E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD atF. show that ∆ABEsimilar∆CFD. solve
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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.
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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