Two adjacent angles are equal. Is it necessary that each of these angles will be a right angle? Justify your answer.

In $\Delta ABC$,the sides $AB$ and $AC$ are produced to $D$ and $E$ respectively,so that exterior angles $\angle CBD$ and $\angle BCE$ are formed. If bisectors of $\angle CBD$ and $\angle BCE$ intersect at point $O$,prove that $\angle BOC = 90^{\circ} - \frac{1}{2} \angle A$.

An exterior angle of a triangle is $105^{\circ}$ and its two interior opposite angles are equal. Each of these equal angles is

$\angle A$ and $\angle B$ are supplementary angles. If $\angle A = 3x - 10^{\circ}$ and $\angle B = 2x + 30^{\circ}$,then find $\angle A$ and $\angle B$.

In the given figure,in $\Delta PQR$,$\angle Q > \angle R$. $PM \perp QR$ and $PA$ is the bisector of $\angle QPR$. Prove that $\angle APM = \frac{1}{2}(\angle Q - \angle R)$.

In the figure,$OD$ is the bisector of $\angle AOC$,$OE$ is the bisector of $\angle BOC$,and $OD \perp OE$. Show that the points $A, O$,and $B$ are collinear.