Question

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

Answer 1

Given BD is Bisector of $\angle ABC$ and CO is Bisector of $\angle ACB$ produce CB to D

In $\Delta ABC$

$AB=AC$

$\angle ACB=\angle ABC$

$\frac{1}{2}\angle ACB=\frac{1}{2}\angle ABC$

$\angle OCB=\angle OBC$

in $\Delta BOC$

$\angle OBC+\angle OCB+\angle BDC={180}^{\circ }$

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

$\angle ABC+\angle BOC={180}^{\circ }$

$\angle ABC+\angle OBA={180}^{\circ }$

$\angle DBA=\angle BOC$