Let $\delta = \angle CRS = \angle CAS$, $\beta = \angle ACR = \angle ASR$, $\gamma = \angle QPA = \angle QCA$, $\alpha = \angle PQC = \angle PAC$. Then the angle between $PQ$ and $CA$ is $\alpha - \gamma$ and the angle between $PQ$ and $RS$ is $\beta-\delta$. Since, $PQ$ and $RS$ are parallel, we have $\alpha-\gamma = \beta-\delta$ or $\alpha+\delta = \gamma+\beta$, implies $\angle DAB=\angle DCB$.
We also have the angle of circumference of $PQ$ on the circumcircle of $ABC$ is $QA+AB+BC+CP$ and it is equal to $\alpha + \beta + \gamma + \delta$. Also, the angle of circumference of $SR$ on the circumcircle of $ADC$ is $RA+AD+DC+CS$ and it is equal to $\alpha + \beta + \gamma + \delta$. Since $PQ=SR$, this implies that the two circumcircle's radii is equal. Therefore, $\angle ADC=\angle CBA$ and $ABCD$ is parallelogram.
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