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AAA Similarity

E is a point ...

Question

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB

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Solution

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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