Let $E$ be the midpoint of $DS$. Since $DCS$ is isosceles, we have that $\angle CED = 90^\circ$. Also we have that $DHE\sim DCS$. Since $CED$ is right triangle, we have that $H$ is the circumcentre of $(CED)$. Therefore, $DH=HC=HE$. Also, we have that $SDA\sim SEN$, hence $EN=\frac{AD}{2}=DH=HE$. We have \begin{align*} \angle HEN&=\angle HED+\angle DEN\\&=15^\circ+180^\circ-\angle NES\\&=15^\circ+180^\circ-\angle ADS\\&=15^\circ+180^\circ-(90^\circ-\angle CDS)\\&=15^\circ+180^\circ-(90^\circ-15^\circ)\\&=120^\circ. \end{align*}Thus, $\angle NHE=\angle HNE=30^\circ$. Therefore, $\angle NHC = \angle NHE+\angle EHC=30^\circ+30^\circ=60^\circ$.

An auxiliary point solution

PS. I would say that this was the hardest geo problem of the last 20 years of Austria Beginners +Junior Regional constests

I think that previous solutions are too hard. I came up with this. Let K be the midpoint of BS. Then MN || AB || CD and MN = AB/2 = CH, so NMCH is a parallelogram, so <NHC = 180 - 120 = 60