In $\Delta PQR$,$\angle Q = \angle R$ and $PQ = 6.5 \, cm$,then find $PR$. (in $, cm$)

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 $AP \parallel BC$.

$ABC$ is a right-angled triangle in which $\angle A = 90^o$ and $AB = AC$. Find $\angle B$ and $\angle C$.

In $\Delta ABC$,$AD$ is a median. $P$ and $Q$ are the midpoints of $AB$ and $AD$ respectively. If $\operatorname{ar}(\Delta ABC) = 72 \, \text{cm}^2$,then $\operatorname{ar}(\Delta APQ) = \dots \text{cm}^2$.

$D$ is any point on side $AC$ of a $\triangle ABC$ with $AB = AC$. Show that $CD < BD$.

$S$ is any point in the interior of $\triangle PQR$. Show that $SQ + SR < PQ + PR$.