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Question
If ABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C, prove that PA is angle bisector of ∠BPC.
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Solution
Given: ΔABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C.
To prove: PA is an angle bisector of ∠BPC.
Construction: Join PB and PC.
Proof: Since, ΔABC is an equilateral triangle.
∠3 = ∠4 = 60°
Now, ∠1 = ∠4 = 60° ...(i) [Angles in the same segment AB]
∠2 = ∠3 = 60° ...(ii) [Angles in the same segment AC]
∴ ∠1 = ∠2 = 60°
Hence, PA is the bisector of ∠BPC.
Hence proved.
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