16题过程详解.

1个回答

  • 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)