Question

Question 8
E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that $\Delta ABE\sim \Delta CFB$.

Answer 1

In $\Delta ABE$ and $\Delta CFB$,
$\angle A=\angle C$ (Opposite angles of a parallelogram)
$\angle AEB=\angle CBF$ (Alternate interior angles as AE || BC)
$\therefore \Delta ABE\sim \Delta CFB$ (By AAA similarity criterion)