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.

Answer 1

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.