4. In $\triangle A B C$, $\angle A=30^{\circ}, \angle B=105^{\circ}$, a line $D E$ is drawn through a point $D$ on side $A C$, intersecting side $A B$ or side $B C$ at point $E$, such that $\angle C D E=60^{\circ}$, and $D E$ bisects the area of $\triangle A B C$. Then $\left(\frac{C D}{A C}\right)^{2}=$ $\qquad$ -
4. $\frac{\sqrt{3}}{6}$.

As shown in Figure 2, if $\angle C D B=60^{\circ}$, then
$$
\begin{array}{l}
\angle C=45^{\circ}, \angle C B D=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}, \\
\angle D B A=30^{\circ}=\angle A, G D>B D=A D, \\
\text { thus, } S_{\triangle B C D}>S_{\triangle A B D} .
\end{array}
$$

Thus, $S_{\triangle B C D}>S_{\triangle A B D}$.
Therefore, point $E$ is on side $B C$.
Assume $A C=1, C D=k$.
Notice that
$S_{\triangle A B C}=\frac{1}{2} a b \sin c=\frac{b^{2} \sin A \cdot \sin C}{2 \sin B}=\frac{\sin 30^{\circ} \cdot \sin 45^{\circ}}{2 \sin 100^{\circ}}$.
Similarly, $S_{\triangle C D E}=\frac{k^{2} \sin 60^{\circ} \cdot \sin 45^{\circ}}{2 \sin 75^{\circ}}$.
Therefore, $\frac{k^{2} \sin 60^{\circ} \cdot \sin 45^{\circ}}{2 \sin 75^{\circ}}=\frac{1}{2} \times \frac{\sin 30^{\circ} \cdot \sin 45^{\circ}}{2 \sin 105^{\circ}}$.
Solving for $k^{2}=\frac{\sqrt{3}}{6}$, we get $\left(\frac{C D}{A C}\right)^{2}=\frac{\sqrt{3}}{6}$.