In the given 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
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.