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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 $$\triangle$$ ABE ~ $$\triangle$$ CFB.

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 $$\triangle$$ ABE and $$\triangle$$ CFB,

$$\angle$$ A = $$\angle$$ C
(Opposite angles of a parallelogram)
$$\angle$$ AEB = $$\angle$$ CBF
(Alternate interior angles as AE || BC)

$$\triangle$$ ABE ~ $$\triangle$$ CFB (AA similarity criterion)