6. In the Cartesian coordinate system, with the origin $O$ as the center, draw $\odot O$. Let $\odot O$ intersect the positive half-axis of the $x$-axis at point $P$, and $D(6,8)$ lies on $\odot O$. Points $E$ and $F$ lie on the line segment $OP$ (not coinciding with points $O$ and $P$). Connect $DE$ and $DF$ and extend them to intersect $\odot O$ at points $B$ and $C$, respectively. The line $BC$ intersects the $x$-axis at point $G$. If $DE = DF$, then the value of $\sin \angle CGO$ is ( ).
(A) $\frac{3}{5}$
(B) $\frac{4}{5}$
(C) $\frac{3}{4}$
(D) $\frac{2}{3}$
6. A.

As shown in Figure 2, draw $DM \perp EF$ at point $M$, extend $DM$ to intersect $\odot O$ at point $N$, and connect $ON$ to intersect $BC$ at point $Q$.

Then $DM$ bisects $\angle EDF$.
Thus, $\overparen{CN}=\overparen{BN} \Rightarrow ON \perp BC$.
It is easy to see that points $G, M, Q, N$ are concyclic.
Therefore, $\angle CGO=\angle ONM$.
Given $MN=DM=8, OM=6$, then
$ON=\sqrt{OM^2+MN^2}=10$.
Hence $\sin \angle CGO=\sin \angle ONM=\frac{OM}{ON}=\frac{3}{5}$.