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Question
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.
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Solution
Given: We have a circle whose centre is O and chords AB and AC are equal. AM is the bisector of ∠BAC.
To prove: Centre O lies on the bisector of ∠BAC.
Construction: Join BM and CM.
Proof: In ΔBAM and ΔCAM,
AB = AC ...[Given]
∠BAM = ∠CAM ...[Given]
AM = AM ...[Common]
∴ ΔBAM ≅ ΔCAM ...[By SAS congruency]
`\implies` BM = CM [By C.P.C.T.] ...(i)
And ∠BMA = ∠CMA [By C.P.C.T.] ...(ii)
In ΔBOM and ΔCOM,
BM = CM ...[By (i)]
OM = OM ...[Common]
∠BMO = ∠CMO ...[By (ii)]
∴ ΔBOM and ΔCOM ...[By SAS congruency]
`\implies` ∠BOM = ∠COM [By C.P.C.T.] ...(iii)
Since, ∠BOM + ∠COM = 180° ...(iv)
∴ By (iii) and (iv), ∠BOM = ∠COM = 90°
So, AM is the perpendicular bisector of the chord BC.
Thus, bisector of ∠BAC i.e., AM passes through the centre O.
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