Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the bisector of angle BAC.
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.
In ΔAPB and ΔAPC,
AB = AC (Given)
∠ BAP = ∠CAP (Given)
AP = AP (Common)
∴ Δ APB ≅ 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.
Video Tutorials For All Subjects
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