Let $ABC$ be an equilateral triangle . Let  point $D$ lie on side $AB,E$  lie on side $AC, D_1$ and $E_1$ lie on side BC such that $AB=DB+BD_1$ and $AC=CE+CE_1$. Calculate the smallest angle between the lines $DE_1$ and $ED_1$.
1. **Understanding the Problem:**
   We are given an equilateral triangle \(ABC\) with points \(D\) on \(AB\), \(E\) on \(AC\), \(D_1\) on \(BC\), and \(E_1\) on \(BC\). The conditions \(AB = DB + BD_1\) and \(AC = CE + CE_1\) imply that \(D_1\) and \(E_1\) are collinear with \(B\) and \(C\) respectively.

2. **Positioning Points:**
   Since \(D_1\) and \(E_1\) lie on \(BC\), we can assume without loss of generality that \(D_1\) is closer to \(B\) and \(E_1\) is closer to \(C\). This means \(D_1\) and \(E_1\) divide \(BC\) into segments.

3. **Using Symmetry:**
   Given the symmetry of the equilateral triangle, we can use geometric properties to find the angles. The key is to recognize that the smallest angle between lines \(DE_1\) and \(ED_1\) will be influenced by the symmetry and equal lengths of the sides.

4. **Calculating the Angles:**
   Since \(ABC\) is equilateral, each internal angle is \(60^\circ\). The lines \(DE_1\) and \(ED_1\) will form angles with the sides of the triangle. We need to find the smallest angle between these lines.

5. **Using Vector Geometry:**
   Let’s place the triangle in a coordinate system with \(A\) at the origin \((0,0)\), \(B\) at \((1,0)\), and \(C\) at \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\). Points \(D\) and \(E\) can be parameterized as \(D = (x, 0)\) and \(E = \left(\frac{1}{2}y, \frac{\sqrt{3}}{2}y\right)\) for \(0 \leq x, y \leq 1\).

6. **Finding \(D_1\) and \(E_1\):**
   Since \(D_1\) and \(E_1\) lie on \(BC\), we can parameterize them as \(D_1 = \left(\frac{1}{2}(1-t), \frac{\sqrt{3}}{2}(1-t)\right)\) and \(E_1 = \left(\frac{1}{2}(1-s), \frac{\sqrt{3}}{2}(1-s)\right)\) for \(0 \leq t, s \leq 1\).

7. **Calculating Vectors:**
   The vectors \(\overrightarrow{DE_1}\) and \(\overrightarrow{ED_1}\) are:
   \[
   \overrightarrow{DE_1} = \left(\frac{1}{2}(1-s) - x, \frac{\sqrt{3}}{2}(1-s)\right)
   \]
   \[
   \overrightarrow{ED_1} = \left(\frac{1}{2}(1-t) - \frac{1}{2}y, \frac{\sqrt{3}}{2}(1-t) - \frac{\sqrt{3}}{2}y\right)
   \]

8. **Dot Product and Angle Calculation:**
   The angle \(\theta\) between two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by:
   \[
   \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}
   \]
   Using the vectors \(\overrightarrow{DE_1}\) and \(\overrightarrow{ED_1}\), we calculate the dot product and magnitudes.

9. **Simplifying the Expression:**
   Given the symmetry and equal lengths, the smallest angle between \(DE_1\) and \(ED_1\) will be when \(D\) and \(E\) are at the midpoints of \(AB\) and \(AC\) respectively, making \(D_1\) and \(E_1\) also at the midpoints of \(BC\).

10. **Final Calculation:**
    The smallest angle between \(DE_1\) and \(ED_1\) in an equilateral triangle is \(60^\circ\).

The final answer is \(\boxed{60^\circ}\).