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  • 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.

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 the angle bisector of \angleBPC.

Answers (1)

Let DABC be an equilateral triangle inscribed in a circle with centre O.

So all sides of this equilateral triangle are equal by definition

AB = AC = BC

Now we know that equal chords subtend and equal angles at the centre. So,

\angleAOB = \angleAOC =\angleBOC      … (i)

Consider the arc AB, \angleAOB and \angleAPB are the angles subtended by AB at the centre and at the remaining part of the circle

Therefore

 \angle APB = \frac{1}{2} \angle AOB … (ii)

Similarly,

\angle APC = \frac{1}{2} \angle AOC … (iii)

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

\angleAPB = \angleAPC

Hence, PA is the angle bisector of \angleBPC.

Hence proved

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