3. As shown in Figure 4, points $A$ and $D$ are on the semicircle with diameter $BC$, and $D$ is the midpoint of arc $\overparen{AC}$. $AC$ intersects $BD$ at point $E$. If $AE=3, CD=2\sqrt{5}$, then $AB=$ $\qquad$ ,$BC=$ $\qquad$
3. 6, 10.

As shown in Figure 7, extend $BA$ and $CD$ to intersect at point $F$.

It is easy to prove that Rt $\triangle CDE \sim$ Rt $\triangle CAF$.
Thus, $\frac{CE}{CF} = \frac{CD}{AC}$
$$
\begin{array}{l}
\Rightarrow \frac{CE}{4 \sqrt{5}} = \frac{2 \sqrt{5}}{CE + 3} \\
\Rightarrow CE = 5 .
\end{array}
$$

By the Pythagorean theorem, $AF = \sqrt{CF^2 - AC^2} = 4$.
By the secant theorem,
$$
\begin{array}{l}
FA \cdot FB = FD \cdot FC \\
\Rightarrow 4(AB + 4) = 2 \sqrt{5} \times 4 \sqrt{5} \\
\Rightarrow AB = 6 .
\end{array}
$$

By the Pythagorean theorem, $BC = \sqrt{AB^2 + AC^2} = 10$.