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In the adjoining figure, line DE || line GF, ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively. Prove that,

i. ∠DEG = (1/2) ∠EDF 

ii. EF = FG

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Best answer

i. ∠DEG = ∠FEG = x° ….(i) [Ray EG bisects ∠DEF] 

∠GFD = ∠GFM = y° …..(ii) [Ray FG bisects ∠DFM] 

line DE || line GF and DF is their transversal. [Given]

∴ ∠EDF = ∠GFD [Alternate angles] 

∴ ∠EDF = y° ….(iii) [From (ii)] 

line DE || line GF and EM is their transversal. [Given]

∴ ∠DEF = ∠GFM [Corresponding angles] 

∴ ∠DEG + ∠FEG = ∠GFM [Angle addition property] 

∴ x° + x° = y° [From (i) and (ii)] 

∴ 2x° = y° 

∴ x° = (1/2) y° 

∴ ∠DEG = (1/2) ∠EDF [From (i) and (iii)]

ii. line DE || line GF and GE is their transversal. [Given]

∴ ∠DEG = ∠FGE …(iv) [Alternate angles] 

∴ ∠FEG = ∠FGE ….(v) [From (i) and (iv)] 

∴ In ∆FEG, 

∠FEG = ∠FGE [From (v)] 

∴ EF = FG [Converse of isosceles triangle theorem]

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