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Question
Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the bisector of angle BAC.
Solution
Given: AB and AC are two equal chords of C (O, r).
To prove: Centre, O lies on the bisector of ∠BAC.
Construction: Join BC. Let the bisector of ∠BAC intersects BC in P.
Proof:
In ΔAPB and ΔAPC,
AB = AC ...(Given)
∠BAP = ∠CAP ...(Given)
AP = AP ...(Common)
∴ ΔAPB ≅ ΔAPC ...(SAS congruence criterion)
`=>` BP = CP and ∠APB = ∠APC ...(CPCT)
∠APB + ∠APC = 180° ...(Linear pair)
`=>` 2∠APB = 180° ...(∠APB = ∠APC)
`=>` ∠APB = 90°
Now, BP = CP and ∠APB = 90°
∴ AP is the perpendicular bisector of chord BC.
`=>` AP passes through the centre, O of the circle.
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