Question

Abc is an equilateral triangle p is a point on the minor arc bc of the circumcircle of triangle abc show that pa = pb +pc

Answer 1

Answer:

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

⇒ AB = AC = BC [∵△ABC is equilateral]

∠AOB = ∠AOC = ∠BOC [equal chords subtend equal angles at centre]

⇒ ∠AOB = ∠AOC ...i)

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

Therefore, ∠APB = 1 / 2 ∠AOB ...ii)

Similarly, ∠APC = 1 / 2 ∠AOC ...iii)

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

∠APB = ∠APC

Hence, PA is angle bisector of ∠BPC.