Question

ABC IS AN EQUIATERAL TRIANGLE INSCRIBED IN A CIRCLE P IS ANY POINT ON ARC BC SHOW THAT BP+CP=AP

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.

Step-by-step explanation: