$P$ is the midpoint of the side $CD$ of a parallelogram $ABCD$. A line through $C$ parallel to $PA$ intersects $AB$ at $Q$ and $DA$ produced at $R$. Prove that $\mathrm{DA}=\mathrm{AR}$ and $\mathrm{CQ}=\mathrm{QR}$.

Three angles of a quadrilateral are $75°, 90°$ and $75°.$ The fourth angle is

If bisectors of $\angle A$ and $\angle B$ of a quadrilateral $ABCD$ intersect each other at $P\mathit{,}$ bisectors of $\angle B$ and $\angle C$ at $Q,$ bisectors of $\angle C$ and $\angle D$ meet at $R$ and bisectors of $\angle D$ and $\angle A$ meet at $S$, then $PQRS$ is a