in∆ DEF, angle E=2 angle F,DM is the angle bisector of angle EDF that intersect EF at M if DM=MF then prove that angle EDF=72°
Answers
∠EDF = 72°
Step-by-step explanation:
Let say ∠ F = α
∠E = 2∠F
=> ∠E = 2α
∠D + ∠E + ∠F = 180°
=> ∠D + 2α + α = 180°
=> ∠D = 180° - 3α
=> ∠EDF = 180° - 3α
DM is the angle bisector of ∠EDF
=> ∠FDM = (180° - 3α)/2
DM = FM
=> ∠FDM = ∠DFM
=> ∠FDM = ∠F
=> (180° - 3α)/2 = α
=> 180° - 3α = 2α
=> 180° = 5α
=> α = 36°
∠EDF = 180° - 3α = 180° - 3(36°) = 72°
=> ∠EDF = 72°
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Answer:
Let say ∠ F = x
∠E = 2∠F
=> ∠E = 2x
∠D + ∠E + ∠F = 180°
=> ∠D + 2x+ x = 180°
=> ∠D = 180° - 3x
=> ∠EDF = 180° - 3x
DM is the angle bisector of ∠EDF
=> ∠FDM = (180° - 3x)/2
DM = FM
=> ∠FDM = ∠DFM
=> ∠FDM = ∠F
=> (180° - 3x)/2 = α
=> 180° - 3x = 2α
=> 180° = 5x
=> α = 36°
∠EDF = 180° - 3α = 180° - 3(36°) = 72°
=> ∠EDF = 72°