Point D is from AC of triangle ABC so that 2AD=DC. Let DE be perpendicular to BC and AE
intersects BD at F. It is known that triangle BEF is equilateral. Find <ADB?
1. **Define the problem and given conditions:**
   - Point \( D \) is on \( AC \) of triangle \( \triangle ABC \) such that \( 2AD = DC \).
   - \( DE \) is perpendicular to \( BC \).
   - \( AE \) intersects \( BD \) at \( F \).
   - \( \triangle BEF \) is equilateral.
   - We need to find \( \angle ADB \).

2. **Introduce angle \( \alpha \):**
   - Let \( \angle EAD = \alpha \).

3. **Apply the Law of Sines in \( \triangle ADE \):**
   \[
   \frac{DE}{\sin \alpha} = \frac{AD}{\sin 30^\circ}
   \]
   Since \( \sin 30^\circ = \frac{1}{2} \):
   \[
   \frac{DE}{\sin \alpha} = \frac{AD}{\frac{1}{2}} = 2AD
   \]

4. **Relate \( DE \) and \( DC \):**
   - Given \( 2AD = DC \), we have:
   \[
   \frac{DE}{\sin \alpha} = DC
   \]
   - This implies:
   \[
   \sin \alpha = \frac{DE}{DC}
   \]

5. **Relate \( \alpha \) and \( \angle ACE \):**
   - Since \( \sin \alpha = \sin \angle ACE \), we have:
   \[
   \alpha = \angle ACE \quad \text{(since \( \alpha + \angle ACE \neq 180^\circ \))}
   \]

6. **Use the equilateral triangle property:**
   - Since \( \triangle BEF \) is equilateral, \( \angle AEB = 60^\circ \).
   - Therefore:
   \[
   \alpha + \angle ACE = 60^\circ
   \]
   - Given \( \alpha = \angle ACE \):
   \[
   2\alpha = 60^\circ \implies \alpha = 30^\circ
   \]

7. **Find \( \angle ADF \) in \( \triangle ADF \):**
   - In \( \triangle ADF \):
   \[
   \angle ADF = 180^\circ - \alpha - \angle AFD
   \]
   - Given \( \alpha = 30^\circ \) and \( \angle AFD = 60^\circ \):
   \[
   \angle ADF = 180^\circ - 30^\circ - 60^\circ = 90^\circ
   \]

8. **Conclusion:**
   - Since \( \angle ADF = 90^\circ \), \( \angle ADB = 90^\circ \).

The final answer is \( \boxed{90^\circ} \)