Example 8
If a square $A B C D$ with side length $a$ can cover an equilateral triangle $E F G$ with side length 1, how large must $a$ be at least?
We first determine the positional relationship between $\triangle E F G$ and the square $A B C D$ in several steps, and then find the value of $a$.
(1) By translation, we can assume that one vertex of the square $A B C D$ coincides with a vertex of $\triangle E F G$. The method is as follows: Among $E, F, G$, let $E$ be the leftmost point, and $F$ be the rightmost point. One of the points $E, F$ must be the highest or the lowest. Without loss of generality, let $E$ be the lowest point. Translate the square $A B C D$ so that $B$ coincides with $E$ (Figure 1.12(1)). At this time, the square $A B C D$ still covers $\triangle E F G$.
(2) Let $B$ coincide with $E$. Connect $B D$. Since $\angle F B G=60^{\circ} > \angle D B A=45^{\circ}$, $F$ and $G$ are not on the same side of $B D$. If $\angle F B D \neq \angle D B G$, without loss of generality, assume $\angle F B D > 30^{\circ} > \angle D B G$. Rotate $\triangle F B G$ around point $B$ to $\triangle F^{\prime} B G^{\prime}$, so that $\angle F^{\prime} B D=30^{\circ}$ (Figure 1.12(2)).

We claim that $F^{\prime}$ is still inside $\triangle B C D$. To verify this, let the rays $B F$ and $B F^{\prime}$ intersect $C D$ at $H$ and $H^{\prime}$, respectively. Since $\angle F^{\prime} B C > \angle F B C$, $B H^{\prime} > B H \geqslant B F = B F^{\prime}$, thus point $F^{\prime}$ is inside the segment $B H^{\{\prime}$, and therefore inside $\triangle B C D$.
$G^{\prime}$ is symmetric to $F^{\prime}$ with respect to $B D$, so $G^{\prime}$ is also inside the square $A B C D$.
Therefore, we can assume $\angle F B D = \angle D B G = 30^{\circ}$, otherwise, we use $\triangle F^{\prime} B G^{\prime}$ to replace $\triangle F E G$ (or, keeping $\triangle E F G$ fixed, rotate the square $A B C D$ around $B$ in the opposite direction so that $\angle F B D = 30^{\circ}$).
(3) Let $B$ coincide with $E$, and $\angle F B D = \angle D B G = 30^{\circ}$. If point $F$ is not on $C D$, draw $C^{\prime} D^{\prime} \parallel C D$ through $F$, and $A^{\prime} D^{\prime} \parallel A D$ through $G$, to form the square $A^{\prime} B C^{\prime} D^{\prime}$ (Figure 1.12(3)). This is a smaller square than $A B C D$, and it still covers $\triangle E F G$.

Combining (1), (2), and (3), we can conclude that every square that covers $\triangle E F G$ is not smaller than the square $A B C D$ in Figure 1.13.

Therefore, the side length $a$ of the square $A B C D$ in the figure is the smallest.

Finding the side length $a$ at this time is not difficult. By congruent triangles, we get $\alpha = \beta = \frac{1}{2}(90^{\circ} - 60^{\circ}) = 15^{\circ}$, so
$$
a = \cos 15^{\circ} = \sqrt{\frac{1 + \cos 30^{\circ}}{2}} = \frac{\sqrt{2 + \sqrt{3}}}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}.
$$