 Hello friends, welcome to the session, I am Atta. Let's discuss the given question. E is a point on the side AB produced of a parallelogram ABCD and BE intersect CD at F. Show that triangle ABE is similar to triangle CFB. According to the question of figure is, ABCD is a parallelogram and E is a point on the side AB, AB produced and BE intersect CD at point F. Now let's start with the solution. We are given ABCD is a parallelogram and we have to show that triangle ABE is similar to triangle CFB. That is ABE is similar to triangle CFB. Now let's start with the proof in triangle ABE angle CFB. We have angle AEB equal to angle CBF. We can see from the figure that angle AEB is to angle CB alternate angles. Angle A equal to angle C. Angle A is equal to angle C since they are opposite angles of parallelogram which are equal. Therefore by A, A criteria of similarity triangle ABE is similar to triangle CFB which is proved. Hope you understood the solution and enjoyed the session. Goodbye and take care.