IMO Team Selection Test 1

Since $O$ is the circumcentre, we get that $$\begin{array}{rl}\angle BAO& =\frac{1}{2}(180^{\circ }-\angle AOB)\\ & =90^{\circ }-\angle ACB\\ & =90^{\circ }-\angle EFA\\ & =\angle FAR,\end{array}$$ 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 $△POB\sim △BAE$ 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 $△AEF\sim △ABC$. Under this similarity, the altitude $AR$ is mapped to the altitude $AD$. (Alternatively, one can simply show that $△ARE\sim △ADB$.) It follows that $\frac{|AR|}{|AD|}=\frac{|AE|}{|AB|}$.