Question

E is the point on the side AD produced of a parallelogram ABC and BE intersects CD at F. Show that TRIANGLE ABE~TRIANGLE CFB.

Answer 1

Draw a parallelogram  ABCD. According to the question to the question to the question, extend AD to E and join E point to B. Let. EB intersect at CD at F point.
In ΔAEB and ΔCBF
∠E=∠B    (Alternate Angles)
∠A=∠C    (Opposite angles of parallelogram are equal)
∴ΔAEB=ΔCBF

Answer 2

OK this will help you