In the plane $\boldsymbol{\alpha}$, there is an equilateral triangle $\mathrm{A}^{\prime} \mathrm{BC}$, the line $D E / / B C$, and it intersects $A^{\prime} B$ and $A^{\prime} C$ at $D$ and $E$ respectively. Along $\mathrm{DE}$, fold $\triangle \mathrm{A}^{\prime} \mathrm{DE}$ so that the plane containing $\mathrm{A}^{\prime} \mathrm{DE}$ is perpendicular to the plane $\boldsymbol{\alpha}$. At this time, the position of point $A^{\prime}$ is at $A$, and connect $\mathrm{AB}$. Ask where the line $\mathrm{DE}$, which is parallel to $\mathrm{BC}$, should be taken so that $\mathrm{AB}$ is the shortest.
Solution 1
Let the equilateral triangle
have a side length of 1,
$F$ be the midpoint of $DE$, and $G$ be the midpoint of $BC$. Connect $AF, AG, FG$, then $\mathrm{AF} \perp \mathrm{DE}, \mathrm{AF} \perp \alpha, \mathrm{AG} \perp \mathrm{BC}, \mathrm{AF} \perp \mathrm{FG}$.
$$
\begin{array}{l}
\text { Let } \mathrm{AF}=\mathrm{x}, \text { then } \mathrm{FG}=\frac{\sqrt{3}}{2}-x . \\
\text { Therefore, } \mathrm{AB}^{2}=\mathrm{BG}^{2}+\mathrm{AG}^{2}=\left(\frac{1}{2}\right)^{2}+\mathrm{x}^{2}+ \\
\left(\frac{\sqrt{3}}{2}-x\right)^{2}=2 \mathrm{x}^{2}-\sqrt{3} \mathrm{x}+1 .
\end{array}
$$

When $x=\frac{\sqrt{3}}{4}$, $\mathrm{AB}^{2}$, i.e., $\mathrm{AB}$, reaches its minimum value, at this time $\mathrm{DE}$ is the midline of the equilateral triangle.

Solution 2 Let the side length of the equilateral triangle be $1, \mathrm{BD}$ be $\mathrm{x}, \angle \mathrm{ADB}=\theta$. In $\triangle \mathrm{ABD}$, by the cosine rule we get
$$
\begin{array}{l}
\mathrm{AB}^{2}=\mathrm{x}^{2}+(1-\mathrm{x})^{2}-2 \mathrm{x} \cdot(1-\mathrm{x}) \cos \theta \\
=2(1+\cos \theta)\left(x-\frac{1}{2}\right)^{2}+\frac{1}{2}(1-\cos \theta)
\end{array}
$$

Therefore, when $x=\frac{1}{2}$, i.e., $\mathrm{DE}$ is the midline of $\triangle \mathrm{A}^{\prime} \mathrm{BC}$, $\mathrm{AB}^{2}$, i.e., $\mathrm{AB}$, reaches its minimum value.

Solution 3 Let the side length of the equilateral triangle be $1, \angle \mathrm{ADB}=\theta$. In $\triangle \mathrm{ABD}$,
$$
\begin{array}{l}
\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}-2 \mathrm{AD} \cdot \mathrm{BD} \cdot \cos \theta \\
=(\mathrm{AD}+\mathrm{BD})^{2}-2 \mathrm{AD} \cdot \mathrm{BD}(1+\cos \theta) \\
=1-2 \mathrm{AD} \cdot \mathrm{BD}(1+\cos \theta) .
\end{array}
$$
$\because A D+B D=1$ (constant), $\therefore$ when and only when $A D=B D$, $A D \cdot B D$ is maximized, at this time $A B^{2}$, i.e., $A B$, is minimized.