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14 KiB
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1 line
14 KiB
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{"embedding_dim": 1024, "data": [{"__id__": "chunk-4ed981658aa2a32c616f4cf86c4efbaa", "__created_at__": 1753235662, "content": "三角形三边关系的证明\n证明方法如下：\n作下图所示的三角形ABC。在三角形ABC中，[三角不等式](https://zhida.zhihu.com/search?content_id=248217850&content_type=Article&match_order=1&q=%E4%B8%89%E8%A7%92%E4%B8%8D%E7%AD%89%E5%BC%8F&zd_token=eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJpc3MiOiJ6aGlkYV9zZXJ2ZXIiLCJleHAiOjE3NTIzNzg0NDAsInEiOiLkuInop5LkuI3nrYnlvI8iLCJ6aGlkYV9zb3VyY2UiOiJlbnRpdHkiLCJjb250ZW50X2lkIjoyNDgyMTc4NTAsImNvbnRlbnRfdHlwZSI6IkFydGljbGUiLCJtYXRjaF9vcmRlciI6MSwiemRfdG9rZW4iOm51bGx9.rH6r8SvGmu-I9piEsmZfg2HjbXzUduYclZ2jfA3jZRs&zhida_source=entity)可以表示为|AB|+|BC|＞|AC|。\n\nheight=\"1.90625in\"}\n①延长直线AB至点D，并使|BD|=|BC|，连接|DC|，那么三角形BCD为等腰三角形。所以∠BDC=∠BCD。\n②记它们均为α，根据[欧几里得第五公理](https://zhida.zhihu.com/search?content_id=248217850&content_type=Article&match_order=1&q=%E6%AC%A7%E5%87%A0%E9%87%8C%E5%BE%97%E7%AC%AC%E4%BA%94%E5%85%AC%E7%90%86&zd_token=eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJpc3MiOiJ6aGlkYV9zZXJ2ZXIiLCJleHAiOjE3NTIzNzg0NDAsInEiOiLmrKflh6Dph4zlvpfnrKzkupTlhaznkIYiLCJ6aGlkYV9zb3VyY2UiOiJlbnRpdHkiLCJjb250ZW50X2lkIjoyNDgyMTc4NTAsImNvbnRlbnRfdHlwZSI6IkFydGljbGUiLCJtYXRjaF9vcmRlciI6MSwiemRfdG9rZW4iOm51bGx9.ltcWsMYJv-ZzcuBaSjYN69JC8hnIyPMFsfhIlum4yqc&zhida_source=entity)，∠ACD大于角∠ADC(α)。\n③由于∠ACD的对边为AD，∠ADC(α)的对边为AC，所以根据大角对大边([几何原本](https://zhida.zhihu.com/search?content_id=248217850&content_type=Article&match_order=1&q=%E5%87%A0%E4%BD%95%E5%8E%9F%E6%9C%AC&zd_token=eyJhbGciOiJIUzI1NiIsInR5cCI6IkpXVCJ9.eyJpc3MiOiJ6aGlkYV9zZXJ2ZXIiLCJleHAiOjE3NTIzNzg0NDAsInEiOiLlh6DkvZXljp_mnKwiLCJ6aGlkYV9zb3VyY2UiOiJlbnRpdHkiLCJjb250ZW50X2lkIjoyNDgyMTc4NTAsImNvbnRlbnRfdHlwZSI6IkFydGljbGUiLCJtYXRjaF9vcmRlciI6MSwiemRfdG9rZW4iOm51bGx9.Q1rCY0S2bj5Dwp3Fg7xb_VSFESz2_pCUETDybnHANvo&zhida_source=entity)中的命题19)就可以得到|AB|+|BC|=|AB|+|BD|=|AD|＞|AC|。\n求证：在三角形ABC中，P为其内部任意一点。请证明：∠BPC > ∠A。\n证明过程：\n\n延长BP交AC于D\n∵∠BPC是△PCD的一个外角，∠PDC是△BAD的一个外角\n∴∠BPC=∠PCD+∠PDC，∠PDC=∠DBA+∠A\n∴∠BPC=∠PCD+∠DBA+∠A\n∴∠BPC＞<43>", "full_doc_id": "JiHe.docx", "file_path": "JiHe.docx"}, {"__id__": "chunk-f24190139d17b1c4d499b9516ffb8a4f", "__created_at__": 1753235662, "content": "mediaimage2.png)\n延长BP交AC于D\n∵∠BPC是△PCD的一个外角，∠PDC是△BAD的一个外角\n∴∠BPC=∠PCD+∠PDC，∠PDC=∠DBA+∠A\n∴∠BPC=∠PCD+∠DBA+∠A\n∴∠BPC＞∠A", "full_doc_id": "JiHe.docx", "file_path": "JiHe.docx"}], "matrix": "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