Circle Theorem Proofs (Edexcel GCSE Maths: Higher): Revision Note

Circle theorem proofs

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 $x$ and $y$.

  • The angle you are trying to prove is 90° is now $x+y$

• STEP 3
  Label the remaining angles in each of the isosceles triangles with algebraic expressions in terms of $x$ and $y$

  • Angles at the base of an isosceles triangle are equal

    • Therefore the two remaining angles at the circumference are also $x$ and $y$

  • Angles in a triangle add up to 180°

    • Therefore each angle at the centre will be labelled with the expression $180-2x$ and $180-2y$

• STEP 4
  The angles at the centre lie on a diameter, which is a straight line, therefore $180-2x+180-2y=180$

  • Rearrange this equation to show that $x+y=90°$

• In STEP 2 you already labeled the angle at the circumference as $x+y$, 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 $\theta +180-2x+180-2y=360$

  • Rearrange this equation to show that $\theta =2\left(x+y\right)$

• In STEP 2 you already labelled the angle at the circumference as $x+y$, so this proves that the angle at the centre is twice the angle at the circumference

  • Give clear reasons throughout your proof, using the keywords 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 $2x$

• By the circle theorem "The angle and the centre is twice the angle at the circumference", any angle at the circumference can be labelled $x$

• 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 $2x$ and $2y$

  • The angles $2x$ and $2y$ are at a point, so they add up to 360°

  • Therefore $2x+2y=360°$ which can be simplified to $x+y=180°$

• 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 $\frac{1}{2}\left(2x\right)=x$ and $\frac{1}{2}\left(2y\right)=y$

  • We have already shown that $x+y=180°$

• 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°'