Example 8 As shown in Figure 2.10.8, in $\triangle A B C$, $B C=5, A C=$ $12, A B=13$, points $D, E$ are taken on sides $A B, A C$ respectively, such that line segment $D E$ divides $\triangle A B C$ into two parts of equal area. Try to find the minimum length of such a line segment $D E$.
Obviously, $\angle A C B=90^{\circ}$, draw the median $C F$ of side $A B$, then $S_{\triangle A C F}=\frac{1}{2} S_{\triangle A B C}=15$, which means $A C \cdot A F=78$.

Take any point $D$ on $B F$, connect $C D$, draw $F E / / C D$ intersecting $A C$ at $E$, connect $D E$, then $D E$ divides $\triangle A B C$ into two parts with equal area.

Let $B D=x$, in $\triangle B C D$, by the cosine rule, we have $C D=\sqrt{25+x^{2}-\frac{50 x}{13}}$. Also, $A D: A F=A C: A E$, then
$$
A E=\frac{A C \cdot A F}{A D}=\frac{78}{13-x}
$$

In $\triangle A E D$, by the cosine rule, we have
$$
D E^{2}=\left(\frac{78}{13-x}\right)^{2}+(13-x)^{2}-144
$$

Let $(13-x)^{2}=y, y>0$, then we have
$$
y^{2}-\left(144+D E^{2}\right) y+78^{2}=0
$$

By $\Delta=\left(144+D E^{2}\right)^{2}-4 \cdot 78^{2} \geqslant 0$, we know the minimum length of $D E$ is $2 \sqrt{3}$.