Example 11. As shown in Figure 11, $AB = CD = 1, \angle ABC$ $= 90^{\circ}, \angle CBD = 30^{\circ}$. Find $AC$. (1986, Canadian Competition)
Let $A C=x$. By the Angle Bisector Theorem, we have
$$
\frac{A D}{D C}=\frac{A B \cdot \sin 120^{\circ}}{B C \cdot \sin 30^{\circ}},
$$

which means $\frac{x+1}{1}=\frac{1 \cdot \frac{\sqrt{3}}{2}}{B C \cdot \frac{1}{2}}$.
Therefore, $B C=\frac{\sqrt{3}}{x+1}$.
Applying the Pythagorean Theorem in $\mathrm{Rt} \triangle A B C$, we get
$$
1+\frac{3}{(x+1)^{2}}=x^{2}.
$$

Rearranging gives $(x+2)\left(x^{3}-2\right)=0$.
According to the problem, $x=\sqrt[3]{2}$, i.e., $A C=\sqrt[3]{2}$.