Solution:-

From the figure, ABCD is a parallelogram,

Then, E is a point on AD and produced and BE intersects CD at F.

We have to prove that ∆ABE ~ ∆CFB

Consider ∆ABE and ∆CFB

∠A = ∠C [opposite angles of a parallelogram]
∠ABE = ∠BFC [alternate angles are equal]
∆ABE ~ ∆CFB