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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.

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