MathNet 2019-part-01 — 184 problems (EN–VI) Question 35

Let $ABCD$ be a square. The equilateral triangle $BCS$ is constructed on the exterior of the side $BC$. Let $N$ denote the midpoint of the line segment $AS$ and let $H$ be the midpoint of the side $CD$.
Prove: $\angle NHC = 60^{\circ}$.