#### theorem

**Theorem :** The perpendicular from the centre of a circle to a chord bisects the chord.

Draw a circle. Let O be its centre. Draw a chord AB. Let the crease cut AB at the point M. Then, ∠ OMA = ∠ OMB = 90° or OM is perpendicular to AB. So MA = MB.**Theorem :** The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Let AB be a chord of a circle with centre O and O is joined to the mid -point M of AB.**To prove:** OM⊥ AB.

**Proof:** Join OA and OB

In triangles OAM and OBM,

OA =OB (radii of the same circle)

AM =BM (M is the midpoint of PQ)

OM =OM (common)

Therefore, ∆OAM ≅ ∆OBM (SSS congruence rule)

This gives , ∠OMA = ∠OMB = 90°

(Corresponding parts of congruent triangles are congruent).