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Question
In the parallelogram ABCD, side AD is produced at point E, and BE intersects CD at point F. Prove that ΔABE ∼ ΔCFB.
Theorem
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Solution
In ΔABE and ΔCFB
∠BAE = ∠FCB .....(Opposite angles of a parallelogram are equal.)
∠AEB = ∠CBF .....(Alternal interior angles since AE || BC.)
Therefore, ΔАВЕ ∼ ΔCFB .....(By AA similarity criterion.)
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