Diagonal DF of a parallelogram DEFG bisects angle D. Show that it bisects angle F.Alsoshow that DEFG is a rhombus
Answers
Given : Diagonal DF of a parallelogram DEFG bisects angle D
To Find : show that it bisects angle F
show that DEFG is a rhombus
Solution:
DEFG is a parallogram
Hence
DE || FG
EF || DG
Diagonal DF bisects angle D
=> ∠FDG = ∠FDE = (1/2) ∠D
DE || FG and DF is transversal
=> ∠FDE = ∠DFG
=> ∠DFG = (1/2) ∠D
EF || DG and DF is transversal
=> ∠FDG = ∠DFE
=> ∠DFE = (1/2) ∠D
Hence ∠DFG = ∠DFE = (1/2) ∠D
∠DFG = ∠DFE
∠DFG + ∠DFE = ∠F
=> ∠DFG = ∠DFE = (1/2) ∠F
=>DF bisects angle F
∠FDG = ∠DFG
=> DG = FG
∠DFE = ∠FDE
=> DE = EF
in Δ GDF & Δ EDF
DF = DF common
∠FDG = ∠FDE
∠DFG = ∠DFE
Δ GDF ≅ Δ EDF (ASA )
=> DG = DE
DG = FG
DE = EF
DG = DE
Hence DG = FG = DE = EF
Hence parallelogram DEFG is a rhombus
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