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

In quadrilateral $ABCD$,$AM$ and $CN$ are altitudes on diagonal $BD$ drawn from $A$ and $C$ respectively. Prove that,$\operatorname{ar}(ABCD) = \frac{1}{2} \times BD \times (AM + CN)$.

$(1)$ Area of a rhombus $= \frac{1}{2} \times \ldots \ldots \ldots$
$(2)$ Area of a triangle $= \ldots \ldots \ldots$

In the given figure,$P$ is a point in the interior of parallelogram $ABCD$. Show that,
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$

In parallelogram $ABCD$,$AB = 12 \, cm$. Altitudes $DM$ and $DN$ correspond to bases $AB$ and $BC$ respectively. If $DM = 5 \, cm$ and $DN = 6 \, cm$,then find the length of $BC$ in $cm$.

The diagonals of a parallelogram $ABCD$ intersect at a point $O$. Through $O$,a line is drawn to intersect $AD$ at $P$ and $BC$ at $Q$. Show that $PQ$ divides the parallelogram into two parts of equal area.