In $\Delta ABC$, $\angle B = \angle C$, $AB = 5 \text{ cm}$, and $BC = 8 \text{ cm}$. Then, the perimeter of $\Delta ABC$ is: (in $\text{ cm}$)

$P$ is a point on the bisector of $\angle ABC$. If the line through $P$,parallel to $BA$,meets $BC$ at $Q$,prove that $\triangle BPQ$ is an isosceles triangle.

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,$\angle B < \angle A$ and $\angle C < \angle D$. Show that $AD < BC$.

In $\Delta PQR$,$PQ = PR$. If $\angle Q = 48^{\circ}$,then find $\angle P$ and $\angle R$.

In $\Delta ABC$,$AD$ is a median and $AM$ is an altitude. The side $BA$ of $\Delta ABC$ is produced to any point $E$,so that $AB = AE$. If $BC = 16\, cm$ and $AM = 8\, cm$,then find the area of $\Delta EBD$ in $cm^2$.