If in $\Delta ABC, AC > AB > BC$,then find the greatest angle of the triangle.

In $\Delta ABC$,$AB = AC$. If $\angle A = 50^{\circ}$,then find $\angle B$ and $\angle C$.

$\angle ACD$ is an exterior angle of $\Delta ABC$. If $AB = AC$ and $\angle B = 70^{\circ}$,then $\angle ACD = \dots$ (in $^{\circ}$)

In the figure,$PR > PQ$ and $PS$ bisects $\angle QPR$. Prove that $\angle PSR > \angle PSQ$.

In $\Delta ABC$,$\angle B = 70^{\circ}$ and $\angle C = 80^{\circ}$. The bisector of $\angle A$ intersects $BC$ at $D$. Find $\angle ADB$ and $\angle ADC$.

In $\Delta ABC$,if $\angle A > \angle B$,then $AC \ldots \ldots BC$.