Bisectors of the angles B and C of an isosceles ΔABC with AB = AC intersect each other at O. Show that external angle adjacent to ∠ABC is equal to
∠BOC.
Given $\triangle A B C$ is an isosceles triangle in which $A B=A C, B O$ and $C O$ are the bisectors of
$\angle A B C$ and $\angle A C B$ respectively intersect at $O$.
To show $\angle D B A=\angle B O C$
Construction Line CB produced to $D$.
Proof In $\triangle A B C$, $A B=A C$ [given]
$\angle A C B=\angle A B C$
[angles opposite to equal sides are equal]
$\Rightarrow \quad \frac{1}{2} \angle A C B=\frac{1}{2} \angle A B C \quad$ [on dividing both sides by 2]
$\Rightarrow \quad \angle O C B=\angle O B C$ ......(i)
$[\because B O$ and $C O$ are the bisectors of $\angle A B C$ and $\angle A C B]$
In $\triangle B O C, \angle O B C+\angle O C B+\angle B O C=180^{\circ}$ [by angle sum property of a triangle]
$\Rightarrow$ $\angle O B C+\angle O B C+\angle B O C=180^{\circ}$ [from Eq. (i)]
$\Rightarrow$ $2 \angle O B C+\angle B O C=180^{\circ}$
$\Rightarrow$ $\angle \mathrm{ABC}+\angle \mathrm{BOC}=180^{\circ}$ $[\because B O$ is the bisector of $\angle A B C]$
$\Rightarrow$ $180^{\circ}-\angle D B A+\angle B O C=180^{\circ}$ $[\because D B C$ is a straight line]
$\Rightarrow$ $-\angle D B A+\angle B O C=0$
$\Rightarrow$ $\angle D B A=\angle B O C$