6. In $\odot O$, chord $CD$ is perpendicular to diameter $AB$ at point $E$, chord $DF$ bisects $BC$ and intersects $BC$ at point $G$, chord $AF$ intersects $OC$ at point $H$. Then the value of $\frac{GH}{AB}$ is ( ).
(A) $\frac{1}{3}$
(B) $\frac{1}{4}$
(C) $\frac{1}{5}$
(D) $\frac{2}{5}$
6. B.

As shown in Figure 2, connect $C F$.
By the perpendicular diameter theorem, we have
$$
\overparen{A C}=\overparen{A D} \text {, }
$$

Thus, $\angle B=\angle A F D$.
Since $O C=O B$, we get
$$
\angle B=\angle O C B \text {. }
$$

Therefore, $\angle A F D=\angle O C B$.
Hence, $C, H, G, F$ are concyclic.
Thus, $\angle C G H=\angle A F C=\angle B$.
Consequently, $G H \parallel O B$.
In $\triangle O C B$, since $C G=G B$, we have $C H=H O$.
By the midline theorem of a triangle, we get
$$
G H=\frac{1}{2} O B=\frac{1}{4} A B \Rightarrow \frac{G H}{A B}=\frac{1}{4} .
$$