In the figure,$ABCDE$ is any pentagon. $BP$ is drawn parallel to $AC$ and meets $DC$ produced at $P$,and $EQ$ is drawn parallel to $AD$ and meets $CD$ produced at $Q$. Prove that $\operatorname{ar}(ABCDE) = \operatorname{ar}(APQ)$.

(N/A) Given: $BP \parallel AC$ and $AD \parallel EQ$.
Since triangles on the same base and between the same parallels are equal in area:
$1$. For $\triangle ABC$ and $\triangle APC$,they are on the same base $AC$ and between the same parallels $BP$ and $AC$. Therefore,$\operatorname{ar}(\triangle ABC) = \operatorname{ar}(\triangle APC) \dots(1)$
$2$. For $\triangle ADE$ and $\triangle ADQ$,they are on the same base $AD$ and between the same parallels $AD$ and $EQ$. Therefore,$\operatorname{ar}(\triangle ADE) = \operatorname{ar}(\triangle ADQ) \dots(2)$
Adding equations $(1)$ and $(2)$,we get:
$\operatorname{ar}(\triangle ABC) + \operatorname{ar}(\triangle ADE) = \operatorname{ar}(\triangle APC) + \operatorname{ar}(\triangle ADQ)$
Adding $\operatorname{ar}(\triangle ACD)$ to both sides:
$\operatorname{ar}(\triangle ABC) + \operatorname{ar}(\triangle ADE) + \operatorname{ar}(\triangle ACD) = \operatorname{ar}(\triangle APC) + \operatorname{ar}(\triangle ADQ) + \operatorname{ar}(\triangle ACD)$
Observing the figure,the left side is the sum of the areas of the triangles forming the pentagon $ABCDE$,and the right side is the sum of the areas forming $\triangle APQ$.
Thus,$\operatorname{ar}(ABCDE) = \operatorname{ar}(\triangle APQ)$.