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 $\angle BPC$

Answer 1

Here, $△ABC$ is an equilateral triangle inscribed in a circle with centre O.

$\Rightarrow AB=AC=BC$ [∵△ABC is equilateral]

$\angle AOB=\angle AOC=\angle BOC$ [equal chords subtend equal angles at centre]

$\Rightarrow \angle AOB=\angle AOC...i)$

Now, $\angle AOB$ and $\angle APB$ are angles subtended by an arc AB at centre and at remaining part of the circle by same arc.

Therefore, $\angle APB=\frac{\angle AOB}{2}$ ...ii)

Similarly, $\angle APC=\frac{\angle AOC}{2}$...iii)

Using (i),(ii) and (iii), we have

$\angle APB=\angle APC$

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