Hrvatska 2011

Let $S$ be the center of the given semicircle, $C$ the intersection of the lines $AK$ and $BL$, and $T$ the intersection of the lines $AB$ and $KL$. From the given condition it follows that the quadrilateral $ABUV$ is cyclic so by the power of a point theorem we have $$|TU|\cdot |TV|=|TA|\cdot |TB|.$$

The quadrilateral $ABLK$ is also cyclic, so we have $$|TA|\cdot |TB|=|TK|\cdot |TL|.$$ Since $\overline{AL}$, $\overline{BK}$ and $\overline{CN}$ are altitudes of the triangle $ABC$ and $S$ is the midpoint of $AB$, points $K$, $L$, $N$ and $S$ belong to the nine points circle of that triangle. Therefore the quadrilateral $NSLK$ is also cyclic, so $$|TU|\cdot |TV|=|TA|\cdot |TB|=|TK|\cdot |TL|=|TS|\cdot |TN|.$$ Thereby the quadrilateral $NSUV$ is cyclic, and hence $\angle NVU=180^{\circ }-\angle NSU=90^{\circ }$.