初一几何证明题.如图.

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/06 09:32:18

初一几何证明题.如图.
初一几何证明题.如图.

初一几何证明题.如图.
证明:
(1)直接证明:
∵BO平分∠ABC,CO平分∠ACB
∴∠OBC=1/2∠ABC,∠OCB=1/2∠ACB
∴∠BOC
=180°-∠OBC-∠OCB
=180°-1/2∠ABC-1/2∠ACB
=180°-1/2(∠ABC+∠ACB)
=180°-1/2(180°-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
(2)延长BO交AC于点D
∵∠BOC是△OCD的外角
∴∠BOC=∠OCD+∠ODC
∵∠ODC是△ABD的外角
∴∠ODC=∠ABD+∠A
∵BO平分∠ABC,CO平分∠ACB
∴∠ABD=1/2∠ABC,∠OCD=1/2∠ACB
∴∠BOC
=∠OCD+∠ODC
=∠OCD+∠ABD+∠A
=1/2∠ACB+1/2∠ABC+∠A
=1/2(∠ACB+∠ABC)+∠A
=1/2(180°-∠A)+∠A
=90°-1/2∠A+∠A
=90°+1/2∠A
(3)连结AO并延长与BC交于点E
∵∠BOE是△ABO的外角
∴∠BOE=∠ABO+∠BAO
∵∠COE是△ACO的外角
∴∠COE=∠ACO+∠CAO
∵BO平分∠ABC,CO平分∠ACB
∴∠ABO=1/2∠ABC,∠ACO=1/2∠ACB
∴∠BOC
=∠BOE+∠COE
=∠ABO+∠BAO+∠ACO+∠CAO
=1/2∠ABC+1/2∠ACB+∠BAO+∠CAO
=1/2(∠ABC+∠ACB)+∠A
=1/2(180°-∠A)+∠A
=90°-1/2∠A+∠A
=90°+1/2∠A

1、∵∠BOC=180°-(1/2∠ABC+1/2∠ACB) 而∠A+∠ABC+∠ACB=180°
∴1/2∠ABC+1/2∠ACB=90°-1/2∠A代入上式
∴∠BOC=1/2∠A+90°
2、延长BO交AC于D,∵∠ADB=∠A+1/2∠ABC,、∠BOC=∠ADB+1/2∠ACB=∠A+1/2∠ABC+1/2∠ACB, 1/2∠ABC+1/2∠ACB=...

全部展开

1、∵∠BOC=180°-(1/2∠ABC+1/2∠ACB) 而∠A+∠ABC+∠ACB=180°
∴1/2∠ABC+1/2∠ACB=90°-1/2∠A代入上式
∴∠BOC=1/2∠A+90°
2、延长BO交AC于D,∵∠ADB=∠A+1/2∠ABC,、∠BOC=∠ADB+1/2∠ACB=∠A+1/2∠ABC+1/2∠ACB, 1/2∠ABC+1/2∠ACB=90°-1/2∠A代入上式
∴∠BOC=∠A+90°-1/2∠A=1/2∠A+90°
3、连接AO并延长交BC于E,∠BOE=1/2∠A+1/2∠ABC,,∠COE=1/2∠A+1/2∠ACB
∴∠BOC=∠BOE+∠COE=1/2∠A+1/2∠ABC+1/2∠A+1/2∠ACB=∠A+90°-1/2∠A=1/2∠A+90°

收起

看图:

BOC=180·-1/2(ABC+ACB)
=180·-1/2(180·-A)
=180·-90·+1/2A
=90·+1/2A