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HuangHai 2 weeks ago
parent f5192a0b0d
commit 402c9596e8
13 changed files with 506 additions and 8 deletions
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393
dsLightRag/Topic/JiHe/graph_chunk_entity_relation.graphml
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@ -0,0 +1,393 @@
<?xml version='1.0' encoding='utf-8'?>
<graphml xmlns="http://graphml.graphdrawing.org/xmlns" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://graphml.graphdrawing.org/xmlns http://graphml.graphdrawing.org/xmlns/1.0/graphml.xsd">
<key id="d11" for="edge" attr.name="created_at" attr.type="long" />
<key id="d10" for="edge" attr.name="file_path" attr.type="string" />
<key id="d9" for="edge" attr.name="source_id" attr.type="string" />
<key id="d8" for="edge" attr.name="keywords" attr.type="string" />
<key id="d7" for="edge" attr.name="description" attr.type="string" />
<key id="d6" for="edge" attr.name="weight" attr.type="double" />
<key id="d5" for="node" attr.name="created_at" attr.type="long" />
<key id="d4" for="node" attr.name="file_path" attr.type="string" />
<key id="d3" for="node" attr.name="source_id" attr.type="string" />
<key id="d2" for="node" attr.name="description" attr.type="string" />
<key id="d1" for="node" attr.name="entity_type" attr.type="string" />
<key id="d0" for="node" attr.name="entity_id" attr.type="string" />
<graph edgedefault="undirected">
<node id="Triangle ABC">
<data key="d0">Triangle ABC</data>
<data key="d1">category</data>
<data key="d2">Triangle ABC is the primary geometric figure used in the proof of the triangle inequality and angle relationships.</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="Point D">
<data key="d0">Point D</data>
<data key="d1">category</data>
<data key="d2">Point D is constructed by extending line AB and adding segment BD equal to BC, forming an isosceles triangle.</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="Point P">
<data key="d0">Point P</data>
<data key="d1">category</data>
<data key="d2">Point P is an arbitrary interior point of triangle ABC, used to demonstrate angle relationships.</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="Euclid's Fifth Postulate">
<data key="d0">Euclid's Fifth Postulate</data>
<data key="d1">category</data>
<data key="d2">Euclid's Fifth Postulate is referenced to justify angle comparisons in the geometric proof.</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="Proposition 19 of the Elements">
<data key="d0">Proposition 19 of the Elements</data>
<data key="d1">category</data>
<data key="d2">Proposition 19 from Euclid's Elements is cited to establish the relationship between angles and opposite sides in the proof.</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="三角形ABC">
<data key="d0">三角形ABC</data>
<data key="d1">category</data>
<data key="d2">三角形ABC是证明三角形三边关系的核心几何图形，用于展示边与角的几何性质。</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="三角不等式">
<data key="d0">三角不等式</data>
<data key="d1">category</data>
<data key="d2">三角不等式是几何学中描述三角形边长关系的基本定理，形式为|AB|+|BC|＞|AC|。</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="欧几里得第五公理">
<data key="d0">欧几里得第五公理</data>
<data key="d1">category</data>
<data key="d2">欧几里得第五公理是几何学基础公理之一，用于证明角的大小关系。</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="几何原本">
<data key="d0">几何原本</data>
<data key="d1">category</data>
<data key="d2">《几何原本》是欧几里得的经典数学著作，包含命题19等核心几何定理。</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="命题19">
<data key="d0">命题19</data>
<data key="d1">category</data>
<data key="d2">命题19指出‘大角对大边’，是三角形边角关系的关键依据。</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="点D">
<data key="d0">点D</data>
<data key="d1">category</data>
<data key="d2">点D是通过延长AB并添加BD=BC构造的辅助点，形成等腰三角形BCD。</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="点P">
<data key="d0">点P</data>
<data key="d1">category</data>
<data key="d2">点P是三角形ABC内部的任意点，用于证明角∠BPC与角∠A的关系。</data>
<data key="d3">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="BP">
<data key="d0">BP</data>
<data key="d1">geo</data>
<data key="d2">BP is a geometric line segment intersecting AC at point D.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="AC">
<data key="d0">AC</data>
<data key="d1">geo</data>
<data key="d2">AC is a geometric line segment intersected by BP at point D.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="D">
<data key="d0">D</data>
<data key="d1">geo</data>
