3. As shown in Figure 4, it is known that quadrilateral $ABCD$ is inscribed in $\odot O$, $AB$ is the diameter, $AD=DC$, and $BA$, $CD$ are extended to intersect at point $E$, $BF \perp EC$ intersects the extension of $EC$ at point $F$. If $AE=AO, BC=6$, then the length of $CF$ is $\qquad$.
3. $\frac{3 \sqrt{2}}{2}$.

As shown in Figure 11, connect $A C$, $B D$, and $O D$.

Since $A B$ is the diameter of $\odot O$, we have
$$
\begin{array}{l}
\angle B C A \\
=\angle B D A \\
=\angle B F C \\
=90^{\circ} .
\end{array}
$$

Also, since quadrilateral $A B C D$ is an inscribed quadrilateral of $\odot O$, we get
$$
\angle B C F=\angle B A D \text {. }
$$

Therefore, $\triangle B C F \backsim \triangle B A D \Rightarrow \frac{B C}{C F}=\frac{B A}{A D}$.
Since $O D$ is the radius and $A D=C D$,
$\Rightarrow O D$ perpendicularly bisects $A C \Rightarrow O D / / B C$
$$
\Rightarrow \frac{D E}{D C}=\frac{O E}{O B}=2 \text {. }
$$

By the secant theorem, we have
$$
\begin{array}{l}
D E \cdot E C=A E \cdot B E \Rightarrow B A=2 \sqrt{2} A D \\
\Rightarrow C F=\frac{B C}{2 \sqrt{2}}=\frac{3 \sqrt{2}}{2} .
\end{array}
$$