Given :
ABCD is a parallelogram,
Diagram :
To Prove :
BF = BC
Proof :
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 and opposite sides of a parallelogram are equal and parallel)
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))
