Question 15 Marks
Bisectors of the angles B and C of an isosceles $\triangle\text{ABC}$ with AB = AC intersect each other at O. Show that external angle adjacent to $ \angle\text{ABC} $ is equal to $\angle\text{BOC.}$
Answer
View full question & answer→Given Lines, BO and CO are angle bisectore isosceles such that AB = AC which $ \angle\text{ABC} $ and $\angle\text{ACB}$ respectively at O.
$\angle\text{DBA}=\angle\text{BOC}$
In $\triangle\text{ABC},$
$\text{AB}=\text{AC}$
$\angle\text{ACB}=\angle\text{ABC}$
$\Rightarrow \frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{ABC}$
$\Rightarrow \angle\text{OCB}=\angle\text{OBC}$
In $\triangle\text{OCB},$
$\angle\text{OCB}+\angle\text{OCB}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow\angle\text{OBC}+\angle\text{OBC}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow 2\angle\text{OBC}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow \angle\text{ABC}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow 180^{\circ}-\angle\text{DBA}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow\angle\text{DBA}+\angle\text{BOC}=0$
$\Rightarrow \angle\text{DBA}=\angle\text{BOC}$
$\angle\text{DBA}=\angle\text{BOC}$
In $\triangle\text{ABC},$
$\text{AB}=\text{AC}$
$\angle\text{ACB}=\angle\text{ABC}$
$\Rightarrow \frac{1}{2}\angle\text{ACB}=\frac{1}{2}\angle\text{ABC}$
$\Rightarrow \angle\text{OCB}=\angle\text{OBC}$
In $\triangle\text{OCB},$$\angle\text{OCB}+\angle\text{OCB}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow\angle\text{OBC}+\angle\text{OBC}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow 2\angle\text{OBC}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow \angle\text{ABC}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow 180^{\circ}-\angle\text{DBA}+\angle\text{BOC}=180^{\circ}$
$\Rightarrow\angle\text{DBA}+\angle\text{BOC}=0$
$\Rightarrow \angle\text{DBA}=\angle\text{BOC}$
On adding eq.(i) and (ii),