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)