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

In ΔABE and ΔCFB,

∠BAE = ∠FCB           ...(Opposite angles of a parallelogram)

∠AEB = ∠CBF          ...(Alternate interior angles as AE || BC)

∴ ΔABE ∼ ΔCFB       ...(By AA similarity criterion)

Answer 2

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

To Prove: ΔABE ~ ΔCFB.

Proof: In parallelogram ABCD, 

opposite angles are equal,

Hence, ∠A = ∠C     ...(1)

Also, In parallelogram ABCD opposite sides are parallel, AD || BC

Now since AE is AD extended,

AE || BC

and BE is the traversal

∴ ∠AEB = ∠CBF      ...(Alternate Angles)   ...(2)

Now in Δ ABE & Δ CFB

∠A = ∠C                ...[From (1)]

∠AEB = ∠CBF       ...[From (2)]

∴ ΔABE ~ ΔCFB           ...(AA similarity criterion)

Hence proved.