#### E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. show that Δ ABE ∼ Δ CFB.

Anonymous

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Krishna

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Step 1: Examine the given figure

Step 2: Identify the parallel lines and transversals in the given figure

NOTE: AD || BC and AB || DC

BE is a transversal

Step 3: Locate the equal angles in the figure

NOTE: ∠ABE = ∠CFB (Alternate angles)

∠AEB = ∠CBF (Alternate angles)

Step 4: Prove that \triangle ABE and \triangle CFE are similar

NOTE: According to AAA theorem we can say \triangle ABE and \triangle CFE

are similar

Hence; Δ ABE ∼ Δ CFB proved (AAA criterion)