If $ABC$ is an equilateral triangle inscribed in a circle and $P$ is any point on the minor arc $BC$ which does not coincide with $B$ or $C$,prove that $PA$ is the angle bisector of $\angle BPC$.
(N/A) Since equal chords of a circle subtend equal angles at the centre,we have:
Chord $AB = $ chord $AC$ (Given,as $ABC$ is an equilateral triangle).
So,$\angle AOB = \angle AOC$ $...(1)$
Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle:
$\angle APC = \frac{1}{2} \angle AOC$ $...(2)$
$\angle APB = \frac{1}{2} \angle AOB$ $...(3)$
From $(1)$,$(2)$,and $(3)$,we get:
$\angle APC = \angle APB$
Hence,$PA$ is the angle bisector of $\angle BPC$.
Chord $AB = $ chord $AC$ (Given,as $ABC$ is an equilateral triangle).
So,$\angle AOB = \angle AOC$ $...(1)$
Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle:
$\angle APC = \frac{1}{2} \angle AOC$ $...(2)$
$\angle APB = \frac{1}{2} \angle AOB$ $...(3)$
From $(1)$,$(2)$,and $(3)$,we get:
$\angle APC = \angle APB$
Hence,$PA$ is the angle bisector of $\angle BPC$.
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