Question 8:
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ∼ ∆CFB
Answer:
In ∆ABE and ∆CFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ∆ABE ∼ ∆CFB (By AA similarity criterion)
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