FINAL ROUND

Note that all right-angled triangles $AB{A}_1$, $AH{C}_1$, $CB{C}_1$, $CH{A}_1$ are pairwise similar. Therefore, since $L$, $N$, $K$, $M$ are the midpoints of the corresponding sides, we see that $△ALB\sim △ANH\sim △CKB\sim △CMH$. It follows that $\angle LAB=\angle NAH=\angle KCB=\angle MCH=\phi$. Let $\angle CAB=\alpha$. Then $$\begin{array}{rl}\angle RCA+\angle RAC& =(\angle RC{C}_1+\angle C_{1}CA)+(\angle CAB-\angle BAL)=\\ & =(\phi +90^{\circ }-\alpha )+(\alpha -\phi )=90^{\circ },\end{array}$$ so $\angle ARC=90^{\circ }$. Similarly we have $\angle AQC=90^{\circ }$.

It follows that $A$, $R$, $Q$, $C$ lie on the same circle, and $AC$ is the diameter of this circle. So $DC=DQ$ and $\angle PQR=\angle CAP$. Therefore, $△DQC$ is an isosceles triangle. So $$\begin{array}{rl}\angle RQD& =180^{\circ }-\angle PQR-\angle DQC=\\ & =180^{\circ }-\angle CAP-\angle ACP=\angle APC=\angle RPQ,\end{array}$$ which yields that the line $DQ$ touches the circumcircle of the triangle $PQR$ (at point $Q$).