Problem 1443

1. Define the problem setup:
- Let $ABCD$ be a square.
- Draw an equilateral triangle $BCD$ on the outside of the square.
- Let $S$ be the vertex of the equilateral triangle opposite $C$.
- Let $N$ be the midpoint of segment $AS$.
- Let $H$ be the midpoint of side $CD$.

2. Identify key properties and relationships:
- Since $BCD$ is an equilateral triangle, $\angle BCD = 60^\circ$.
- $H$ is the midpoint of $CD$, so $H$ divides $CD$ into two equal segments.
- $N$ is the midpoint of $AS$, so $N$ divides $AS$ into two equal segments.

3. Establish the relationship between the points:
- Let $E$ be the midpoint of $DS$. Since $DS$ is a side of the equilateral triangle $BCD$, $E$ is equidistant from $D$ and $S$.
- Since $DCS$ is an isosceles triangle with $DC = DS$, $\angle CED = 90^\circ$.

4. Use similarity and congruence:
- Since $\triangle DHE \sim \triangle DCS$, we have that $\angle HED = \angle CED = 90^\circ$.
- Since $\triangle CED$ is a right triangle, $H$ is the circumcenter of $\triangle CED$. Therefore, $DH = HC = HE$.

5. Analyze the angles:
- Since $SDA \sim SEN$, we have that $EN = \frac{AD}{2} = DH = HE$.
- We need to find $\angle HEN$:
$$\angle HEN = \angle HED + \angle DEN = 15^\circ + 180^\circ - \angle NES$$
Since $\angle NES = \angle ADS$ and $\angle ADS = 90^\circ - \angle CDS$, we have:
$$\angle HEN = 15^\circ + 180^\circ - (90^\circ - 15^\circ) = 15^\circ + 180^\circ - 75^\circ = 120^\circ$$

6. Conclude the angle calculation:
- Thus, $\angle NHE = \angle HNE = 30^\circ$.
- Therefore, $\angle NHC = \angle NHE + \angle EHC = 30^\circ + 30^\circ = 60^\circ$.

$\blacksquare$