Prove that if a chord (which is not a diameter) of a circle is bisected by a straight line drawn from the center of the circle, then that line is perpendicular to the chord.

Q.Prove that if a chord (which is not a diameter) of a circle is bisected by a straight line drawn from the center of the circle, then that line is perpendicular to the chord.

Let AB be a chord of a circle with center O, such that AB is not a diameter. Let D be the midpoint of AB, i.e., AD = DB. To Prove: OD ⊥ AB, i.e., OD is perpendicular to chord AB. Construction: Join OA and OB. Proof: In triangles ∆OAD and ∆OBD: - OA = OB [radii of the same circle] - AD = DB [given, since D is the midpoint of AB] - OD is the common side ∴ ∆OAD ≅ ∆OBD [by S-S-S congruence rule] ∴ ∠ODA = ∠ODB [corresponding angles of congruent triangles] Since OD stands on chord AB and forms equal angles on both sides, ∴ ∠ODA = ∠ODB = right angle ∴ OD ⊥ AB (Proved)