4. Given is an isosceles triangle $\triangle A B C$. A square $A C D E$ is drawn over the leg $\overline{A C}$, and an equilateral triangle $\triangle B F C$ is drawn over the leg $\overline{B C}$. The diagonals $\overline{E C}$ and $\overline{A D}$ of the square intersect at point $S$, and the line $A F$ intersects the leg $\overline{B C}$ at point $G$. The line $S G$ intersects the leg $\overline{A C}$ at point $H$. If $|\angle D A F|=85^{\circ}$, what is the measure of the angle $\angle A H S$?
First method:

The diagonals of a square bisect each other and are of equal length, i.e., $|S A|=|S C|$. We have proven that $|G A|=|G C|$. For points $S$ and $G$, it holds that they are equidistant from the endpoints of the segment $\overline{A C}$, so they lie on the perpendicular bisector of the segment $\overline{A C}$. Since the perpendicular bisector of a segment is perpendicular to the segment, it follows that $S G \perp \overline{A C}$, i.e., $|\angle A H S|=90^{\circ}$.