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In ABC, E and F are mid points of sides AB and AC respectively then prove that 

i) EF // BC and 

ii) EF = 1/2 BC

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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.

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