E is a point on the side AD produced of aparallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB
In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles because AE || BC)
Therefore,
ΔABE
ΔCFB (By AA similarity)