The perpendicular from the centre of a circle to a chord bisect the ch

1.	The perpendicular from the centre of a circle to a chord bisect the chord. Prove it.
Answer» Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.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\xa0(Since OB ⊥ AC)OA = OC (Radii of the same circle)OB = OB (Common side)ΔOBA\xa0≅ 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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