# E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB

To Prove: Δ ABE ∼ Δ CFB
Given: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. As shown in the figure.
Proof:
In ΔABE and ΔCFB,

A = C                     (Opposite angles of a parallelogram are equal)

AEB = CBF              (Alternate interior angles are equal  because AE || BC)

Therefore,

ΔABE ΔCFB (By AA similarity)

Hence, Proved.

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