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

(N/A) Through $P$ and $Q$,draw $PR$ and $QS$ parallel to $AB$. Now $PQRS$ is a parallelogram and its base $PQ = \frac{1}{3} BC$.
$\operatorname{ar}(APD) = \frac{1}{2} \operatorname{ar}(ABCD)$ [Same base $BC$ and $BC \parallel AD$]....$(1)$
$\operatorname{ar}(AQD) = \frac{1}{2} \operatorname{ar}(ABCD)$....$(2)$
From $(1)$ and $(2)$,we get
$\operatorname{ar}(APD) = \operatorname{ar}(AQD)$....$(3)$
Subtracting $\operatorname{ar}(AOD)$ from both sides,we get
$\operatorname{ar}(APD) - \operatorname{ar}(AOD) = \operatorname{ar}(AQD) - \operatorname{ar}(AOD)$
$\operatorname{ar}(APO) = \operatorname{ar}(OQD)$....$(4)$
Adding $\operatorname{ar}(OPQ)$ on both sides in $(4)$,we get
$\operatorname{ar}(APO) + \operatorname{ar}(OPQ) = \operatorname{ar}(OQD) + \operatorname{ar}(OPQ)$
$\operatorname{ar}(APQ) = \operatorname{ar}(DPQ)$
Since,$\operatorname{ar}(APQ) = \frac{1}{2} \operatorname{ar}(PQRS)$,therefore
$\operatorname{ar}(DPQ) = \frac{1}{2} \operatorname{ar}(PQRS)$
Now,$\operatorname{ar}(PQRS) = \frac{1}{3} \operatorname{ar}(ABCD)$
Therefore,$\operatorname{ar}(APQ) = \operatorname{ar}(DPQ) = \frac{1}{2} \operatorname{ar}(PQRS) = \frac{1}{2} \times \frac{1}{3} \operatorname{ar}(ABCD) = \frac{1}{6} \operatorname{ar}(ABCD)$.