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{"embedding_dim": 1024, "data": [{"__id__": "chunk-75b23a7e22383153b011bd3d121184f0", "__created_at__": 1752211461, "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=\"1.91044072615923in\"}\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", "full_doc_id": "doc-744d2f4e81528499ae55a82849ed415b", "file_path": "unknown_source"}, {"__id__": "chunk-e2c7bd24a26246e194d4d56ab2ed22f1", "__created_at__": 1752211461, "content": "2d6b6c62c9b4c41/media/image2.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": "doc-744d2f4e81528499ae55a82849ed415b", "file_path": "unknown_source"}], "matrix": "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