E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that. ∆ABE ∼ ∆CFB

From the given information in the question,the figure looks like below;

From the figure above;

In ∆ABE and ∆CFB , <ABE = < CFB ( As Alternate angles are always equal).

Also, <BAE = BCD (As opposite angles in a parallelogram are always equal).

According to Angle-Angle criterion: If two angles of one triangle are equal to two angles of another triangle, then the two triangles are similar.

Therefore, ∆ABE ∼ ∆CFB

Hence, we concluded that, ∆ABE ∼ ∆CFB