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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Since angles BPA and BCA are on the same segment BA, they're equal.
Similarly angles APC and ABC being on the same segment AC, they're equal too.
Since triangle ABC is equilateral, BCA = ABC.
Combining the three equations, we have BPA = APC.
Therefore, PA is the angle bisector of angle BPC. (QED)
Similarly angles APC and ABC being on the same segment AC, they're equal too.
Since triangle ABC is equilateral, BCA = ABC.
Combining the three equations, we have BPA = APC.
Therefore, PA is the angle bisector of angle BPC. (QED)
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