Question

If in Triangle ABC , bisector of angle B and C intersect each other at point O then show that angle BOC = 90 + 1/2 angle A

Need Help!
First answer Get brainleast marks

Answer 1

mark it as a brainliest answer

please, please, please, please, please, please

Answer 2

$\huge{\bf{\underline{\red{Solution :}}}}$

Given,

• B and C is bisector of ∠B and ∠C and intersect at point "O"

To Proof :

• ∠BOC = 90° + $\bold{\frac{1}{2}}$∠A

Proof :

⇒B and C is bisector of ∠B and ∠C

∴ $\bold{\angle{OBC = \bold{\frac{\angle B}{2} }}}$

$\bold{\angle OCB = \bold{\frac{\angle C}{2}}}$

In ΔOBC ,

⇒∠OBC +  ∠OCB +  ∠BOC = 180°

$\implies \bold{\frac{\angle B}{2} + \frac{\angle C}{2} + \angle BOC = \bold{180^{\circ}}}$

$\implies \bold{\angle BOC = \bold{ 180^{\circ} - [ \frac{\angle B + \angle C }{2} ]}}$

In ΔABC ,

⇒∠A + ∠B +∠C  = 180°

$\implies \bold{\angle B + \angle C = \bold{180 - \angle A}}$

$\implies \bold{\angle BOC = \bold{180^{\circ} - [ \frac{180^{\circ} - \angle A}{2}}} ]$

$\implies \bold{\angle BOC = \bold{180^{\circ} - \frac{180^{\circ}}{2} + \frac{\angle A}{2}}}$

$\implies \bold{\angel BOC = \bold{ 90^{\circ} + \frac{\angle A }{2} }}$

Hence Proved

$\rule{200}2$