Given,

ABCD is a parallelogram,

In ΔACE

D and O are the mid points of AE and AC

DO ⎸⎸EC

OB ⎸⎸CF and AB = BF → (i)

DC = BF (AB = DC as ABCD is a parallelogram)

In ΔEDC and ΔCBF

DC = BC

∠EDC = ∠CBF

∠ECD = ∠CFB

So, by ASA congruency

ΔEDC≅ΔCBF

DE = BC

DC = BC

AB = BC

BF = BC ( AB = BF from (i) )