20 (1) (2)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/17 01:37:57
解题思路: 三角形内角和性质
解题过程:
证明:(1)延长BO交AC于D
∵O点是∠ABC和∠ACB的角平分线的交点
∴∠ABO=∠ABC/2,∠ACO=∠ACB/2
∵∠BDC=∠A+∠ABO,∠BOC=∠BDC+∠ACO
∴∠BOC=∠A+∠ABO+∠ACO
=∠A+(∠ABC+∠ACB)/2
=∠A+(180-∠A)/2
=90+∠A/2
(2)结论:∠BOC= 1 2∠A,
理由如下:
∵BO和CO分别是∠ABC和∠ACD的角平分线,
∴∠1= 1 2∠ABC,∠2= 1 2∠ACD,
又∵∠ACD是△ArC的一外角,
∴∠ACD=∠A+∠ABC,
∴∠2= 1 2(∠A+∠ABC)= 1 2∠A+∠1,
∵∠2是△BOC的一外角,
∴∠BOC=∠2-∠1= 1 2∠A+∠1-∠1= 1 2∠A;
(3)∠OBC= 1 2(∠A+∠ACB),∠OCB= 1 2(∠A+∠ABC),
∠BOC=180°-∠0BC-∠OCB,
=180°- 1 2(∠A+∠ACB)- 1 2(∠A+∠ABC),
=180°- 1 2∠A- 1 2(∠A+∠ABC+∠ACB),
结论∠BOC=90°- 1 2∠A.
最终答案:略
解题过程:
证明:(1)延长BO交AC于D
∵O点是∠ABC和∠ACB的角平分线的交点
∴∠ABO=∠ABC/2,∠ACO=∠ACB/2
∵∠BDC=∠A+∠ABO,∠BOC=∠BDC+∠ACO
∴∠BOC=∠A+∠ABO+∠ACO
=∠A+(∠ABC+∠ACB)/2
=∠A+(180-∠A)/2
=90+∠A/2
(2)结论:∠BOC= 1 2∠A,
理由如下:
∵BO和CO分别是∠ABC和∠ACD的角平分线,
∴∠1= 1 2∠ABC,∠2= 1 2∠ACD,
又∵∠ACD是△ArC的一外角,
∴∠ACD=∠A+∠ABC,
∴∠2= 1 2(∠A+∠ABC)= 1 2∠A+∠1,
∵∠2是△BOC的一外角,
∴∠BOC=∠2-∠1= 1 2∠A+∠1-∠1= 1 2∠A;
(3)∠OBC= 1 2(∠A+∠ACB),∠OCB= 1 2(∠A+∠ABC),
∠BOC=180°-∠0BC-∠OCB,
=180°- 1 2(∠A+∠ACB)- 1 2(∠A+∠ABC),
=180°- 1 2∠A- 1 2(∠A+∠ABC+∠ACB),
结论∠BOC=90°- 1 2∠A.
最终答案:略
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