Q. 104.4( 50 Votes )

# 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.

Answer :

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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