $A D$ . Prove that they are equal.

Angles ABD and ACD are in the same segment because they stand on the same chord AD.

Draw the radii $O A$ and $O D$ .
The angle at the centre standing on chord $A D$ is $∠AOD\angle AOD$ .
By the angle at the centre theorem, $∠AOD=2∠ABD\angle AOD = 2\angle ABD$ .
By the same theorem, $∠AOD=2∠ACD\angle AOD = 2\angle ACD$ .
Since both angles are half of the same central angle, $∠ABD=∠ACD\angle ABD = \angle ACD$ .

5. Opposite angles in a cyclic quadrilateral

Definition

Cyclic quadrilateral

A cyclic quadrilateral is a four-sided shape whose four vertices all lie on the circumference of a circle.

The theorem says: opposite angles in a cyclic quadrilateral add to 180°.

Example

Proving opposite angles add to 180°

A quadrilateral $A B C D$ is drawn inside a circle, with its vertices on the circumference. Prove that $∠ABC+∠ADC=180∘\angle ABC + \angle ADC = 180^\circ$ .

In a cyclic quadrilateral, opposite angles such as ABC and ADC add to 180°.

Join $O A$ and $O C$ .
Angle $∠ABC\angle ABC$ stands on the arc from $A$ to $C$ that does not contain $B$ .
Angle $∠ADC\angle ADC$ stands on the other arc from $A$ to $C$ .
The two central angles standing on these two arcs add to 360° because they make a full turn around $O$ .
Each circumference angle is half its matching central angle.
Therefore $∠ABC+∠ADC\angle ABC + \angle ADC$ is half of 360°, so $∠ABC+∠ADC=180∘\angle ABC + \angle ADC = 180^\circ$ .

Common Mistake

Only opposite angles

In a cyclic quadrilateral, opposite angles add to 180°. Adjacent angles do not necessarily add to 180°.

6. The tangent and chord theorem

The tangent-chord theorem is often called the alternate segment theorem.

It says: the angle between a tangent and a chord equals the angle in the opposite segment.

You also need the fact that a radius meets a tangent at 90°.

Example

Proving the alternate segment theorem

A tangent touches a circle at $C$ . Points $A$ and $B$ are also on the circumference, and chord $B C$ is drawn. Prove that the angle between the tangent and $B C$ equals $∠BAC\angle BAC$ .

The radius to the point of contact is perpendicular to the tangent, and the tangent-chord angle matches the angle in the opposite segment.

Join $O C$ and $O B$ .
The radius $O C$ is perpendicular to the tangent, so the angle between $O C$ and the tangent is 90°.
Let $∠OCB=y\angle OCB = y$ . Then the angle between the tangent and chord $B C$ is $90∘−y90^\circ - y$ .
Since $O B = O C$ , triangle $O B C$ is isosceles, so $∠OBC=y\angle OBC = y$ .
In triangle $O B C$ , $∠BOC=180∘−2y\angle BOC = 180^\circ - 2y$ .
By the angle at the centre theorem, $∠BAC\angle BAC$ is half of $∠BOC\angle BOC$ .
So $∠BAC=90∘−y\angle BAC = 90^\circ - y$ .
Therefore the angle between the tangent and chord $B C$ equals $∠BAC\angle BAC$ .

Exam technique

In the exam

Draw in the centre and any missing radii before starting the proof.
Use named basic facts clearly: “radii are equal”, “base angles in an isosceles triangle are equal”, and “angles in a triangle add to 180°”.
If you use a theorem already proved, state it precisely and make sure it applies to the correct chord or arc.

Self review

Check yourself

Can you prove the angle at the centre theorem without quoting any circle theorem?
In a cyclic quadrilateral, which pair of angles must add to 180°?
Why does drawing a radius to the point of contact help in a tangent proof?

Proof of the Circle Theorems

What you'll learn

1. The basic ingredients

2. Angle at the centre theorem

3. Angle in a semicircle

4. Angles in the same segment

5. Opposite angles in a cyclic quadrilateral

6. The tangent and chord theorem

Recap questions

You've reached the end

Proof of the Circle Theorems

What you'll learn

1. The basic ingredients

2. Angle at the centre theorem

3. Angle in a semicircle

4. Angles in the same segment

5. Opposite angles in a cyclic quadrilateral

6. The tangent and chord theorem

Recap questions

You've reached the end

Proof of the Circle Theorems

What you'll learn

1. The basic ingredients

2. Angle at the centre theorem

3. Angle in a semicircle

4. Angles in the same segment

5. Opposite angles in a cyclic quadrilateral

6. The tangent and chord theorem

Recap questions

You've reached the end

Proof of the Circle Theorems Revision Guide

Proof of the Circle Theorems

What you'll learn

1. The basic ingredients

2. Angle at the centre theorem

3. Angle in a semicircle

4. Angles in the same segment

5. Opposite angles in a cyclic quadrilateral

6. The tangent and chord theorem

Recap questions

You've reached the end

Proof of the Circle Theorems Revision Guide