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

(N/A) Through $P$,draw a line parallel to $AB$ which intersects $BC$ at $Q$ and $AD$ at $R$.
Now,in quadrilateral $ABQR$,
$AB \parallel QR$ (By construction)
$BQ \parallel AR$ (In parallelogram $ABCD, BC \parallel AD$)
$\therefore$ Quadrilateral $ABQR$ is a parallelogram. Similarly,$DCQR$ is a parallelogram.
$\Delta APB$ and parallelogram $ABQR$ are on the same base $AB$ and between the same parallels $AB$ and $QR$.
$\therefore \operatorname{ar}(APB) = \frac{1}{2} \operatorname{ar}(ABQR) \quad \dots(1)$
Similarly,$\Delta PCD$ and parallelogram $DCQR$ are on the same base $DC$ and between the same parallels $DC$ and $QR$.
$\therefore \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(DCQR) \quad \dots(2)$
Adding $(1)$ and $(2)$,
$\operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABQR) + \frac{1}{2} \operatorname{ar}(DCQR)$
$\therefore \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} [\operatorname{ar}(ABQR) + \operatorname{ar}(DCQR)]$
$\therefore \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD) \quad \dots(3)$
Now,through $P$,draw a line parallel to $AD$ which intersects $AB$ at $S$ and $CD$ at $T$.
Then,as above,it can be proved that $\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \frac{1}{2} \operatorname{ar}(ABCD) \quad \dots(4)$
From $(3)$ and $(4)$,
$\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$