3. As shown in Figure 2, $\triangle A B C$ is inscribed in $\odot O$, and $A B=A C$, diameter $A D$ intersects $B C$ at point $E, F$ is the midpoint of $O E$. If $B F \parallel$ $C D, B C=2 \sqrt{5}$, then $C D=$ $\qquad$
3. $\sqrt{6}$.

Since $A B=A C, A D$ is the diameter, so $C E=B E=\sqrt{5}$.
Also, because $B F / / C D$, we have $E F=D E$. And since $F$ is the midpoint of $O E$, then $O F=E F=D E$.
Let $D E=x$, then $O A=O D=3 x, A E=5 x$.
According to the intersecting chords theorem, we have $C E \cdot B E=D E \cdot A E$, which is $(\sqrt{5})^{2}=x \cdot 5 x$.
Solving this, we get $x= \pm 1$ (the negative value is discarded).
Thus, $D E=1$. Therefore, $C D=\sqrt{C E^{2}+D E^{2}}=\sqrt{6}$.