Given : B and F are mid points of AB and AC.

R.T.P. : i) EF // BC, ii) EF = 1/2 BC
Construction: GC // AB, extend EF upto G.
Proof:
ΔAEF ΔCGF
∠AFE = ∠CFG (Vertically opposite angles)
AF = FC
∠EAF = ∠GCF (Alternate angles)
∴ ΔAEF ≅ ΔCGF
∴ CG = BE and CG // BF (Construction)
∴ EBCG is a parallelogram.