Given is a parallelogram $ABCD$, with $AB <AC <BC$. Points $E$ and $F$ are selected on the circumcircle  $\omega$ of $ABC$ so that the tangenst to $\omega$ at these points pass through point $D$ and the segments $AD$ and $CE$ intersect. 
It turned out that $\angle ABF = \angle DCE$. Find the angle $\angle{ABC}$.
A. Yakubov, S. Berlov
1. **Denote the angles:**
   Let $\angle ABC = x$, $\angle CBF = y$, and $\angle ACE = z$.

2. **Express $\angle DCE$:**
   Since $D$ lies on the tangent to the circumcircle $\omega$ at $E$ and $F$, we have $\angle DCE = x + y$.

3. **Express $\angle ACB:**
   Using the fact that the sum of angles in $\triangle ABC$ is $180^\circ$, we get:
   \[
   \angle ACB = 180^\circ - \angle BAC - \angle ABC = 180^\circ - 2x - y - z
   \]

4. **Express $\angle BCF:**
   Since $F$ lies on the circumcircle $\omega$ of $\triangle ABC$, we have:
   \[
   \angle BCF = x + z
   \]

5. **Express $\angle FCD:**
   Since $D$ lies on the tangent to the circumcircle $\omega$ at $F$, we have:
   \[
   \angle FCD = 180^\circ - z
   \]

6. **Express $\angle CFD:**
   Since $D$ lies on the tangent to the circumcircle $\omega$ at $F$, we have:
   \[
   \angle CFD = y
   \]

7. **Express $\angle AED:**
   Since $E$ lies on the circumcircle $\omega$ of $\triangle ABC$, we have:
   \[
   \angle AED = 180^\circ - z
   \]

8. **Express $\angle EAD:**
   Since $E$ lies on the circumcircle $\omega$ of $\triangle ABC$, we have:
   \[
   \angle EAD = \angle BAE - \angle BAD = 180^\circ - \angle BCE - \angle BAD = 3x + y - 180^\circ
   \]

9. **Equate $\angle EAD$ and $\angle CFD$:**
   For $x = 60^\circ$, we have:
   \[
   \angle EAD = \angle CFD
   \]

10. **Equate $\angle FCD$ and $\angle AED$:**
    We have:
    \[
    \angle FCD = \angle AED
    \]

11. **Prove $\triangle FCD \sim \triangle AED$:**
    To prove $\triangle FCD \sim \triangle AED$, we need to show:
    \[
    \frac{AE}{ED} = \frac{FC}{CD}
    \]

12. **Use the Sine Law in $\triangle FCD$:**
    Using the Sine Law in $\triangle FCD$ and the tangents $DE = DF$, we have:
    \[
    \frac{DE}{CD} = \frac{DF}{CD} = \frac{\sin z}{\sin y}
    \]

13. **Use the Sine Law in the circumcircle of $\triangle ABC$:**
    Using the Sine Law in the circumcircle of $\triangle ABC$, we have:
    \[
    \frac{AE}{FC} = \frac{\sin \angle ACE}{\sin \angle CBF} = \frac{\sin z}{\sin y}
    \]

14. **Conclusion:**
    Since $\frac{AE}{ED} = \frac{FC}{CD}$, we have $\triangle FCD \sim \triangle AED$, which completes the proof.

The final answer is $\boxed{60^\circ}$