$D$ is a point on the side $BC$ of a $\triangle ABC$ such that $AD$ bisects $\angle BAC$. Then

In the figure,$PQ > PR$ and $QS$ and $RS$ are the bisectors of $\angle Q$ and $\angle R$,respectively. Show that $SQ > SR$.

If $AB = QR$,$BC = PR$,and $CA = PQ$,then

In the given figure,$XP = XS$,$XQ = XR$ and $\angle PXR = \angle SXQ$. Prove that $PQ = SR$.

Write the measures of its sides in ascending order in each of the following triangles:
$(1)$ In $\Delta ABC$,$\angle B = 70^{\circ}$ and $\angle C = 20^{\circ}$.

$\angle ABD$ and $\angle ACE$ are exterior angles of $\Delta ABC$. If $\angle ABD > \angle ACE$,then prove that $AC > AB$.