Circle Theorems - Mathematics

The perpendicular bisector of a chord always goes through the centre of the circle.

• Draw the perpendicular from centre, O, to the chord AB.
• Draw a radius to A and another to B to form two isosceles triangles.
• Their hypotenuses are the same length (because they are both radii) and they share another edge so the triangles are congruent by the ‘RHS’ rule.
• Therefore AM = BM and so the perpendicular splits the chord exactly in half.

Angle at centre is double the angle at the circumference

• Split the shape into two isosceles triangles and label the angles.
• The angle at the centre is 360° - x - y and the angle at the circumference is a + b.
• Split the shape into two isosceles triangles and label the angles.
• The angle at the centre is 360° - x - y and the angle at the circumference is a + b.

Angle in a semicircle is 90°

• Split the triangle into two triangles which are both isosceles since they both have two sides which are radii.
• Mark one of the angles at the centre x.
• y = 1⁄2(180° - x) since the triangle is isosceles and all angles add up to 180°.
• Similarly z = 1⁄2(180° - (180° - x)) = 1⁄2x
• Therefore the angle at the circumference is z + y = 1⁄2 × 180° = 90° as required.

• Alternatively, using the previous theorem we see that the angle at the centre is twice the angle at the circumference.
• So 180° is twice the angle at the circumference so the angle is 90°.

• Opposite angles in a cyclic quadrilateral add up to 180°.
  • Here x + y = 180°

• Label the opposite angles, then draw two radii from the other corners.
• Use ‘angle at centre twice that at the circumference’ to get the angles at the centre as 2x, 2y.
• Since angles at a point add up to 360°:
  • 2x + 2y = 360° and so x + y = 180°.

Two tangents to the same circle from the same point (C) will be the same length. Here AC = BC

• These are congruent triangles (by using the 'RHS' rule).
• They share a common side (length x) and another side which is a radius.
• Therefore the remaining side, the tangent, is the same length for both.

All triangles drawn from a chord in the same segment will have the same angle at the circumference.

• Draw in radii from the ends of the chord.
• Label the angle at the centre.
• Using ‘angle at centre twice that at the circumference’ we see that 
• z = 2x, z = 2y and so x = y.

The angle between a tangent and a chord is always equal to the angle at the circumference of any triangle formed from the chord.

• Draw in a diameter from the point where the tangent touches. Connect it to the other end of the chord.
• Using ‘radius and tangent meet at a right angle’ we get a right angle where the diameter meets the tangent.
• Using ‘angle in a semicircle is 90°’ we get that the angle at the other end of the chord is a right angle too.
• We find that the angle at the bottom of the diameter is the same as x.
• Using ‘angles in same segment’ we get that x = y.