Question

The bisectors of $\angle B$ and $\angle C$ of an isosceles $\Delta ABC$ with$AB=AC$ intersect each other at point $O.$ Shows that the exterior angle adjacent to $\Delta ABC$ is equal to $\Delta BOC$

Answer 1

Considering the $△ABC$
It is given that $AB=AC$
So we get
$\angle ABC=\angle ACB$
Dividing by $2$ both sides
$\frac{1}{2}\angle ABC=\frac{1}{2}\angle ACB$
So we get
$\angle OBC=\angle OCB\dots \dots \left(1\right)$
By using the angle sum property in $△BOC$
$\angle BOC+\angle OBC+\angle OCB={180}^{\circ }$
Substituting equation (1)
$\angle BOC+2\angle OBC={180}^{\circ }$
So we get
$\angle BOC+\angle ABC={180}^{\circ }$
From the figure we know that $\angle ABC$ and $\angle ABP$ form a linear pair of angles so we get
$\angle ABC+\angle ABP={180}^{\circ }$
$\angle ABC={180}^{\circ }-\angle ABP$
By substituting the value in the above equation we get
$\angle BOC+({180}^{\circ }-\angle ABP)={180}^{\circ }$
On further calculation
$\angle BOC+{180}^{\circ }-\angle ABP={180}^{\circ }$
By subtraction
$\angle BOC-\angle ABP={180}^{\circ }-{180}^{\circ }$
$\angle BOC-\angle ABP=0$
$\angle BOC=\angle ABP$
Therefore, it is proved that the exterior angle adjacent to $\angle ABC$ is equal to $\angle BOC$.