Let $R = AC \cap BD$ $\textbf{Claim 1)}$ $PX, AC, QY$ are concurrent. $\textbf{Proof 1)}$ Let $S = AC \cap PX, S' = AC \cap QY$ $\frac{AS}{SC}=\frac{PA}{PC}=\frac{\sin \angle PCA}{\sin \angle PAC} = \frac{\sin \angle ACQ}{\sin \angle CAQ}=\frac{AQ}{QC}=\frac{AS'}{S'C}$, so $S = S'$. $\textbf{Claim 2)}$ $\angle PSQ = 90$ $\textbf{Proof 2)}$ Simple angle chasing $\textbf{Claim 3)}$ $PQ \parallel BC$ $\textbf{Proof 3)}$ Let $X, Y$ be the midpoint of $PQ, AC$ respectively. Newton-Gauss Theorem in $PBDQ$ gives $X, Y, R$ are collinear. Therefore $X$ lies on $AC$, and $PQ \parallel BC$. $\textbf{Claim 4)}$ $\angle PXR = 90$ $\textbf{Proof 4)}$ well known By $\textbf{Claim 2}$ and $\textbf{Claim 4}$, we get $XRSY$ is rectangle. Construct a rectangle $BUDV$ such that $BU \parallel PX$ and $DU \parallel QY$. Since $PQ \parallel BD$, then $U, V$ lies on $AC$. Therefore, triangle $XSY$ and $BTD$ are homothetic, so $XY \parallel BD$.