Question:

1. Identify the parallel sides of the parallelogram\:
\[AB \parallel CD \quad\text{and}\quad AD \parallel BC.\]

2. Locate the points\:
- \(E\) lies on the extension of \(AD\) beyond \(D\).
- \(BE\) meets \(CD\) at \(F\).

3. Show that \(\angle ABE = \angle CFB\)\:
- \(\angle ABE\) is the angle formed by the line \(AB\) and the transversal \(BE\).
- Since \(AB \parallel CD\) and \(CF\) lies on \(CD\), the angle between \(AB\) and \(BE\) equals the angle between \(CF\) and \(FB\) (alternate interior angles).\
Hence, \(\angle ABE = \angle CFB\).

4. Show that \(\angle AEB = \angle CBF\)\:
- \(AE\) is a continuation of \(AD\); therefore \(AE\) is collinear with \(AD\).
- Because \(AD \parallel BC\), the angle made by \(AE\) (or \(AD\)) with \(BE\) equals the angle made by \(BC\) with \(BF\) (alternate interior angles).\
Hence, \(\angle AEB = \angle CBF\).

5. Apply the AA similarity criterion\:
- From steps 3 and 4 we have two pairs of equal angles:
\[\angle ABE = \angle CFB \quad\text{and}\quad \angle AEB = \angle CBF.\]
- Therefore, by the AA (Angle‑Angle) criterion, the triangles \(\Delta ABE\) and \(\Delta CFB\) are similar.

6. Conclusion\:
\[\boxed{\Delta ABE \sim \Delta CFB}.\]

Remark: The similarity also implies the proportionality of corresponding sides:
\[\frac{AB}{CF}=\frac{BE}{FB}=\frac{AE}{CB}.\]