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
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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
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
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
∵∠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
∵∠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+...
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∵∠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
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∠BDC=180-1/2(∠CBE+∠BCF)
=180°-1/2(2∠A+∠ABC+∠ACB)
=180°-1/2*∠A-1/2(∠A+∠ABC+∠ACB)
=180°-1/2*∠A-1/2*180°
=180°-1/2∠A-90°
=90°-∠A/2
∵∠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+...
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∵∠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
收起