证明:
∵∠A+∠ABC+∠C+∠ADC=360,∠A=∠C=90
∴∠ABC+∠ADC=360-(∠A+∠C)=180
∵BE平分∠ABC
∴∠ABE=∠ABC/2
∴∠BED=∠A+∠ABE=90+∠ABC/2
∵DF平分∠ADC
∴∠ADF=∠ADC/2
∴∠BED+∠ADF=90+∠ABC/2+∠ADC/2=90+(∠ABC+∠ADC)/2=90+90=180
∴BE∥DF (同旁内角互补,两直线平行)
证明:
∵∠A+∠ABC+∠C+∠ADC=360,∠A=∠C=90
∴∠ABC+∠ADC=360-(∠A+∠C)=180
∵BE平分∠ABC
∴∠ABE=∠ABC/2
∴∠BED=∠A+∠ABE=90+∠ABC/2
∵DF平分∠ADC
∴∠ADF=∠ADC/2
∴∠BED+∠ADF=90+∠ABC/2+∠ADC/2=90+(∠ABC+∠ADC)/2=90+90=180
∴BE∥DF (同旁内角互补,两直线平行)