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, then prove that PA is angle bisector of `angleBPC`.

Explanation

Given `DeltaABC` 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 ` angleBPC`.
Construction Join PB and PC.

Proof Since, `DeltaABC ` is an equilateral triangle.
`angle3=angle4=60^(@)`
Now, ` angle1=angle4=60^(@)` ...(i)
[angles in the same segment AB]
`angle2=angle3=60^(@)` ....(ii)
[angles in the same segment AC]
`:. angle1=angle2 = 60^(@)`
Hence, PA is the bisector of ` angleBPC`.