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The bisector of ∠B and ∠C intersect each other at a point O. Prove that ∠BOC = 90° + ½ ∠A​

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\huge{\underline{\underline{\sf{\orange{Question:-}}}}}

The bisector of ∠B and ∠C intersect each other at a POINT O. Prove that ∠BOC = 90° + ½ ∠A

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{\blue{\sf\underline{Given}}}

A ∆ABC in which BO and CO are the bisector of ∠B and ∠C respectively.

{\pink{\sf\underline{To\:Prove}}}

\angle \: BOC \: = (90\degree + \frac{1}{2} \angle \: a). \\

{\blue{\sf\underline{Proof}}}

We know that the SUM of the ANGLES of a triangle is 180°.

\therefore \: \angle \: A + \angle \: B + \angle \: C = 180\degree

⇒ \frac{1}{2} \angle \: A + \frac{1}{2} \angle \: B + \frac{1}{2} \angle \: C = 90\degree \\

⇒ \frac{1}{2} \angle \: A + \angle \: OBC + \angle \: OCB = 90\degree \\

⇒ \angle \: OBC + \angle \: OCB = (90\degree - \frac{1}{2} \angle \: A) \: \: \: \: \: \: \: \: \: ...(i) \\

Now, in ∆OBC,we have

\angle \: OBC + \angle \: OCB + \angle \: BOC \: = 180\degree \\ \: \: \: \: \: \: \: \: \: \: (sum \: of \: the \: angles \: of \: a \: triangle)

⇒ (90\degree - \frac{1}{2} \angle \: A) + \angle \: BOC = 180\degree \: \: \: \: [using \: (i)] \\

⇒ \angle \: BOC \: = 180\degree - (90\degree - \frac{1}{2} \angle \: A) = (90\degree \: + \frac{1}{2} \angle \: A). \\

hence \: \angle \: BOC = (90\degree + \frac{1}{2} \angle \: A).

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