Since $O$ is the center of the circumscribed circle, we calculate that

$$\begin{aligned} \angle B A O & =\frac{1}{2}\left(180^{\circ}-\angle A O B\right) \\ & =90^{\circ}-\angle A C B \\ & =90^{\circ}-\angle E F A \\ & =\angle F A R, \end{aligned}$$

where we also used the fact that $B C E F$ is a cyclic quadrilateral due to Thales, and that $R$ is the orthogonal projection of $A$ onto $E F$. From this, we conclude that $A, R$, and $O$ are collinear. To show that $D R$ and $O Q$ are parallel, it is sufficient to show that $\frac{|A R|}{|A D|}=\frac{|A O|}{|A Q|}$.

Due to the cyclic quadrilateral $B C E F$, we have that $\triangle A E F \sim \triangle A B C$. Under this similarity, the altitude $A R$ maps to the altitude $A D$. (Alternatively, one can simply show that $\triangle A R E \sim \triangle A D B$.) From this, it follows that $\frac{|A R|}{|A D|}=\frac{|A E|}{|A B|}$.

On the other hand, we know that

$$\angle P O B=\frac{1}{2} \angle C O B=\angle C A B=\angle E A B$$

due to symmetry and the fact that $O$ is the center of the circumscribed circle. Also, $\angle O B P=90^{\circ}=\angle A E B$. Thus, we find that $\triangle P O B \sim \triangle B A E$, which implies that $\frac{|A E|}{|A B|}=\frac{|B O|}{|O P|}$.

Furthermore, the lines $A O$ and $P Q$ are parallel because they are both perpendicular to $E F$, and the lines $A Q$ and $O P$ are parallel because they are both perpendicular to $B C$. This means that $A O P Q$ is a parallelogram, and in particular, that $|O P|=|A Q|$. We also know that $|B O|=|A O|$. Therefore, we conclude that

$$\frac{|A R|}{|A D|}=\frac{|A E|}{|A B|}=\frac{|B O|}{|O P|}=\frac{|A O|}{|A Q|}$$

which implies that $D R$ and $O Q$ are parallel.