In the figure,$AP || BQ || CR$. Prove that $\operatorname{ar}(AQC) = \operatorname{ar}(PBR)$.

(N/A) Given: $AP || BQ || CR$.
Step $1$: Since $BQ || CR$,$\Delta BCQ$ and $\Delta BQR$ are on the same base $BQ$ and between the same parallels $BQ$ and $CR$.
Therefore,$\operatorname{ar}(\Delta BCQ) = \operatorname{ar}(\Delta BQR)$ --- $(1)$
Step $2$: Since $AP || BQ$,$\Delta ABQ$ and $\Delta PBQ$ are on the same base $BQ$ and between the same parallels $AP$ and $BQ$.
Therefore,$\operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta PBQ)$ --- $(2)$
Step $3$: Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\Delta BCQ) + \operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta BQR) + \operatorname{ar}(\Delta PBQ)$
From the figure,$\operatorname{ar}(\Delta BCQ) + \operatorname{ar}(\Delta ABQ) = \operatorname{ar}(\Delta AQC)$ and $\operatorname{ar}(\Delta BQR) + \operatorname{ar}(\Delta PBQ) = \operatorname{ar}(\Delta PBR)$.
Thus,$\operatorname{ar}(\Delta AQC) = \operatorname{ar}(\Delta PBR)$.
Hence proved.