The circle theorems can be proved using knowledge of basic angle facts and properties of 2D shapes. If you are asked to prove a circle theorem, adding in any radii and finding all equal angles will help. There are two types of proofs that can be used for the different circle theorems. 

How do I prove circle theorems using radii to form isosceles triangles?

• This type of proof can be used to prove the following circle theorems
  • The angle in a semicircle is always 90°
  • The angle at the centre is twice the angle at the circumference
  • Angles in the same segment are equal
  • Opposite angles in a cyclic quadrilateral add up to 180°

How do I prove that the angle in a semicircle is 90°?

• This circle theorem states that an angle subtended at the circumference of a semicircle is always a right angle
• Although it is a special case of the angle at the centre and circumference circle theorem, it can be proved without using any other circle theorems
• STEP 1
  Draw a radius from the centre of the circle to the angle subtended at the circumference
  • This will form two  isosceles triangles
• STEP 2
  Label the two angles formed at the angle subtended at the circumference and 
• The angle you are trying to prove is 90° is now 
• STEP 3
  Label the remaining angles in each of the isosceles triangles with algebraic expressions in terms of and 
• Angles at the base of an isosceles triangle are equal
  • Therefore the two remaining angles at the circumference are also  and 
• Angles in a triangle add up to 180°
  • • Therefore each angle at the centre will be labelled with the expression  and 

• STEP 4
  The angles at the centre lie on a diameter, which is a straight line, therefore 
• Rearrange this equation to show that 
• In STEP 2 you already labeled the angle at the circumference as , so this proves that the angle at the circumference equals 90°
• Give clear reasons throughout your proof, using the key words given in bold
  

How do I prove that the angle at the centre is twice the angle at the circumference?

• This circle theorem states that an angle subtended at the centre of a circle is twice the angle subtended at the circumference of a circle from the same arc
• It does not need any other circle theorems to prove it
• STEPS 1, 2 and 3 are the same as above

• STEP 4
  The three angles formed at the centre lie at a point, therefore they will add to 360°
  • Label the third angle at the centre θ, then you can form the equation
  • Rearrange this equation to show that 
• In STEP 2 you already labeled the angle at the circumference as , so this proves that the angle at the centre is twice the angle at the circumference
• Give clear reasons throughout your proof, using the key words given in bold
• This circle theorem is also a more generic version of the circle theorem the angle in a semicircle is always 90° 
  • You could be asked to prove either without the use of any circle theorems

How do I prove that angles at the circumference from the same arc are equal?

• This circle theorem states that any angles subtended at the circumference of a circle from the same arc are equal 
• This theorem is proved using the circle theorem an angle subtended at the centre of a circle is twice the angle subtended at the circumference of a circle
• Draw the radii from the centre of the circle to the points on the circumference forming the arc which the angles are subtended from
  • This will form an angle at the centre, label this angle 
• By the circle theorem “The angle and the centre is twice the angle at the circumference”, any angle at the circumference can be labelled

• You do not need to prove the circle theorem you have used in this proof, but you must give clear reasons

How do I prove that opposite angles in a cyclic quadrilateral add up to 180°?

• This theorem is proved using the circle theorem “An angle subtended at the centre of a circle is twice the angle subtended at the circumference of a circle”
• STEP 1 Draw the radii from the centre of the circle to any two of the vertices of the cyclic quadrilateral that are opposite each other
  • This will form two angles at the centre, label these angles and 
  • The angles  and  are at a point, so they add up to 360°
  • Therefore  which can be simplified to
• STEP 2 By the circle theorem “The angle and the centre is twice the angle at the circumference”, the two angles at the circumference can be labelled and 
  • We have already shown that

• You do not need to prove the circle theorem you have used in this proof, but you must give clear reasons

How do I prove circle theorems involving chords and tangents?

• This type of proof can be used to prove the following circle theorems
  • The perpendicular from the centre of a circle bisects a chord
  • The tangent to a circle meets the radius at 90°
  • The alternate segment theorem
• These proofs can be more tricky and will use other circle theorems within them
• The questions will often guide you through the proof
• The circle theorem “The perpendicular from the centre of a circle bisects a chord’ can be proved using congruent triangles
• If the radius is perpendicular to the chord, two right-angled triangles will be formed  
  • Prove that these two triangle are congruent using the RHS (right angle, hypotenuse, side) rule
    • The chord and the radius are perpendicular, therefore both triangles have a right angle
    • The hypotenuse is the line from the centre to the circumference, therefore both triangles have an equal hypotenuse
    • The line joining the chord to the centre is shared between both triangles, therefore this is a same side in both triangles
    • Therefore, by RHS, the two triangles are congruent

• • If the two triangles are congruent, then all three sides will be the same and so the perpendicular must bisect the chord
• The proof for the circle theorem “The tangent to a circle meets the radius at a right angle” uses proof by contradiction and involves assuming that they do not meet at 90° and proving that this is not possible
• The proof for the alternate segment theorem uses the circle theorems ‘the angle in a semicircle is always 90°’ and ‘the tangent to a circle meets the radius at 90°’