Question

In the figure 3.13, line DE || line GF ray EG and Ray FG are bisectors of DEF and DFM respectively. Prove that DEG is 1/2 EDF. EF=FG​

Answer 1

Answer:

Given data:

DE//GF

EG is the bisector of ∠DEF and FG is the bisector of ∠DFM

Please refer to the figure attached below.

Case (i): To prove ∠DEG = 1/2 ∠EDF

Since it is given,

EG is the bisector of ∠DEF

∴ ∠DEG = ∠GEF = ½ * ∠DEF …… (i)

And,

FG is the bisector of ∠DFM

∴ ∠DFG = ∠GFM = ½ * ∠DFM …… (ii)

Also, since DE//GF,

∴ ∠EDF = ∠DFG ….. [∵ alternate angles] …. (iii)

In ∆ DEF,  

Since the exterior angle of a triangle is equal to the sum of opposite interior angles.

∴ ∠DFM = ∠EDF + ∠DEF

⇒  2 * ∠EDF = ∠ EDF + ∠DEF ….. [from (ii) & (iii)]

⇒ 2∠EDF – ∠EDF = ∠DEF

⇒ ∠EDF = ∠DEF

⇒ ∠EDF = 2 * ∠DEG …. [from (i)]

⇒ ∠DEG = ½ * ∠EDF

Hence proved  

Case (ii): To prove EF = FG

Since DE//FG

∴ ∠DEG = ∠EGF …. [∵ alternate angles] ….. (iv)

From (i) & (iv), we get

∠EGF = ∠GEF

∴ EF = FG ….. [∵ sides opposite to equal angles are also equal]

Hence proved  

Answer 2

Step-by-step explanation:

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