Question

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

Answer 1

Given: ABCD is a parallelogram, where BE intersects CD at F. E is a point on the side AD produced.

To Prove: ΔABE ~ ΔCFB

Proof: In ∆ABE and ∆CFB,  
$\angle$A = $\angle$C (Opposite angles of a parallelogram)  
$\angle$AEB = $\angle$CBF (Alternate interior angles as AE || BC)  
∴ ∆ABE ∼ ∆CFB (By AA similarity criterion)

Hence Proved