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, then prove that PA is angle bisector of $\angle BPC$.

Answer 1

Given
$\Delta$ 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.
To Prove that PA is an angle bisector $\angle BPC$
Construction : Join PB and PC

Proof
Given, $\Delta$ ABC is an equilateral triangle
$\angle 3=\angle 4={60}^{\circ }...\left(i\right)$

Now, $\angle 1=\angle 4={60}^{\circ }...\left(ii\right)$
[angles in the same segment AB]

$\angle 2=\angle 3={60}^{\circ }...\left(iii\right)$
[angles in the same segment AC]

From (i), (ii) and (iii) we get,
$\angle 1=\angle 2={60}^{\circ }$

Hence, PA is the bisector of $\angle BPC$