Question

8, E is a point on the side AD produced of a
parallelogram ABCD and BE intersects CD
at E. Show that AABE – ACFB.​

Answer 1

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)

Hope this will help you.....

Answer 2

Answer:

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