Bisectors of the angles $B$ and $C$ of an isosceles triangle $ABC$ with $AB = AC$ intersect each other at $O$. Show that the external angle adjacent to $\angle ABC$ is equal to $\angle BOC$.
(N/A) In $\triangle ABC$,we have $AB = AC$.
Therefore,$\angle ABC = \angle ACB$ [Since angles opposite to equal sides of a triangle are equal].
Since $BO$ and $CO$ are bisectors of $\angle B$ and $\angle C$ respectively,we have $\angle OBC = \frac{1}{2} \angle ABC$ and $\angle OCB = \frac{1}{2} \angle ACB$.
Since $\angle ABC = \angle ACB$,it follows that $\angle OBC = \angle OCB$.
In $\triangle OBC$,the sum of angles is $180^{\circ}$,so $\angle BOC + \angle OBC + \angle OCB = 180^{\circ}$.
Substituting $\angle OCB = \angle OBC$,we get $\angle BOC + 2 \angle OBC = 180^{\circ}$,which implies $\angle BOC = 180^{\circ} - 2 \angle OBC$.
Now,consider the external angle adjacent to $\angle ABC$. Let $D$ be a point on the extension of $AB$. The external angle is $\angle CBD = 180^{\circ} - \angle ABC$.
Since $\angle ABC = 2 \angle OBC$,we have $\angle CBD = 180^{\circ} - 2 \angle OBC$.
Comparing the two expressions,we conclude that $\angle BOC = \angle CBD$.
Therefore,$\angle ABC = \angle ACB$ [Since angles opposite to equal sides of a triangle are equal].
Since $BO$ and $CO$ are bisectors of $\angle B$ and $\angle C$ respectively,we have $\angle OBC = \frac{1}{2} \angle ABC$ and $\angle OCB = \frac{1}{2} \angle ACB$.
Since $\angle ABC = \angle ACB$,it follows that $\angle OBC = \angle OCB$.
In $\triangle OBC$,the sum of angles is $180^{\circ}$,so $\angle BOC + \angle OBC + \angle OCB = 180^{\circ}$.
Substituting $\angle OCB = \angle OBC$,we get $\angle BOC + 2 \angle OBC = 180^{\circ}$,which implies $\angle BOC = 180^{\circ} - 2 \angle OBC$.
Now,consider the external angle adjacent to $\angle ABC$. Let $D$ be a point on the extension of $AB$. The external angle is $\angle CBD = 180^{\circ} - \angle ABC$.
Since $\angle ABC = 2 \angle OBC$,we have $\angle CBD = 180^{\circ} - 2 \angle OBC$.
Comparing the two expressions,we conclude that $\angle BOC = \angle CBD$.
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