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来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/04/30 00:21:12
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解题思路: 内心一个性质如试题:已知三角形ABC,A为顶点,在三角形ABC中,I是三角形的内心,连接AI并延长,交BC于点E。求证:AI/IE=AB/BE=AC/EC 证明:连接BI,CI 由正弦定理 AB/sin∠AIB = AI/sin∠ABI AB/AI = sin∠AIB/sin∠ABI BE/sin∠BIE = BI/sin∠EBI BE/BI = sin∠BIE/sin∠EBI I为内心,BI为∠ABC平分线 ∠ABI = ∠EBI sin∠ABI = sin∠EBI 又∠AIB和∠BIE互为补角 sin∠AIB = sin∠BIE 所以 AB/AI = BE/BI 即 AI/IE = AB/BE 同理可证 AI/IE = AC/EC 所以 AI/IE = AB/BE = AC/EC
解题过程:
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最终答案:略