dsProject/dsLightRag/Topic/JiHe/kv_store_doc_status.json

{
  "doc-744d2f4e81528499ae55a82849ed415b": {
    "status": "processed",
    "chunks_count": 2,
    "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![](./Images/868c80b8ff2547adb2d6b6c62c9b4c41/media/image1.png)\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![](./Images/868c80b8ff2547adb2d6b6c62c9b4c41/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",
    "content_summary": "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.eyJpc3MiOiJ6aGlkYV9z...",
    "content_length": 1975,
    "created_at": "2025-07-11T05:24:21.601678+00:00",
    "updated_at": "2025-07-11T05:25:09.796687+00:00",
    "file_path": "unknown_source"
  }
}