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"chunk-d4c4f366a47f3e13da193e0b600addae": {
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"tokens": 1200,
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"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∴∠",
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"chunk_order_index": 0,
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"full_doc_id": "doc-2fa5acefbe7fd21e1f33cb1c739119ea",
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"file_path": "unknown_source"
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},
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"chunk-618195be0ad6c05ea189e352a4c2e7bb": {
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"tokens": 106,
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"content": "c35908e/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",
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"chunk_order_index": 1,
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"full_doc_id": "doc-2fa5acefbe7fd21e1f33cb1c739119ea",
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"file_path": "unknown_source"
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