The median of a triangle divides it into two:

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

In $\Delta ABC$,medians $AD$,$BE$,and $CF$ intersect at point $G$. Prove that,$ar(GAB) = ar(GBC) = ar(GCA) = \frac{1}{3} ar(ABC)$.

In the figure,$ABCD$ is a parallelogram. Points $P$ and $Q$ on $BC$ trisect $BC$ into three equal parts. Prove that $\operatorname{ar}(APQ) = \operatorname{ar}(DPQ) = \frac{1}{6} \operatorname{ar}(ABCD)$.

Write True or False and justify your answer:
$ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If $\text{ar}(AXCD) = 24 \text{ cm}^2$,then $\text{ar}(ABC) = 24 \text{ cm}^2$.

Write True or False and justify your answer:
$PQRS$ is a parallelogram whose area is $180 \, cm^{2}$ and $A$ is any point on the diagonal $QS$. The area of $\triangle ASR = 90 \, cm^{2}$.