1. The problem states: In parallelogram ABCD, point E lies on CD, and we need to find angle $\angle ABC$.\n\n2. Recall that in a parallelogram, opposite angles are equal and adjacent angles are supplementary (sum to 180°). Also, the sum of angles in any triangle is 180°.\n\n3. Given the figure, angle $\angle A = 25^\circ$ and angle $\angle C = 80^\circ$. Since ABCD is a parallelogram, $\angle A = \angle C$ and $\angle B = \angle D$. However, here $\angle A \neq \angle C$, so these must be angles inside the figure, not the parallelogram's interior angles.\n\n4. Since $\angle A = 25^\circ$ and $\angle C = 80^\circ$ are given, and E lies on CD, we consider triangle $\triangle B C E$ or parallelogram properties.\n\n5. In parallelogram ABCD, adjacent angles are supplementary, so $\angle A + \angle B = 180^\circ$. Given $\angle A = 25^\circ$, then\n$$\angle B = 180^\circ - 25^\circ = 155^\circ.$$\n\n6. Therefore, $\angle ABC = 155^\circ$.\n\nFinal answer: \n$$\boxed{155^\circ}$$