Given: A circle with centre O. AC is a chord and OB ⊥ AC.
To prove: AB = BC.
Construction: Join OA and OC.
Proof: In triangles OBA and OBC,
∠OBA = ∠OBC = 90o (Since OB ⊥ AC)
OA = OC (Radii of the same circle)
OB = OB (Common side)
ΔOBA ≅ ΔOBC (By RHS congruence rule)
⇒ AB = BC (Corresponding sides of congruent triangles)
Thus, OB bisects the chord AC.
Hence, the theorem is proved.