In the figure,$l, m,$ and $n$ are straight lines such that $l \parallel m$ and $n$ intersects $l$ at $P$ and $m$ at $Q$. $ABCD$ is a quadrilateral such that its vertex $A$ is on $l$. The vertices $C$ and $D$ are on $m$ and $AD \parallel n$. Show that $\operatorname{ar}(ABCQ) = \operatorname{ar}(ABCDP).$

(N/A) We are given that $AD \parallel n$. The triangles $\triangle APD$ and $\triangle AQD$ lie on the same base $AD$ and are between the same parallel lines $AD$ and $n$.
Therefore,$\operatorname{ar}(\triangle APD) = \operatorname{ar}(\triangle AQD) \dots(1)$
Now,adding $\operatorname{ar}(ABCD)$ to both sides of equation $(1)$,we get:
$\operatorname{ar}(\triangle APD) + \operatorname{ar}(ABCD) = \operatorname{ar}(\triangle AQD) + \operatorname{ar}(ABCD)$
From the figure,$\operatorname{ar}(\triangle APD) + \operatorname{ar}(ABCD) = \operatorname{ar}(ABCDP)$ and $\operatorname{ar}(\triangle AQD) + \operatorname{ar}(ABCD) = \operatorname{ar}(ABCQ)$.
Thus,$\operatorname{ar}(ABCDP) = \operatorname{ar}(ABCQ)$.