QQuestion
Prove that the perpendicular drawn to a chord which is not a diameter, from the center of the circle, bisects the chord.
✓Answer
Given : AB is a chord of the circle with its centre at O, which is not a diameter, and OD, is perpendicular on the chord AB.
To prove : OD, bisects the chord AB i.e., AD = DB.
Construction : O, A and O, B are joined.
Proof : In △ODA and △ODB,
∠ODA = ∠ODB (Each is right-angle)
OA = OB [radii of same circle] and
OD = OD (common side)
∴ △ODA ≅ △ODB [By R-H -S axiom of congruency]
∴ AD = DB [Corresponding sides of the congruent triangles] [proved]
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