Question

In $△ABC$, the bisector of interior angle $A$ and the bisector of exterior angle $C$ meet at point $O$.If $\angle AOC=\frac{1}{m}\angle B$.Find $m$

Answer 1

Exterior angle in a triangle=sum of two opposite interior angles.

$⟹In△ABC,$

$\angle BCT=\angle A+\angle B$

$⟹2\alpha =2\theta +\angle B$

$=>\alpha =\theta +\frac{\angle B}{2}-\left(1\right)$

$In△OAC$

using the same angle property,

$=>\angle OCT=\angle OAC+\angle AOC$

$=>\alpha =\theta +\angle AOC-\left(2\right)$

comparing (1)& (2)

$=>\theta +\angle AOC=\theta +\frac{\angle B}{2}$

$\angle AOC=\frac{\angle B}{2}=\frac{\angle B}{m}$

$\therefore m=2$