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)


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