如图,BD,CD分别是三角形ABC的两个外角,角CBE和角BCF的平分线,试探索角BOD与角A之间的数量关系

4个回答

  • ∵∠A+∠ABC+∠ACB=180

    ∴∠ABC+∠ACB=180-∠A

    ∵∠CBE=180-∠ABC,BD平分∠CBE

    ∴∠CBD=∠CBE/2=(180-∠ABC)/2=90-∠ABC/2

    ∵∠BCF=180-∠ACB,CD平分∠BCF

    ∴∠BCD=∠BCF/2=(180-∠ACB)/2=90-∠ABC/2

    ∴∠BDC=180-(CBD+∠BCD)

    =180-(90-∠ABC/2+90-∠ACB/2)

    =∠ABC/2+∠ACB/2

    =(∠ABC+∠ACB)/2

    =(180-∠A)/2

    =90-∠A/2