In figure 3.13, 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.
Given: line DE || line GF
Ray EG and ray FG are bisectors of 
and 
respectively
To Prove: i. 
ii. 
Proof: Ray EG and ray FG are bisectors of 
and 
respectively.
So, ∠DEG = ∠GEF = 1/2 ∠DEF ……………..(1)
∠DFG = ∠GFM = 1/2 ∠DFM ………..(2)
Also, ∠EDF = ∠DFG …..(3) [Alternate interior angles]
In ΔDEF
∠DFM = ∠DEF + ∠EDF
From (2) and (3)
2∠EDF = ∠DEF + ∠EDF
⇒ ∠EDF = ∠DEF
From (1)
⇒ ∠EDF = 2∠DEG
⇒ ∠DEG = 1/2 ∠EDF
Hence, (i) is proved.
Line DE || line GF
From alternate interior angles
∠DEG = ∠EGF …….(4)
From (1)
∠GEF = ∠EGF
Since, in the ΔEGF sides opposite to equal angles are equal.
∴ EF = FG
Hence, (ii) is proved.
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