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03/06/2021 12:41 pm
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E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that ΔABE ~ ΔCFB.
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03/06/2021 12:42 pm
Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,
In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (AA similarity criterion)
This post was modified 4 years ago by Raavi Tiwari