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प्रश्न
In the given figure, line DE || line GF, ray EG and ray FG are bisectors of ∠DEF and ∠DFM respectively. Prove that,
- ∠DEG = `1/2∠"EDF"`
- EF = FG
उत्तर
(i) ∠DEG = ∠FEG = x° ...(i)[Ray EG bisects ∠DEF]
∠GFD = ∠GFM = y° ...(ii)[Ray FG bisects ∠DFM]
line DE || line GF and DF is their transversal. ...[Given]
∴ ∠EDF = ∠GFD ...[Alternate angles]
∴ ∠EDF = y° ...(iii) [From (ii)]
line DE || line GF and EM is their transversal. ...[Given]
∴ ∠DEF = ∠GFM ...[Corresponding angles]
∴ ∠DEG + ∠FEG = ∠GFM ...[Angle addition property]
∴ x° + x° = y° ...[From (i) and (ii)]
∴ 2x° = y°
∴ x° = `1/2`y°
∴ ∠DEG = `1/2`∠EDF ...[From (i) and (iii)]
(ii) line DE || line GF and GE is their transversal. ...[Given]
∴ ∠DEG = ∠FGE ...(iv)[Alternate angles]
∴ ∠FEG = x° ...(v)[From (i) and (iv)]
∴ In ∆FEG,
∠FEG = ∠FGE ...[From (v)]
∴ EF = FG ...[Converse of isosceles triangle theorem]
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