Answer: $3 + \sqrt{2}$.

Denote $\angle EDB = \alpha$ (Fig. 32). The similarity of $ADB$ and $BEC$ yields $\angle DAB = \angle EBC$ and $\angle BDA = \angle CEB$; the second of which gives $\angle DEB = 180^\circ - \angle CEB = 180^\circ - \angle BDA = \angle EDB = \alpha$. Therefore $DB = BE$. Hence the similarity of $ADB$ and $BEC$ yields $\frac{BE}{AD} = \frac{EC}{DB} = \frac{EC}{BE}$; thus $\left(\frac{BE}{AD}\right)^2 = \frac{BE}{AD} \cdot \frac{EC}{BE} = \frac{EC}{AD} = 2$, which gives $\frac{BE}{AD} = \sqrt{2}$ or $BE = \sqrt{2}AD$.

The isosceles triangle $BDE$ gives $\angle DBE = 180^\circ - 2\alpha$. On the other hand,
$$\begin{aligned} \angle DBE &= 120^\circ - \angle ABD - \angle EBC = 120^\circ - \angle ABD - \angle DAB \\ &= 120^\circ - (\angle ABD + \angle DAB) = 120^\circ - \alpha. \end{aligned}$$
The equation $180^\circ - 2\alpha = 120^\circ - \alpha$ yields $\alpha = 60^\circ$. So the triangle $BDE$ is equilateral, which yields $DE = BE = \sqrt{2}AD$.

Therefore $\frac{AC}{AD} = \frac{AD+DE+EC}{AD} = \frac{AD+\sqrt{2}AD+2AD}{AD} = 1 + \sqrt{2} + 2 = 3 + \sqrt{2}$.

The similarity of $ADB$ and $BEC$ yields $\angle DAB = \angle EBC$ and $\angle ABD = \angle BCE$. So the triangles $ADB$ and $BEC$ are also similar to $ABC$. Therefore $\angle ADB = \angle BEC = \angle ABC = 120^\circ$. But this yields $\angle BDE = \angle DEB = 180^\circ - 120^\circ = 60^\circ$, meaning that the triangle $BDE$ is equilateral.

The similarity of $ADB$ and $ABC$ yields $\frac{AD}{AB} = \frac{AB}{AC}$. The similarity of $BEC$ and $ABC$ yields $\frac{EC}{BC} = \frac{BC}{AC}$. Combining these with $2AD = EC$, we see that

Fig. 32

$$BC^2 = EC \cdot AC = 2AD \cdot AC = 2AB^2.$$
So $BC = \sqrt{2}AB$, meaning that the scale factor between triangles $ADB$ and $BEC$ is $\sqrt{2}$. Thus $DE = BD = \sqrt{2}AD$.

$$\text{Therefore } \frac{AC}{AD} = \frac{AD+DE+EC}{AD} = \frac{AD+\sqrt{2}AD+2AD}{AD} = 1+\sqrt{2}+2 = 3+\sqrt{2}.$$