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