In trapezium $ABCD$,$AB \parallel DC$ and $L$ is the mid-point of $BC$. Through $L$,a line $PQ \parallel AD$ has been drawn which meets $AB$ in $P$ and $DC$ produced in $Q$. Prove that $\operatorname{ar}(ABCD) = \operatorname{ar}(APQD)$.

(N/A) Given: In trapezium $ABCD$,$AB \parallel DC$ and $L$ is the mid-point of $BC$. $PQ \parallel AD$ meets $AB$ at $P$ and $DC$ produced at $Q$.
To prove: $\operatorname{ar}(ABCD) = \operatorname{ar}(APQD)$.
Proof:
In $\triangle CLQ$ and $\triangle BLP$:
$1$. $\angle QCL = \angle LBP$ (Alternate interior angles,since $AB \parallel DQ$)
$2$. $CL = LB$ (Given,$L$ is the mid-point of $BC$)
$3$. $\angle CLQ = \angle BLP$ (Vertically opposite angles)
Therefore,$\triangle CLQ \cong \triangle BLP$ by the $ASA$ congruence rule.
Since the triangles are congruent,their areas are equal: $\operatorname{ar}(\triangle CLQ) = \operatorname{ar}(\triangle BLP)$ ... $(1)$
Now,adding $\operatorname{ar}(APLCD)$ to both sides of equation $(1)$:
$\operatorname{ar}(\triangle CLQ) + \operatorname{ar}(APLCD) = \operatorname{ar}(\triangle BLP) + \operatorname{ar}(APLCD)$
This simplifies to:
$\operatorname{ar}(APQD) = \operatorname{ar}(ABCD)$
Hence,$\operatorname{ar}(ABCD) = \operatorname{ar}(APQD)$.