To Prove: Δ ABE ∼ Δ CFB
Given: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. As shown in the figure.
In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram are equal)
∠AEB = ∠CBF (Alternate interior angles are equal because AE || BC)
ΔABE ΔCFB (By AA similarity)Hence, Proved.
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