$ABCD$ is a parallelogram and $BC$ is produced to a point $Q$ such that $AD = CQ$. If $AQ$ intersects $DC$ at $P$,show that $\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$.

(A) Given: $ABCD$ is a parallelogram,$AD = CQ$,and $AQ$ intersects $DC$ at $P$.
Step $1$: Since $ABCD$ is a parallelogram,$AD \parallel BC$ and $AD = BC$.
Given $AD = CQ$,therefore $BC = CQ$.
Step $2$: Consider $\triangle ADC$ and $\triangle ACQ$. Since $AD \parallel QC$ (as $AD \parallel BC$),$\triangle ADC$ and $\triangle ACQ$ are on the same base $AC$ and between the same parallels $AD$ and $QC$ is not strictly true,but rather $\triangle ADQ$ and $\triangle ADC$ share base $AD$ and height between $AD$ and $QC$. Actually,consider $\triangle ADQ$ and $\triangle ADC$. Since $AD \parallel QC$,$\operatorname{ar}(ADQ) = \operatorname{ar}(ADC)$ because they are on the same base $AD$ and between the same parallels $AD$ and $QC$.
Step $3$: Subtract $\operatorname{ar}(ADP)$ from both sides:
$\operatorname{ar}(ADQ) - \operatorname{ar}(ADP) = \operatorname{ar}(ADC) - \operatorname{ar}(ADP)$
$\operatorname{ar}(DPQ) = \operatorname{ar}(APC) .....(1)$
Step $4$: Now,consider $\triangle APC$ and $\triangle BPC$. They are on the same base $PC$ and between the same parallels $AB$ and $DC$ (since $AB \parallel DC$).
Therefore,$\operatorname{ar}(APC) = \operatorname{ar}(BPC) .....(2)$
Step $5$: From $(1)$ and $(2)$,we get:
$\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$.