E is a point on the side ad produced of the parallelogram ABCD and BE intersectsCD at F. Show that ∆ABE ~ ∆ CFB
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In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ∼ ΔCFB (By AA similarity criterion)
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