On the side $BC$ of the equilateral triangle $ABC$, choose any point $D$, and on the line $AD$, take the point $E$ such that $| B A | = | BE |$. Prove that the size of the angle $AEC$ is  of does not depend on the choice of point $D$, and find its size.
1. **Draw the equilateral triangle \(ABC\)**:
   - Since \(ABC\) is an equilateral triangle, we have \(|AB| = |BC| = |CA|\) and \(\angle BAC = \angle ABC = \angle BCA = 60^\circ\).

2. **Choose any point \(D\) on side \(BC\)**:
   - Let \(D\) be an arbitrary point on \(BC\).

3. **Construct point \(E\) on line \(AD\) such that \(|BA| = |BE|\)**:
   - Since \(|BA| = |BE|\), point \(E\) lies on the circle with center \(B\) and radius \(BA\). This circle passes through points \(A\) and \(C\) because \(|BA| = |BC|\).

4. **Draw the circle with center \(B\) and radius \(BA\)**:
   - This circle intersects line \(AD\) at point \(E\).

5. **Prove that \(\angle AEC\) does not depend on the choice of point \(D\)**:
   - Since \(E\) lies on the circle with center \(B\) and radius \(BA\), and the circle passes through points \(A\) and \(C\), the arc \(AC\) subtends the angle \(\angle AEC\) at point \(E\).
   - By the intercepted arc theorem (or the inscribed angle theorem), the measure of the angle subtended by an arc at any point on the circle is half the measure of the central angle subtended by the same arc.
   - The central angle subtended by arc \(AC\) is \(\angle ABC = 60^\circ\).
   - Therefore, \(\angle AEC = \frac{1}{2} \times 60^\circ = 30^\circ\).

6. **Conclusion**:
   - The size of the angle \(\angle AEC\) is \(30^\circ\) and it does not depend on the choice of point \(D\).

\(\blacksquare\)

The final answer is \( \boxed{ 30^\circ } \)