Q.Prove that if a perpendicular is drawn from the center of a circle to a chord that is not a diameter, then it bisects the chord.

Let AB be a chord of a circle with center O, and let OD be the perpendicular drawn from O to the chord AB. To prove: OD bisects the chord AB, i.e., AD = DB. Construction: Join O to A and O to B. Proof: In triangles △OAD and △OBD: - ∠ODA = ∠ODB [Both are right angles] - OA = OB [Radii of the same circle] - OD is the common side ∴ △OAD ≅ △OBD [By RHS congruence rule] ∴ AD = DB [Corresponding sides of congruent triangles] Proved.
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