3.2.4

Circle Theorems - Perpendicular Bisector

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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.
Jump to other topics
1

Proof

2

Algebra & Functions

2.1

Powers & Roots

2.2

Quadratic Equations

2.3

Inequalities

2.4

Polynomials

2.5

Graphs

2.6

Functions

2.7

Transformation of Graphs

2.8

Partial Fractions (A2 Only)

3

Coordinate Geometry

4

Sequences & Series

5

Trigonometry

6

Exponentials & Logarithms

7

Differentiation

8

Integration

9

Numerical Methods

10

Vectors

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