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

$(1)$ If two figures are congruent,they must have $\ldots \ldots$ areas.
$(2)$ Area of figure $A$ is denoted as $\ldots \ldots$ symbolically.

$ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$:

In the figure,$ABCD$ and $AEFD$ are two parallelograms. Prove that $\operatorname{ar}(\triangle PEA) = \operatorname{ar}(\triangle QFD)$.

$ABCD$ is a parallelogram in which $BC$ is produced to $E$ such that $CE = BC$ $(Fig.)$. $AE$ intersects $CD$ at $F.$ If $\text{ar}(\Delta DFB) = 3 \, \text{cm}^2$,find the area of the parallelogram $ABCD$ (in $\text{cm}^2$).

In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 8\, \text{cm}$ and $BC = 15\, \text{cm}$,then $\text{ar}(\Delta ABC) = \dots \text{cm}^2$.