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  • AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.

AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.

Answers (1)

Given: AB and AC are two equal chords of the circle.

Prove: Centre lies on the bisector of the \angleBAC.

Construction: Join BC

Proof: In \triangleAOB and \triangleAOC

AB = AC                                 (given)

\angleBAO = \angleCAO                     (given)

AO = AO                                 (common)

\therefore \triangleAOB \cong \triangleAOC                  (by SAS congruence)

\Rightarrow BO = CO                             (CPCT)                       …(i)

and \angleAOB = \angleAOC               (CPCT)

\angleAOB + \angleAOC = 180°          (Linear pair)

\therefore 2 \angleAOB = 180°                   (\angleAPB = \angleAPC)

\Rightarrow \angleAOB = 90°

\therefore AO is perpendicular bisector of the chord BC and BC is the diameter as BO = CO

\Rightarrow Hence, AO passes through the centre of the circle, and O is the centre of the circle.

Hence proved

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