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E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that `DeltaA B E ~ DeltaC F B`.

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In △ABE and △CFB,
`∠ABE=∠CFB` (Alternate angles)
`∠BAE=∠BCF` (opposite angles of a parallelogram)
∴ By AA criterion of similarity, △ABE ∼ △CFB
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