10. A. As shown in Figure 6, quadrilateral $ABCD$ is inscribed in $\odot O$, $AB$ is the diameter, $AD=DC$, $BA$ and $CD$ are extended to intersect at point $E$, $BF \perp EC$ is drawn and intersects the extension of $EC$ at point $F$. If $AE=AO, BC=6$, then the length of $CF$ is $\qquad$
10. A. $\frac{3 \sqrt{2}}{2}$.

As shown in Figure 11, connect $A C$, $B D$, and $O D$.
Since $A B$ is the diameter of $\odot O$, we have
$$
\angle B C A = \angle B D A = 90^{\circ}.
$$

From the given conditions, $\angle B C F = \angle B A D$.
Thus, Rt $\triangle B C F \backsim$ Rt $\triangle B A D$
$$
\Rightarrow \frac{B C}{C F} = \frac{B A}{A D}.
$$

Since $O D$ is the radius of $\odot O$ and $A D = C D$, $O D$ is the perpendicular bisector of $A C$, and $O D \parallel B C$.
Therefore, $\frac{D E}{D C} = \frac{O E}{O B} = 2$.
Hence, $D E = 2 C D = 2 A D$, and $C E = 3 A D$.
From $\triangle A E D \backsim \triangle C E B$
$$
\Rightarrow D E \cdot E C = A E \cdot B E
$$
$$
\Rightarrow 2 A D \cdot 3 A D = \frac{B A}{2} \cdot \frac{3}{2} B A
$$
$$
\Rightarrow B A = 2 \sqrt{2} A D.
$$
Thus, $C F = \frac{A D}{B A} \cdot B C = \frac{B C}{2 \sqrt{2}} = \frac{3 \sqrt{2}}{2}$.