# E is point on the side AD produced of a parallelogram ABCD and BE intersects CD at E. Show that $$\triangle$$ ABE ~ $$\triangle$$ CFB

## Solution :

In triangles ABE and CFB, we have,

$$\angle$$ AEB = $$\angle$$ CBF           (alternate $$\angle$$s are equal)

$$\angle$$ A = $$\angle$$ C                      (opposite angles of parallelogram are equal)

$$\therefore$$ By AA similarity,

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