Q26 of 63 Page 156

If an equilateral triangle ABC is inscribed in a circle and P is any point lying on minor arc BC, which is not coincident with B or C, then prove that PA is the bisector of angle BPC.


Given ΔABC is an equilateral triangle inscribed in a circle with centre O.


P is a point lying on minor arc BC.


We have to prove that PA is bisector of ∠BPC.


Construction:


Join OA, OB, OC, BP and PC.


Proof:


Since ΔABC is equilateral,


⇒ AB = BC = CA


We know that equal chords subtend equal angles at centre.


⇒∠AOB = ∠AOC = ∠BOC


Consider ∠AOB = ∠AOC … (1)


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


⇒∠APB = 1/2 ∠AOB … (2)


⇒∠APC = 1/2 ∠AOC … (3)


From (1), (2) and (3),


∠APB = ∠APC


∴ PA is angle bisector of ∠BPC.


Hence proved.


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