<data key="d2">D is the intersection point of BP and AC.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="∠BPC">
<data key="d0">∠BPC</data>
<data key="d1">category</data>
<data key="d2">∠BPC is an angle formed at point P, exterior to triangle PCD.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="∠PCD">
<data key="d0">∠PCD</data>
<data key="d1">category</data>
<data key="d2">∠PCD is an angle within triangle PCD.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="∠PDC">
<data key="d0">∠PDC</data>
<data key="d1">category</data>
<data key="d2">∠PDC is an angle within triangle PCD and exterior to triangle BAD.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="∠DBA">
<data key="d0">∠DBA</data>
<data key="d1">category</data>
<data key="d2">∠DBA is an angle within triangle BAD.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="∠A">
<data key="d0">∠A</data>
<data key="d1">category</data>
<data key="d2">∠A is an angle within triangle BAD.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="△PCD">
<data key="d0">△PCD</data>
<data key="d1">category</data>
<data key="d2">△PCD is a triangle formed by points P, C, and D, with ∠PCD and ∠PDC as its interior angles.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="△BAD">
<data key="d0">△BAD</data>
<data key="d1">category</data>
<data key="d2">△BAD is a triangle formed by points B, A, and D, with ∠DBA and ∠A as its interior angles.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="P">
<data key="d0">P</data>
<data key="d1">geo</data>
<data key="d2">P is a geometric point where BP originates and ∠BPC is formed.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="B">
<data key="d0">B</data>
<data key="d1">geo</data>
<data key="d2">B is a geometric point where BP originates and ∠DBA is formed.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="C">
<data key="d0">C</data>
<data key="d1">geo</data>
<data key="d2">C is a geometric point where AC terminates and ∠PCD is formed.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<node id="A">
<data key="d0">A</data>
<data key="d1">geo</data>
<data key="d2">A is a geometric point where AC originates and ∠A is formed.</data>
<data key="d3">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d4">unknown_source</data>
<data key="d5">1752211508</data>
</node>
<edge source="Triangle ABC" target="Point D">
<data key="d6">8.0</data>
<data key="d7">Point D is constructed from triangle ABC by extending AB and adding segment BD equal to BC.</data>
<data key="d8">geometric construction,triangle extension</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="Triangle ABC" target="Proposition 19 of the Elements">
<data key="d6">9.0</data>
<data key="d7">Proposition 19 is applied to triangle ABC to justify the relationship between angles and sides.</data>
<data key="d8">angle-side relationship,geometric theorem</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="Triangle ABC" target="Point P">
<data key="d6">8.0</data>
<data key="d7">Point P is an interior point of triangle ABC, used to demonstrate angle relationships within the triangle.</data>
<data key="d8">angle proof,interior point</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="Point D" target="Euclid's Fifth Postulate">
<data key="d6">7.0</data>
<data key="d7">Euclid's Fifth Postulate is used to compare angles in triangle ACD, which includes point D.</data>
<data key="d8">angle comparison,geometric proof</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="三角形ABC" target="三角不等式">
<data key="d6">9.0</data>
<data key="d7">三角不等式直接应用于三角形ABC的三边关系证明。</data>
<data key="d8">几何定理,边角关系</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="三角形ABC" target="欧几里得第五公理">
<data key="d6">8.0</data>
<data key="d7">第五公理用于比较三角形ACD中的角度，与三角形ABC的构造相关。</data>
<data key="d8">公理应用,角度推导</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="三角形ABC" target="命题19">
<data key="d6">9.0</data>
<data key="d7">命题19被用于证明三角形ABC中边AD与边AC的关系。</data>
<data key="d8">几何推理,边角逻辑</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="三角形ABC" target="点D">
<data key="d6">7.0</data>
<data key="d7">点D是三角形ABC的延伸构造，用于辅助证明边角不等式。</data>
<data key="d8">几何扩展,辅助构造</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="三角形ABC" target="点P">
<data key="d6">8.0</data>
<data key="d7">点P的存在证明了三角形内部点的角与顶角的关系。</data>
<data key="d8">内部角度,几何性质</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="几何原本" target="命题19">
<data key="d6">10.0</data>
<data key="d7">命题19源自《几何原本》，是证明边角关系的理论来源。</data>
<data key="d8">定理引用,数学经典</data>
