Question

Question 8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB

Class 10 - Math - Triangles Page 140

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)

Answer 2

In a parallelogram opposite angles are equal to each other. Then find the relation between angles of both triangle's to show them similar.

If two angles of a triangle are respectively equal to two angles of another triangle, then Triangles are similar (because by the angle sum property of a triangle their third angle will also be equal ) and it is called AA similarity criterion.
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Solution:

[Fig is in the attachment]

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)

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Hope this will help you.....