15. (25 points) Given a square $A B C D$ with side length 1, points $E, F, G$ are on sides $A D, A B, D C$ respectively (they can coincide with the vertices). If $\triangle E F G$ is an equilateral triangle, find the maximum and minimum values of the area of $\triangle E F G$.
Three, 15. As shown in Figure 6, draw $E M \perp F G$ at point $M$, and connect $A M, D M$.
Since $\angle E D G=\angle E M G=90^{\circ}$
$\Rightarrow E, D, G, M$ are concyclic
$\Rightarrow \angle E D M=\angle E G M=60^{\circ}$.
Similarly, $\angle E A M=60^{\circ}$.

Therefore, $\triangle A M D$ is an equilateral triangle, which means $M$ is a fixed point. When $F G \parallel A D$, the area of $\triangle E F G$ is minimized.
At this time, $F G=A D=1$, and the area of $\triangle E F G$ is $\frac{\sqrt{3}}{4}$.
When point $F$ coincides with $B$ or point $G$ coincides with $C$ (as shown in Figure 7), the area of $\triangle E F G$ is maximized.
It is easy to see that $\triangle C B F \cong \triangle C D E$.
Let $B F=D E=x, A E=A F=1-x$. Then $A E^{2}+A F^{2}=E F^{2}$,

which is $2(1-x)^{2}=1+x^{2}$.
Solving for $x$ gives $x=2 \pm \sqrt{3}$.
Since $x<1$, we have $x=2-\sqrt{3}$.
Thus, $C F=\sqrt{6}-\sqrt{2}$.
Therefore, the area of $\triangle E F G$ is $\frac{\sqrt{3}}{4} C F^{2}=2 \sqrt{3}-3$.
Hence, the maximum area of $\triangle E F G$ is $2 \sqrt{3}-3$, and the minimum area is $\frac{\sqrt{3}}{4}$.