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Circle Theorems

Perpendicular bisector goes through centre

Perpendicular bisector goes through centre

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

Proof

  • Draw the perpendicular from centre, O, to the chord AB.
  • Draw a radius to A and another to B to form two triangles.
Proof continued

Proof continued

  • 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.

Circle Theorems

Proof

Proof

  • 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.
Proof continued

Proof continued

  • Since the angles in a triangle add up to 180°:
    a = 12(180° - y), b = 12(180° - x)
  • So angle at circumference = a + b = 12(360° - x - y)
    = 12 × angle at centre.

Circle Theorems

Proof

Proof

  • 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.
Proof continued

Proof continued

  • y = 12(180° - x) since the triangle is isosceles and all angles add up to 180°.
  • Similarly z = 12(180° - (180° - x)) = 12x
  • Therefore the angle at the circumference is z + y = 12 × 180° = 90° as required.
Alternative proof

Alternative proof

  • 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°.
Jump to other topics
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Numbers

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Geometry

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Vectors & Transformation Geometry

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