Let $H$ be the projection of $B$ onto $AC$. Since $B, P, Q, H, D$ lie on the semicircle of diameter $[BD]$, we obtain
$$\angle PHB = \angle PDB = 90^\circ - \frac{\hat{B}}{2} = \angle QDB = \angle QHB.$$

We will show that if $BH \perp AC$ and $\angle PHB = \angle QHB$, then $AP, CQ, BH$ are concurrent. Let the line passing through $B$ and parallel to $AC$ intersect the lines $PH$ and $QH$ at $K$ and $L$, respectively. As $BH \perp AC$ and $\angle PHB = \angle QHB$, we get $HB \perp KL$ and $\angle KHB = \angle LHB$. Therefore, we have $|KB| = |LB|$.

$KL \parallel AC$ and Thales theorem imply
$$\frac{|KB|}{|CH|} = \frac{|BP|}{|PC|} \quad \text{and} \quad \frac{|LB|}{|AH|} = \frac{|BQ|}{|QA|}$$
Since $|KB| = |LB|$, we obtain $\frac{|AH|\,|PC|\,|BQ|}{|CH|\,|BP|\,|AQ|} = 1$. Ceva theorem on triangle $ABC$ gives that $AP, CQ, BH$ are concurrent. Hence, $X \in BH$ and therefore, $BX \perp AC$. Because of the symmetric structure we similarly get $CY \perp AB$. Finally, the point $Z$, which is the intersection of the lines $BX$ and $CY$, is the orthocenter of the triangle $ABC$ and the result follows.