The difference of any two sides of a triangle is $\ldots \ldots \ldots$ than the third side.

In $\Delta ABC$,$\angle B = 90^{\circ}$,$BC = 8 \, \text{cm}$,and $AC = 17 \, \text{cm}$. $BE$ is a median of the triangle and $M$ is the midpoint of $BE$. Find the area of $\Delta BMC$ in $\text{cm}^2$.

In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 14 \, cm$ and $AC = 50 \, cm$,then find the area of $\Delta ABC$ in $cm^2$.

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 $\Delta ABC$,$AB = AC$ and $AD$ is an altitude to the base $BC$. Prove that $D$ is the midpoint of $BC$.

$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.