Question

Bisectors of the angles $B$ and $C$ of an isosceles triangle with $AB=AC$ intersect each other at $O.$ Show that external angle adjacent to $\angle ABC$ is equal to $\angle BOC$.

Answer 1

Given : $\Delta ABC$ is an isosceles triangle in which AB = AC. BO and BO and CO are the bisectors of $\angle ABC$ and $\angle ACB$ respectively and intersect at $O.$

To show: $\angle DBA=\angle BOC$

Construction: Line CB produced to D

Proof: In $\Delta ABC,AB=AC$ [given]

$\Rightarrow \angle ACB=\angle ABC$ [angles opposite to equal sides are equal]

$\Rightarrow \frac{1}{2}\angle ACB=\frac{1}{2}\angle ABC$ [on dividing both sides by $2$]

$\Rightarrow \angle OCB=\angle OBC\dots \dots \left(i\right)$

$[\because BO$ and $CO$ are the bisectors of $\angle ABC$ and $\angle ACB]$

In $\Delta BOC,\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$

$\Rightarrow \angle OBC+\angle OBC+\angle BOC={180}^{\circ }$ [from $Eq.\left(i\right)$]

$\Rightarrow 2\angle OBC+\angle BOC={180}^{\circ }$

$\Rightarrow \angle ABC+\angle BOC={180}^{\circ }[\because BO$ is the bisector of $\angle ABC]$

$\Rightarrow ({180}^{\circ }-\angle DBA)+\angle BOC={180}^{\circ }[\because DBC$ is a straight line]

$\Rightarrow -\angle DBA+\angle BOC=0$

$\Rightarrow \angle DBA=\angle BOC$