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.
Solution:
Given, ABC is an equilateral triangle inscribed in a circle
P is any point on the minor arc BC which does not coincide with B or C
We have to prove that PA is the angle bisector of ∠BPC.
Join PB and PC
We know that in an equilateral triangle each angle is equal to 60 degrees.
∠3 = ∠4 = 60°
We know that angles in the same segment of a circle are equal.
Considering segment AB
∠1 = ∠4 = 60° ----------------------------- (1)
Considering segment AC
∠2 = ∠3 = 60° --------------------------- (2)
So, ∠1 = ∠2 = 60°
Therefore, PA is the bisector of ∠BPC.
✦ Try This: An arc of length 15 cm subtends an angle of 45° at the centre of a circle. Find in terms of π ,the radius of the circle.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.4 Problem 7
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.
Summary:
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, it is proven that PA is angle bisector of ∠BPC
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