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

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

Write True or False and justify your answer.
The base and the corresponding altitude of a parallelogram are $10 \text{ cm}$ and $3.5 \text{ cm}$,respectively. The area of the parallelogram is $30 \text{ cm}^2$.

In the given figure,$ABED$ is a parallelogram and $DE = EC$. Prove that $\operatorname{ar}(ABF) = \operatorname{ar}(BEC)$.

In the given figure,$ABCD$ is a trapezium with $AB \parallel DC$. $E$ is a point on extended $BC$. Prove that $ar(BDE) = ar(ACED)$.

In $\Delta ABC$,points $P$ and $Q$ are the points of trisection of $BC$. Then,$\operatorname{ar}(\Delta APQ) : \operatorname{ar}(\Delta ABC) = \dots$