Example 5 As shown in Figure 7, the side length of square $ABCD$ is 1, and points $E, F, G$ are on sides $AB, AD, BC$ respectively (they can coincide with the vertices). If $\triangle EFG$ is an equilateral triangle, find the maximum and minimum values of the area of $\triangle EFG$.
(2010, Wuhan Junior High School Mathematics Competition)
Solve as shown in Figure 7, construct $E K \perp F G$. Then $K$ is the midpoint of side $F G$. Connect $A K$ and $B K$.
$$
\begin{array}{l}
\text { By } \angle E K G=\angle E B G=\angle E K F \\
=\angle E A F=90^{\circ},
\end{array}
$$

we get that $E, K, G, B$ and $E, K, F, A$ are respectively concyclic.
Thus, $\angle K B E=\angle E G K=60^{\circ}$,
$$
\angle E A K=\angle E F K=60^{\circ} \text {. }
$$

Therefore, $\triangle A B K$ is an equilateral triangle, i.e., $K$ is a fixed point.
Thus, when $F G / / A B$, the area of $\triangle E F G$ is minimized, at this time,
$$
S_{\triangle E F G}=\frac{\sqrt{3}}{4} F G^{2}=\frac{\sqrt{3}}{4} A B^{2}=\frac{\sqrt{3}}{4} .
$$

When point $F$ coincides with $D$ or point $G$ coincides with $C$, the area of $\triangle E F G$ is maximized, at this time,
$$
\begin{array}{l}
S_{\triangle E F G}=\frac{\sqrt{3}}{4} C K^{2} \\
=\sqrt{3}\left[\left(\frac{1}{2}\right)^{2}+\left(1-\frac{\sqrt{3}}{2}\right)^{2}\right]=2 \sqrt{3}-3 .
\end{array}
$$
3 Constructing the inscribed triangle