Austria2019

Let $P$ be the midpoint of $BS$, see Figure 1.

Since triangles $△SNP$ and $△SAB$ are similar with factor $2$, the segment $NP$ is parallel to $AB$ and half the length of the segment $AB$. Therefore $NPCH$ is a parallelogram.

As $NP$ and $BC$ are orthogonal and $PC$ and $BS$ are orthogonal, we have $\angle NPC=\angle CBP=60^{\circ }$.

Thus we obtain $\angle NHC=\angle NPC=60^{\circ }$.

Figure 1: Problem 2