6. In the Cartesian coordinate system $x O y$, the line $l_{1}$: $4 x+5 y=20$ intersects the $x$-axis and $y$-axis at points $A$ and $B$, respectively. The line $l_{2}$ intersects the line segment $A B$ and $O A$ at points $C$ and $D$, respectively, and bisects the area of $\triangle A O B$. Then the minimum value of $C D^{2}$ is
6. $5 \sqrt{41}-25$.

From the conditions, we know $O A=5, O B=4, A B=\sqrt{41}$. Let $\angle B A O=\theta$. Then $\cos \theta=\frac{5}{\sqrt{41}}$. Given $S_{\triangle A O B}=2 S_{\triangle A C D}$, we get $A C \cdot A D=\frac{1}{2} A B \cdot A O=\frac{5 \sqrt{41}}{2}$.
By the cosine rule, we have
$$
\begin{array}{l}
C D^{2}=A D^{2}+A C^{2}-2 A D \cdot A C \cos \theta \\
\geqslant 2 A D \cdot A C-2 A D \cdot A C \cos \theta \\
=5 \sqrt{41}-25,
\end{array}
$$

When $A D=A C$, the equality holds.
Therefore, the minimum value of $C D^{2}$ is $5 \sqrt{41}-25$.