If $M$ denotes the midpoint of $AB$, then $M,P,Q$ are collinear. Since $PQ \parallel AD \parallel BC$, the condition implies that $\angle MAP = \angle BAC = \angle QAD = \angle AQM$. Hence $AM$ is tangent to the circumcircle of $\triangle APQ$ at $A$, which implies that $MB^2=MA^2=MP \cdot MQ$. But then $BM$ is also tangent to the circumcircle of $\triangle BPQ$, which in turn implies $\angle PBA = \angle PQB = \angle CBD$.

Let $AC\cap BD=L$. Since $AC$ is a symmedian in $\triangle ABD$ we obtain that $\frac{|AB|^{2}}{|AD|^{2}}=\frac{|BL|}{|LD|}=\frac{|BC|}{|AD|}=\frac{|CL|}{|LA|}$. Thus $\frac{|CL|}{|LA|}=\frac{|BC|^{2}}{|AD|^{2}}\cdot \frac{|AD|^2}{|AB|^2}=\frac{|BC|^{2}}{|AB|^{2}}$ and $BD$ is a symmedian in $\triangle ABC$ as desired.