Question

in triangle abc bisector of Angle B and angle C meet at O prove that angle BOC = 90 + angle a ÷2

Answer 1

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Answer 2

$\huge\underline\mathfrak{Answer:}$

Given: A ∆ABC such that the bisectors of $\angle$B and $\angle$C meet at O.

To prove : $\angle$BOC = 90° + $\angle$A / 2.

Proof : In ∆BOC, we have

$\angle$1 + $\angle$2 + $\angle$BOC = 180°

$\implies$ $\angle$BOC = 180° - ($\angle$1 - $\angle$2)

In ∆ABC, we have

$\angle$A + $\angle$B + $\angle$C = 180°

$\implies$ $\angle$A + 2$\angle$1 + 2$\angle$2 = 180°.

$\implies$ 2$\angle$1+ 2$\angle$2 = 180° - $\angle$A.

$\implies$ 2($\angle$1+$\angle$2)= 180° - $\angle$A.

$\implies$ $\angle$1 + $\angle$2 = 90° - $\angle$A/ 2

From (1) and (2), we have

$\angle$BOC = 180° - (90°- $\angle$A/2)

$\implies$ $\angle$BOC = 90° + $\angle$A / 2