Bisectors of the angles \(Y\) and \(Z\) of an isosceles triangle \(XYZ\) with \(XY = XZ\) intersect each other at \(O\). Show that external angle adjacent to \(\angle XYZ\) is equal to \(\angle YOZ\).
Proof:
Given that \(XY =XZ\).
\(\angle XYZ = \angle \) ---- (\(1\)) [Angles opposite to equal sides are equal]
Here, \(OY\) is the bisector of \(\angle Y\).
Then, \(\angle XYO = \angle OYZ\)
Now, \(\angle XYZ= \angle XYO + \angle OYZ\)
\(= \angle OYZ + \angle OYZ\)
\(\angle XYZ= 2 \angle \) ---- (\(2\))
Also, \(OZ\) is the bisector of \(\angle Z\).
\(\angle XZO = \angle OZY\)
And, \(\angle XZY = \angle XZO + \angle OZY\)
\(= \angle OZY + \angle OZY\)
\(\angle XZY = 2 \angle \) ---- (\(3\))
Using equations (\(2\)) and (\(3\)) in (\(1\)), we get:
\(2 \angle OYZ = 2 \angle OZY\)
\(\angle OYZ = \angle \) ---- (\(4\))
Applying angle sum property in \(\triangle YOZ \), we have:
\(\angle OYZ + \angle OZY + \angle YOZ = 180^{\circ}\)
\(\angle OYZ + \angle OYZ+ \angle YOZ = 180^{\circ}\) [Using (\(4\))]
\(2 \angle OYZ + \angle YOZ = 180^{\circ}\)
\(\angle XYZ + \angle YOZ = 180^{\circ}\) ---- (\(5\)) [Using (\(2\))]
Now, \(\angle XYZ+ \angle XYD = 180^{\circ}\) [Linear pair]
\(\angle XYZ = 180^{\circ} - \angle XYD\) ---- (\(6\))
Using equation (\(6\)) in (\(5\)), we get:
\(180^{\circ} - \angle XYD + \angle YOZ = 180^{\circ}\)
\(180^{\circ} - 180^{\circ} - \angle XYD + \angle YOZ = 0\)
\(\angle YOZ = \angle \)
Therefore, the external angle adjacent to \(\angle ABC\) is equal to \(\angle YOZ \).
Hence, we proved.