In plane $\alpha$, there is an equilateral triangle $A^{\prime} B C$ (as shown in the figure). Line $D E / / B C$ intersects $A^{\prime} B$ and $A^{\prime} C$ at $D$ and $E$, respectively. Along $D E$, $\triangle A^{\prime} D E$ is folded up to be perpendicular to plane $\alpha$ at the position of $\triangle A D E$. Question: Where should the line $D E$, parallel to $B C$, be taken so that $A B$ is the shortest?
In $\triangle A^{\prime} B C$, construct the altitude $A^{\prime} M$ on side $B C$, and let $A^{\prime} M$ intersect $D E$ at $N$. Connect $B N$. When $\triangle A^{\prime} B C$ is folded to be perpendicular to plane $\alpha$, $A N \perp$ plane $\alpha$, then $A N \perp B N$. Let the side length of the equilateral triangle $A^{\prime} B C$ be $a$, and $A^{\prime} N=x$. In the right triangle $\triangle N M B$, we have $B N^{2}=N M^{2}+B M^{2}=\left(\frac{\sqrt{3}}{2} a-x\right)^{2}+\left(\frac{a}{2}\right)^{2}$, so
$$
\begin{aligned}
A B^{2} & =A N^{2}+B N^{2}=x^{2}+\left(\frac{\sqrt{3}}{2} a-x\right)^{2}+\left(\frac{a}{2}\right)^{2} \\
& =2\left(x-\frac{\sqrt{3}}{4} a\right)^{2}+\frac{5}{8} a^{2}
\end{aligned}
$$

Therefore, when $x=\frac{\sqrt{3}}{4} a$, $A B$ is the shortest, at which point $D E$ is the line connecting the midpoints of $A^{\prime} B$ and $A^{\prime} C$.