4. In $\triangle A B C$, $A C=2, D$ is the midpoint of $A B, E$ is a point on $C D$, $E D=\frac{1}{3} C D$. If $C E=\frac{1}{3} A B$, and $C E \perp$ $A E$, then $B C=$ $\qquad$.
4. $2 \sqrt{2}$ From the given, $E$ is the centroid of $\triangle ABC$, and $CE=\frac{2}{3} CD$. Also, $CE=\frac{1}{3} AB$, so $CD=\frac{1}{2} AB$, thus $\angle ACB=90^{\circ}$. Extending $AE$ to intersect $BC$ at $F$, then $AE \cdot AF = AC^2 = 4$, hence $CF^2 = EF \cdot AF = \frac{1}{2} AE \cdot AF = 2$, so $BC = 2 CF = 2 \sqrt{2}$.