Parallelogram $ABCD$ is such that angle $B < 90$ and $AB<BC$. Points E and F are on the circumference of $\omega$ inscribing triangle ABC, such that tangents to $\omega$ in those points pass through D. If $\angle EDA= \angle{FDC}$, find $\angle{ABC}$.

1. **Identify the given conditions and setup:**
   - Parallelogram \(ABCD\) with \(\angle B < 90^\circ\) and \(AB < BC\).
   - Points \(E\) and \(F\) are on the circumcircle \(\omega\) of \(\triangle ABC\), and tangents to \(\omega\) at \(E\) and \(F\) pass through \(D\).
   - \(\angle EDA = \angle FDC\).

2. **Introduce point \(G\):**
   - Let \(G\) be the second intersection of \(DA\) with \(\odot (ABC)\).

3. **Reflection argument:**
   - The reflection about the perpendicular bisector of \(EF\) fixes \(\odot (ABC)\) and \(D\), but swaps the lines \(DA\) and \(DC\). Therefore, it swaps points \(G\) and \(C\).

4. **Equilateral triangle argument:**
   - Since the reflection swaps \(G\) and \(C\), we have \(DG = DC\).
   - Given that \(AB = DC\) (since \(ABCD\) is a parallelogram and opposite sides are equal), and \(GC = AB\) (since \(ABCG\) is an isosceles trapezium), we conclude \(DG = DC = GC\).

5. **Conclusion about \(\triangle DGC\):**
   - Since \(DG = DC = GC\), \(\triangle DGC\) is equilateral.
   - Therefore, \(\angle CDG = 60^\circ\).

6. **Determine \(\angle ABC\):**
   - Since \(\angle CDG = 60^\circ\) and \(\angle ABC\) is part of the same configuration, we conclude that \(\angle ABC = 60^\circ\).

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