Bisectors of the angles \(B\) and \(C\) of an isosceles triangle \(ABC\) with \(AB = AC\) intersect each other at \(O\). Show that external angle adjacent to \(\angle ABC\) is equal to \(\angle BOC\).
Proof:
Given that \(AB = AC\).
\(\angle ABC = \angle \) ---- (\(1\)) [Angles opposite to equal sides are equal]
Here, \(OB\) is the bisector of \(\angle B\).
Then, \(\angle ABO = \angle OBC\)
Now, \(\angle ABC = \angle ABO + \angle OBC\)
\(= \angle OBC + \angle OBC\)
\(\angle ABC = 2 \angle \) ---- (\(2\))
Also, \(OC\) is the bisector of \(\angle C\).
\(\angle ACO = \angle OCB\)
And, \(\angle ACB = \angle ACO + \angle OCB\)
\(= \angle OCB + \angle OCB\)
\(\angle ACB = 2 \angle \) ---- (\(3\))
Using equations (\(2\)) and (\(3\)) in (\(1\)), we get:
\(2 \angle OBC = 2 \angle OCB\)
\(\angle OBC = \angle \) ---- (\(4\))
Applying angle sum property in \(\triangle BOC\), we have:
\(\angle OBC + \angle OCB + \angle BOC = 180^{\circ}\)
\(\angle OBC + \angle OBC + \angle BOC = 180^{\circ}\) [Using (\(4\))]
\(2 \angle OBC + \angle BOC = 180^{\circ}\)
\(\angle ABC + \angle BOC = 180^{\circ}\) ---- (\(5\)) [Using (\(2\))]
Now, \(\angle ABC + \angle ABD = 180^{\circ}\) [Linear pair]
\(\angle ABC = 180^{\circ} - \angle ABD\) ---- (\(6\))
Using equation (\(6\)) in (\(5\)), we get:
\(180^{\circ} - \angle ABD + \angle BOC = 180^{\circ}\)
\(180^{\circ} - 180^{\circ} - \angle ABD + \angle BOC = 0\)
\(\angle BOC = \angle \)
Therefore, the external angle adjacent to \(\angle ABC\) is equal to \(\angle BOC\).
Hence, we proved.