A regular polygon is

$ABCD$ is a quadrilateral. A line through $D$, parallel to $AC$, meets $BC$ produced at $P$ as shown in figure. Prove that $ar\left(△ABP\right)=ar\left(\mathrm{quad}.ABCD\right)$.

The given figure of a quadrilateral can be drawn if

In the given figure, $PQRS$ is an isosceles trapezium and $\overline{PQ}\parallel \overline{SR}\parallel \overline{MN}$. If $\angle SPM=70° \text{and} \angle PQR=110°,$ then find $\angle PMN.$