Question

E is a point on the side ad produced of the parallelogram ABCD and BE intersectsCD at F. Show that ∆ABE ~ ∆ CFB​

Answer 1

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Refer the atttachment...

Answer 2

Answer:

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)