In the figure,$P$ is a point in the interior of a parallelogram $ABCD$. Show that:
$(i)$ $\operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(ii)$ $\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
[Hint: Through $P$,draw a line parallel to $AB$.]

(N/A) $(i)$ We have a parallelogram $ABCD$,i.e.,$AB \parallel CD$ and $BC \parallel AD$.
Let us draw a line $EF \parallel AB$ passing through $P$,where $E$ lies on $AD$ and $F$ lies on $BC$.
Since $AB \parallel EF$,$\Delta APB$ and parallelogram $AEFB$ are on the same base $AB$ and between the same parallels $AB$ and $EF$.
Therefore,$\operatorname{ar}(\Delta APB) = \frac{1}{2} \operatorname{ar}(AEFB)$ ...... $(1)$
Also,$\Delta PCD$ and parallelogram $CDEF$ are on the same base $CD$ and between the same parallels $CD \parallel EF$.
Therefore,$\operatorname{ar}(\Delta PCD) = \frac{1}{2} \operatorname{ar}(CDEF)$ ...... $(2)$
Adding $(1)$ and $(2)$,we have:
$\operatorname{ar}(\Delta APB) + \operatorname{ar}(\Delta PCD) = \frac{1}{2} [\operatorname{ar}(AEFB) + \operatorname{ar}(CDEF)]$
$\operatorname{ar}(\Delta APB) + \operatorname{ar}(\Delta PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(ii)$ Let us draw a line $GH \parallel AD$ passing through $P$,where $G$ lies on $CD$ and $H$ lies on $AB$.
$\Delta APD$ and parallelogram $ADGH$ are on the same base $AD$ and between the same parallels $AD$ and $GH$.
Therefore,$\operatorname{ar}(\Delta APD) = \frac{1}{2} \operatorname{ar}(ADGH)$ ...... $(3)$
Similarly,$\Delta PBC$ and parallelogram $BCGH$ are on the same base $BC$ and between the same parallels $BC$ and $GH$.
Therefore,$\operatorname{ar}(\Delta PBC) = \frac{1}{2} \operatorname{ar}(BCGH)$ ...... $(4)$
Adding $(3)$ and $(4)$,we have:
$\operatorname{ar}(\Delta APD) + \operatorname{ar}(\Delta PBC) = \frac{1}{2} [\operatorname{ar}(ADGH) + \operatorname{ar}(BCGH)]$
$\operatorname{ar}(\Delta APD) + \operatorname{ar}(\Delta PBC) = \frac{1}{2} \operatorname{ar}(ABCD)$ ...... $(5)$
From $(i)$,we know $\operatorname{ar}(\Delta APB) + \operatorname{ar}(\Delta PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$ ...... $(6)$
Comparing $(5)$ and $(6)$,we get:
$\operatorname{ar}(\Delta APD) + \operatorname{ar}(\Delta PBC) = \operatorname{ar}(\Delta APB) + \operatorname{ar}(\Delta PCD)$