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Question

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
1385702_48261b68581a41fe8b960e0fd549c8e3.PNG

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Solution

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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