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Diagonal DF of a parallelogram DEFG bisects angle D. Show that it bisects angle F.Alsoshow that DEFG is a rhombus |
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Answer» 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 Learn More: EFGH is a parallelogram.EI bisects angle E and GJ bisects angle G ... Prove that the line drawn through the centre of a circle to BISECT a ... |
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