Distinct points $A$ and $B$ are given on the plane. Consider all triangles $ABC$ in this plane on whose sides $BC,CA$ points $D,E$ respectively can be taken so that
(i) $\frac{BD}{BC}=\frac{CE}{CA}=\frac{1}{3}$;
(ii) points $A,B,D,E$ lie on a circle in this order.

Find the locus of the intersection points of lines $AD$ and $BE$.
1. **Given Conditions and Setup:**
   - Points \(A\) and \(B\) are distinct points on the plane.
   - Consider all triangles \(ABC\) such that points \(D\) and \(E\) can be taken on sides \(BC\) and \(CA\) respectively.
   - The conditions are:
     \[
     \frac{BD}{BC} = \frac{CE}{CA} = \frac{1}{3}
     \]
     \[
     \text{Points } A, B, D, E \text{ lie on a circle in this order.}
     \]

2. **Using the Given Ratios:**
   - From the given ratios, we have:
     \[
     BD = \frac{1}{3}BC \quad \text{and} \quad CE = \frac{1}{3}CA
     \]

3. **Applying Power of a Point:**
   - Since \(A, B, D, E\) lie on a circle, by the Power of a Point theorem, we have:
     \[
     BD \cdot BC = CE \cdot CA
     \]
     Substituting the given ratios:
     \[
     \frac{1}{3}BC \cdot BC = \frac{1}{3}CA \cdot CA
     \]
     Simplifying, we get:
     \[
     \frac{1}{3}BC^2 = \frac{1}{3}CA^2 \implies BC^2 = CA^2
     \]
     This implies:
     \[
     CA = \sqrt{2} \cdot CB
     \]

4. **Locus of Point \(C\):**
   - The ratio \(\frac{CA}{CB} = \sqrt{2}\) is constant, which means \(C\) moves on the Apollonius circle \(\mathcal{O}\) of \(\overline{AB}\) with the given ratio.

5. **Intersection Point \(P\):**
   - Let \(P\) be the intersection of lines \(AD\) and \(BE\).
   - Let \(F\) be the intersection of line \(CP\) with line \(AB\).

6. **Using Ceva's Theorem:**
   - By Ceva's theorem for the concurrent cevians \(AD, BE, CF\), we have:
     \[
     \frac{AF}{FB} = \frac{DC}{BD} \cdot \frac{EA}{CE}
     \]
     Substituting the given ratios:
     \[
     \frac{AF}{FB} = \frac{2}{1} \cdot \frac{2}{1} = 4
     \]

7. **Using Menelaus' Theorem:**
   - For \(\triangle CBF\) cut by \(APD\), we have:
     \[
     \frac{PF}{CP} = \frac{AF}{AB} \cdot \frac{BD}{DC}
     \]
     Substituting the given ratios:
     \[
     \frac{PF}{CP} = \frac{4}{5} \cdot \frac{1}{2} = \frac{2}{5}
     \]
     Hence:
     \[
     \frac{FP}{FC} = \frac{2}{7}
     \]

8. **Conclusion:**
   - Since \(F\) is fixed and the ratio \(\frac{FP}{FC}\) is constant, the locus of \(P\) is the homothetic circle of \(\mathcal{O}\) under homothety with center \(F\) and ratio \(\frac{2}{7}\).

The final answer is the homothetic circle of \( \boxed{ \mathcal{O} } \) under homothety with center \(F\) and ratio \(\frac{2}{7}\).