The best way to solve the question of class 10th math of triangles chapter of exercise 6.3, how i solve this question, because it is very tough E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
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E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB. Q.8
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Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,
In ΔABE and ΔCFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ ΔABE ~ ΔCFB (AA similarity criterion)