Since $O$ is the circumcentre, we get that
$$\begin{align*} \angle BAO &= \frac{1}{2}(180^\circ - \angle AOB) \\ &= 90^\circ - \angle ACB \\ &= 90^\circ - \angle EFA \\ &= \angle FAR, \end{align*}$$
where we also used that $BCEF$ is a cyclic quadrilateral due to Thales and that $R$ is the perpendicular projection of $A$ onto $EF$. From this we conclude that $A$, $R$, and $O$ are collinear. Thus, to show that $DR$ and $OQ$ are parallel, it suffices to show that $\frac{|AR|}{|AD|} = \frac{|AO|}{|AQ|}$.

On the other hand, we know that
$$\angle POB = \frac{1}{2} \angle COB = \angle CAB = \angle EAB$$
due to symmetry and the fact that $O$ is the circumcentre. Also, $\angle OBP = 90^\circ = \angle AEB$. So we find that $\triangle POB \sim \triangle BAE$ from which it follows that $\frac{|AE|}{|AB|} = \frac{|BO|}{|OP|}$.

Furthermore, the lines $AO$ and $PQ$ are parallel because they are both perpendicular to $EF$, and the lines $AQ$ and $OP$ are parallel because they are both perpendicular to $BC$. This means that $AOPQ$ is a parallelogram, and in particular that $|OP| = |AQ|$. Also, of course, we know that $|BO| = |AO|$. So we conclude that
$$\frac{|AR|}{|AD|} = \frac{|AE|}{|AB|} = \frac{|BO|}{|OP|} = \frac{|AO|}{|AQ|}$$
From which it follows that $DR$ and $OQ$ are parallel. $\square$

Because of the cyclic quadrilateral $BCEF$, we have $\triangle AEF \sim \triangle ABC$. Under this similarity, the altitude $AR$ is mapped to the altitude $AD$. (Alternatively, one can simply show that $\triangle ARE \sim \triangle ADB$.) It follows that $\frac{|AR|}{|AD|} = \frac{|AE|}{|AB|}$.