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 $EN=\frac{AD}{2}=DH=HE$.