In $\triangle PQR,$ if $\angle R > \angle Q,$ then

In the figure,$AD$ is the bisector of $\angle BAC$. Prove that $AB > BD$.

Show that in a quadrilateral $ABCD$,$AB + BC + CD + DA < 2(BD + AC)$.

In the given figure,if $PQ = ST$,$QU = TR$,$PQ \perp QT$ and $ST \perp TQ$,then prove that $PR = SU$.

In $\Delta ABC$,$AB = AC$ and $\angle CAD$ is an exterior angle of $\Delta ABC$ at vertex $A$. If ray $AP$ is the bisector of $\angle CAD$,then prove that $AP \parallel BC$.

In $\Delta ABC$,$AD$,$BE$,and $CF$ are its medians. Prove that $AB + BC + CA > AD + BE + CF$.