# E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB

**Solution:**

Let's draw a parallelogram ABCD as per the given question

If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

This is referred to as the AA criterion for two triangles.

In ΔABE and ΔCFB

∠BAE = ∠FCB (opposite angles of a parallelogram)

∠AEB = ∠FBC [AE || BC and EB is a transversal, alternate interior angle]

⇒ ΔABE ~ ΔCFE (AA criterion)

**Video Solution:**

## E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that ΔABE ~ ΔCFB

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.3 Question 8:

E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that ΔABE ~ ΔCFB