In $\Delta ABC$,$AB = 6 \text{ cm}$ and $BC = 9 \text{ cm}$,then $AC < \dots \text{ cm}$.

In $\triangle PQR$,$\angle R = \angle P$,$QR = 4 \, cm$,and $PR = 5 \, cm$. Then the length of $PQ$ is (in $cm$):

In $\Delta ABC$,$AD$ is a median. If $ar(\Delta ABC) = 50 \, cm^2$,then $ar(\Delta ADC) = \dots \dots \dots cm^2$.

In $\triangle PQR$,$\angle P = 70^{\circ}$ and $\angle R = 30^{\circ}$. Which side of this triangle is the longest? Give a reason for your answer.

If the bisector of an angle of a triangle also bisects the opposite side,prove that the triangle is isosceles.

$D$ is a point on side $BC$ of $\Delta ABC$ such that $AD = AC$ (see figure). Show that $AB > AD$.