In $\triangle\text{APB},$ by angle sum property,
$\angle\text{APB}+\angle\text{PAB}+\angle\text{PBA}=180^{\circ}$
$\Rightarrow\angle\text{APB}+\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}=180^{\circ}$
$\Rightarrow\angle\text{APB}=180^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}\Big)$
In $\triangle\text{CRD},$ by angle sum property,
$\angle\text{CRD}+\angle\text{RDC}+\angle\text{RCD}=180^{\circ}$
$\Rightarrow\angle\text{CRD}+\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}=180^{\circ}$
$\Rightarrow\angle\text{CRD}=180^{\circ}-\Big(\frac{1}{2}\angle\text{D}+\frac{1}{2}\angle\text{C}\Big)$
Now, $\angle\text{SPQ}+\angle\text{SRQ}=\angle\text{APB}+\angle\text{CRD}$
$=360^{\circ}-\Big(\frac{1}{2}\angle\text{A}+\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\frac{1}{2}\angle\text{D}\Big)$
$=360^{\circ}=\frac{1}{2}(\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D})$
$=360^{\circ}-\frac{1}{2}\times360^{\circ}$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
Now, $\angle\text{PSR}+\angle\text{PQR}=360^{\circ}-(\angle\text{SPQ}+\angle\text{SRQ})$
$=360^{\circ}-180^{\circ}$
$=180^{\circ}$
Hence, $PQRS$ is a quadrilateral whose opposite angles are supplementary.
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