6. As shown in Figure 3, given that $AB$ and $CD$ are two perpendicular diameters of $\odot O$, $E$ is a point on segment $OB$, and $BE=2OE$. The extension of segment $CE$ intersects $\odot O$ at point $F$, and segment $AF$ intersects $DO$ at point $G$. Then
$$
DG: GC=(\quad) .
$$
(A) $1: 2$
(B) $1: 3$
(C) $2: 3$
(D) $3: 4$
6. B.

As shown in Figure 8, let the radius of $\odot O$ be 3. Then
$$
O E=1, C E=\sqrt{10} .
$$

Connect $D F$. Then
$$
\begin{array}{l}
\angle C F D=90^{\circ} \\
\Rightarrow \triangle E O C \sim \triangle D F C \\
\Rightarrow \frac{E C}{D C}=\frac{O C}{C F} \\
\Rightarrow C F=\frac{9 \sqrt{10}}{5} \\
\Rightarrow E F=C F-C E=\frac{4 \sqrt{10}}{5} \text {. } \\
\end{array}
$$

Draw $F H \perp O B$ at point $H$. Then
$$
\begin{array}{l}
\triangle E O C \sim \triangle E H F \Rightarrow F H=\frac{12}{5}, E H=\frac{4}{5} . \\
\text { Also } \triangle A G O \sim \triangle A F H \\
\Rightarrow \frac{O G}{A O}=\frac{F H}{A H} \\
\Rightarrow O G=\frac{3}{2} . \\
\text { Therefore } \frac{D G}{G C}=\frac{O D-O G}{O C+O G}=\frac{1}{3} \\
\Rightarrow D G: G C=1: 3 .
\end{array}
$$