Question

in the given figure, triangle ABC is an isoscele , inscribed in circle with centre O, Prove that Ap bisect angle bpc​

Answer 1

Answer:

Given –  ∆ABC is an isosceles triangle and O is the centre of its circumcircle

To prove – AP bisects ∠BPC

Proof –

Chord AB subtends ∠APB and chord AC subtends ∠APC at the Circumference of the circle.

But chord AB = Chord AC

∴ ∠APB = ∠APC

∴ AP is the bisector of ∠BPC

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Answer 2

Question :-- in the given figure, triangle ABC is an isoscele , inscribed in circle with centre O, Prove that Ap bisect angle BPC ?

Formula and Concept used :--

→ we know that Equal chord subtand Equal Angle at the Circumference of the circle....

→ Isosceles ∆ has Two Sides Equal in Length .

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Solution :--

Given :--

∆ABC is a isosceles ∆ So, AB = AC .

O = centre of circle.

To Prove :--

→ AP Bisect Angle BPC. ( That means , it divide the angle in two Equal parts ).

Now, Since , AB = AC (Given).

So, They both Subtand Equal Angle at the Circumference of the circle . ( According to Theoram ).

So,

→ Angle APB = Angle APC.

Hence, we can say that, AP is the Angle bisector of Angle BPC.