Mathematics
Two chords AB and AC of a circle are equal. Prove that the center of the circle, lies on the bisector of the angle BAC.
Circles
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Answer
Given,
AB and AC are two equal chords of circle with center O and radius r.
In ∆APB and ∆APC,
⇒ AB = AC [Given]
⇒ ∠BAP = ∠CAP [Given]
⇒ AP = AP [Common]
Hence, ∆APB ≅ ∆APC by SAS axiom.
By CPCT, we get :
BP = CP and ∠APB = ∠APC
From figure,
⇒ ∠APB + ∠APC = 180° [Linear pairs]
⇒ 2∠APB = 180°
⇒ ∠APB = 90°
Since, BP = CP and ∠APB = 90°.
∴ AP is the perpendicular bisector of chord BC.
∵ The perpendicular bisector of a chord passes through the centre of the circle.
∴ AD is the bisector of ∠BAC and passes through the centre O of the circle.
Hence proved that, AD passes through the centre of the circle.
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