Circle theorems

• The angle at the centre is double the angle at the circumference.
• Angles in the same segment are equal.

  
• Every corner of a cyclic quadrilateral must touch the circumference of the circle.
• Opposite angles in a cyclic quadrilateral add up to 180°.
• The angle at the circumference in a semicircle is a right angle.
• Proof of angles in a semi-circle

  • The angle on a straight line is 180°. The angle VOY = 180°.
  • The angle at the centre is double the angle at the circumference.
  • Angle VWY = half of angle VOY = half×180.
  • Angle VWY = 90°

  
• Proof of angles in a cyclic quadrilateral

  • The angle at the centre is double the angle at the circumference.
  • Angle COE = 2𝑦 and the reflex angle COE = 2𝑥.
  • Angles around a point add to 360°.
  • 2y + 2x = 360°
  • $\frac{2y}{2}+$$\frac{2x}{2}=$$\frac{360}{2}$
  • So y + x = 180°

  
• Proof of angles in the same segment

  • Using the circle theorem, the angle at the centre is twice the angle at the circumference.
  • Angle MNQ = 𝑥 and angle MPQ = 𝑥.
  • Therefore angle MNQ = angle MPQ.

  
• Proof of angles at the centre and circumference

  • Angle OGH (𝑦) = angle OHG (triangle GOH is an isosceles).
  • Angle OGK (𝑥) = angle OKG (triangle GOK is also isosceles).
  • Lengths OK, OG and OH are all radii.
  • Angle GOH = 180−2𝑦 (angles in a triangle add up to 180°)
  • Angle GOK = 180−2𝑥 (angles in a triangle add up to 180°)
  • Angle JOH = 2𝑦 (angles on a straight line add up to 180° 180−2𝑦+2𝑦=180)
  • Angle JOK = 2𝑥 (angles on a straight line add up to 180°)
  • The angle at the centre KOH (2𝑦+2𝑥) is double the angle at the circumference KGH (𝑥+𝑦).

  
• The angle between a tangent and a radius is 90°.
• Tangents which meet at the same point are equal in length.
• Proof of tangents are equal length when they meet

  • Draw the line OB. It creates two triangles OCB and OAB. These share the length OB.
  • Triangles OCB and OAB are congruent because of the SAS rule.
  • Two of the sides are the same length: OB = OB and OC = OA
  • One of the angles is equal in size: OCB = OAB
  • Congruent triangles are identical.
  • So length CB = AB.

  
• Proof of angles between tangents and radii is 90°

  • The angle between the tangent and the radius is 90°.
  • Angle BCO = angle BAO = 90°
  • AO and OC are both radii of the circle.
  • Length AO = Length OC

  
• The perpendicular from the centre of a circle to a chord bisects the chord.

  
• Proof of chords

  • Angles OMA and OMB are both right angles.
  • OA is the hypotenuse of triangle OAM.
  • OB is the hypotenuse of triangle OBM.
  • OA = OB as both are radii of the circle.
  • OM is common to both triangles.
  • Therefore, triangles OAM and OMB are congruent by RHS – right angle, hypotenuse, side).
  • Therefore, the remaining sides of the triangles are equal, AM = MB.
  • So, M must be the mid-point of AB, and the chord has been bisected.

  
• The angle between a tangent and a chord is equal to the angle in the alternate segment.

  
• Proof of the alternate segment theorem

  • The angle between a tangent and the radius is 90°.
  • Angle BDO = 90−𝑥
  • Triangle DOB is an isosceles triangle so angle DBO is 90−𝑥.
  • Angles in a triangle add up to 180°.
  • Angle DOB = 180−BDO−DBO
  • Angle DOB = 180−(90−𝑥)−(90−𝑥)=2𝑥
  • The angle at the centre is double the angle at the circumference.
  • Angle DAB = 𝑥
  • Therefore BDC = DAB.