The perpendicular from the centre

Join the centre C to the midpoint M of a chord AB. Something special happens.

Theorem 4: CM is perpendicular to AB.

• triangle CAB is isosceles, so ∠A = ∠B;
• AM = BM and CA = CB;
• so ΔCMA ≅ ΔCMB (SAS), and ∠CMA = ∠CMB;
• they sit on a straight line, so they add to 180°: each is 90°.

Theorem 5, the converse: the perpendicular from the centre to a chord bisects it. If ∠CMA = ∠CMB = 90°, the two right triangles have equal hypotenuses CA = CB and the shared side CM, so they are congruent (RHS) and AM = BM.

So the perpendicular bisector of any chord passes through the centre. (Draw two chords' bisectors and they cross at the centre.)

A right triangle appears every time: half the chord, the distance from the centre, and the radius. With radius 5 cm:

• a 6 cm chord is √(25 – 9) = 4 cm from the centre;
• an 8 cm chord is √(25 – 16) = 3 cm from the centre.