6. As shown in Figure 3, the incircle $\odot O$ of the right triangle $\triangle ABC$ touches the hypotenuse $AB$, the legs $BC$, and $CA$ at points $D$, $E$, and $F$, respectively. Perpendiculars are drawn from point $D$ to $AC$ and $BC$, with the feet of the perpendiculars being $M$ and $N$. If the area of rectangle $CMND$ is 8, then $\frac{1}{AD}+\frac{1}{BD}=(\quad)$.
(A) $\frac{1}{2}$
(B) $\frac{\sqrt{2}}{2}$
(C) $\frac{2}{3}$
(D) $\frac{\sqrt{3}}{2}$
6. A.

From the problem, let $O D=O E=O F=r$,
$$
\begin{array}{l}
A D=A F=m, B D=B E=n, C E=C F=r . \\
\text { Then } S=S_{\triangle A B C}=\frac{1}{2}(r+m)(r+n) \\
=\frac{1}{2}(r(r+m+n)+m n)=\frac{1}{2}(S+m n) \\
\Rightarrow S=m n .
\end{array}
$$

By $D M / / B C \Rightarrow \triangle A D M \backsim \triangle A B C$
$$
\Rightarrow S_{\triangle A D M}=S \cdot \frac{m^{2}}{(m+n)^{2}}=\frac{m^{3} n}{(m+n)^{2}} \text {. }
$$

Similarly, $S_{\triangle B D N}=\frac{m n^{3}}{(m+n)^{2}}$.
Thus, $S_{\text {rectangle } C M D N}$
$$
\begin{array}{l}
=m n-\frac{m^{3} n}{(m+n)^{2}}-\frac{m n^{3}}{(m+n)^{2}} \\
=\frac{2 m^{2} n^{2}}{(m+n)^{2}}=8 \\
\Rightarrow \frac{1}{m}+\frac{1}{n}=\frac{1}{2} \Rightarrow \frac{1}{A D}+\frac{1}{B D}=\frac{1}{2} .
\end{array}
$$