The triangle $ABC$ is equilateral. The ray $[BE$ intersects the segment $[AC]$ at $D$, such that $\angle CBE= 20^o$ and $|DE|=|AB|$. What is the measure of the angle $\angle BEC$ ?
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1. **Identify the given information and draw the diagram:**
   - Triangle \(ABC\) is equilateral, so \(\angle BAC = \angle ABC = \angle BCA = 60^\circ\).
   - Ray \(BE\) intersects segment \(AC\) at \(D\) such that \(\angle CBE = 20^\circ\).
   - \(DE = AB\).

2. **Determine the angles in the triangle:**
   - Since \(\angle CBE = 20^\circ\), we have \(\angle ABE = 60^\circ - 20^\circ = 40^\circ\).
   - Since \(ABC\) is equilateral, \(\angle BAC = 60^\circ\).

3. **Calculate the angles involving point \(D\):**
   - \(\angle ABD = 40^\circ\) (as calculated above).
   - \(\angle BAD = 60^\circ\) (since \(ABC\) is equilateral).
   - Therefore, \(\angle ADB = 180^\circ - \angle ABD - \angle BAD = 180^\circ - 40^\circ - 60^\circ = 80^\circ\).

4. **Determine the angles involving point \(E\):**
   - Since \(DE = AB\) and \(AB = AC\) (as \(ABC\) is equilateral), triangle \(ADE\) is isosceles with \(DE = AD\).
   - \(\angle ADE = \angle ADB = 80^\circ\).
   - Therefore, \(\angle EDC = 180^\circ - \angle ADE = 180^\circ - 80^\circ = 100^\circ\).

5. **Calculate the angle \(\angle BEC\):**
   - Since \(\angle EDC = 100^\circ\) and \(\angle CBE = 20^\circ\), we can use the exterior angle theorem in triangle \(BEC\).
   - \(\angle BEC = \angle EDC - \angle CBE = 100^\circ - 20^\circ = 80^\circ\).

The final answer is \(\boxed{80^\circ}\).