FINAL ROUND

Construct the circumference ${\Gamma }_{BC}$ with the diameter $BC$. ${\Gamma }_{BC}$ passes through $M$ and $L$. Indeed, $\angle AMB=90^{\circ }$ ($AB$ is the diameter of ${\Gamma }_{AB}$) then $\angle BMC=180^{\circ }-\angle AMB=90^{\circ }$, so $M$ belongs to ${\Gamma }_{BC}$. Similarly $$\angle DLC=90^{\circ }\Rightarrow \angle BLC=180^{\circ }-\angle DLC=90^{\circ },$$ so $L$ also belongs to ${\Gamma }_{BC}$. Now construct the circumference ${\Gamma }_{AD}$ with the diameter $AD$. We conclude similarly that $N$ and $K$ belong to ${\Gamma }_{AD}$.

Note that $$\angle NKM=\angle NKA=0.5\sim AN=\angle NDA=\angle BDA$$ and $$\angle KML=\angle CML=0.5\sim CL=\angle CBL=\angle CBD.$$ Since $BC\parallel AD$, we have $\angle BDA=\angle CBD$, and so $\angle NKM=\angle KML$, which gives $NK\parallel ML$, as required.