Prove that the line drawn through the centre of a circle which bisects a chord is perpendicular to the chord

Given, a circle with center O, in which line OM bisect the chord AB, i.e., AM = MB.

To prove: OM ⊥ AB

Construction: Join OA and OB.

Proof:

In triangle OAM and OBM, we have

OA = OB  (radii of same circle)

OM = OM  (common)

Also, AM = BM  (given)

⇒∠OMA = ∠OMB  [c.p.c.t]

Now, ∠OMA + ∠OMB = 180°  [linear pair]

⇒ 2∠OMA = 180°

⇒ ∠OMA = 90°

Hence, OM ⊥ AB.