Question

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

Answer 1

$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

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

━━━━━━━━━━━━━━━━━━━━━━━━━

$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

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

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

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

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

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

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

━━━━━━━━━━━━━━━━━━━━━━━━━