is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ?ABC is similar to ?CFB.
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In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ∼ ΔCFB (By AA similarity criterion)
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ∼ ΔCFB (By AA similarity criterion)
rathorepratham:
I said ∆abc~cfb
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