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)