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](/uploads/image/z/2528271-63-1.jpg?t=%E5%A6%82%E5%9B%BE%2CBD%2CCD%E5%88%86%E5%88%AB%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%A4%E4%B8%AA%E5%A4%96%E8%A7%92%2C%E8%A7%92CBE%E5%92%8C%E8%A7%92BCF%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E8%AF%95%E6%8E%A2%E7%B4%A2%E8%A7%92BOD%E4%B8%8E%E8%A7%92A%E4%B9%8B%E9%97%B4%E7%9A%84%E6%95%B0%E9%87%8F%E5%85%B3%E7%B3%BB)
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