6) $5 \sqrt{41}-25$ Hint: From the conditions, we know that $O A=5, O B=4, A B=\sqrt{41}$. 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 have
$$
A C \cdot A D=\frac{1}{2} A B \cdot A O=\frac{5 \sqrt{41}}{2} .
$$

By the cosine rule, we get
$$
\begin{aligned}
C D^{2} & =A D^{2}+A C^{2}-2 \cdot A D \cdot A C \cdot \cos \theta \\
& \geqslant 2 A D \cdot A C-2 A D \cdot A C \cdot \cos \theta \\
& =5 \sqrt{41}-25,
\end{aligned}
$$

Equality holds when $A D=A C$. Therefore, the minimum value of $C D^{2}$ is $5 \sqrt{41}-25$.
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 segments $A B$ 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 $\qquad$