<data key="d9">chunk-75b23a7e22383153b011bd3d121184f0</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="BP" target="AC">
<data key="d6">8.0</data>
<data key="d7">BP intersects AC at point D, forming geometric relationships.</data>
<data key="d8">geometric construction,intersection</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="BP" target="P">
<data key="d6">7.0</data>
<data key="d7">P is the endpoint of BP where ∠BPC is formed.</data>
<data key="d8">angle formation,point-line connection</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="BP" target="B">
<data key="d6">8.0</data>
<data key="d7">B is the endpoint of BP where it originates.</data>
<data key="d8">line origin,point-line connection</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="AC" target="C">
<data key="d6">8.0</data>
<data key="d7">C is the endpoint of AC where it terminates.</data>
<data key="d8">line termination,point-line connection</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="AC" target="A">
<data key="d6">8.0</data>
<data key="d7">A is the endpoint of AC where it originates.</data>
<data key="d8">line origin,point-line connection</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="∠BPC" target="∠PCD">
<data key="d6">9.0</data>
<data key="d7">∠BPC is the sum of ∠PCD and ∠PDC, showing an exterior angle relationship.</data>
<data key="d8">angle sum,exterior angle theorem</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="∠BPC" target="∠A">
<data key="d6">7.0</data>
<data key="d7">∠BPC is greater than ∠A due to the sum of angles in the geometric proof.</data>
<data key="d8">angle comparison,inequality</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="∠BPC" target="△PCD">
<data key="d6">9.0</data>
<data key="d7">∠BPC is an exterior angle of △PCD, equal to the sum of its non-adjacent interior angles.</data>
<data key="d8">angle sum,exterior angle theorem</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="∠PDC" target="∠DBA">
<data key="d6">9.0</data>
<data key="d7">∠PDC is the sum of ∠DBA and ∠A, demonstrating an exterior angle relationship.</data>
<data key="d8">angle sum,exterior angle theorem</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="∠PDC" target="△BAD">
<data key="d6">9.0</data>
<data key="d7">∠PDC is an exterior angle of △BAD, equal to the sum of its non-adjacent interior angles.</data>
<data key="d8">angle sum,exterior angle theorem</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
<edge source="△PCD" target="△BAD">
<data key="d6">7.0</data>
<data key="d7">Both triangles share point D and are part of the geometric proof involving angle relationships.</data>
<data key="d8">geometric proof,shared vertex</data>
<data key="d9">chunk-e2c7bd24a26246e194d4d56ab2ed22f1</data>
<data key="d10">unknown_source</data>
<data key="d11">1752211508</data>
</edge>
</graph>
</graphml>

14
dsLightRag/Topic/JiHe/kv_store_doc_status.json
Unescape View File

@ -1,12 +1,12 @@
{
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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![](./Images/778094016d3642f2bc43eeaf1074cc38/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|。",
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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![](./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...",
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{
"doc-744d2f4e81528499ae55a82849ed415b": {
"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"
}
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{
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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![](./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",
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"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",
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<div class="example-item" onclick="fillExample('三角形中两边之和大于第三边的证明')">
三角形中两边之和大于第三边的证明？
</div>
<div class="example-item" onclick="fillExample('求证：在三角形ABC中，P为其内部任意一点。请证明：∠BPC > ∠A。')">
求证：在三角形ABC中，P为其内部任意一点。请证明：∠BPC > ∠A。
</div>
<div class="example-item" onclick="fillExample('苏轼的好朋友都有谁?')">苏轼的好朋友都有谁?</div>
<div class="example-item" onclick="fillExample('苏轼的家人都有谁?')">苏轼的家人都有谁?</div>
<div class="example-item" onclick="fillExample('苏轼有哪些有名的诗句?')">苏轼有哪些有名的诗句?</div>

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三角形三边关系的证明
证明方法如下：
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\|。
![](./Images/778094016d3642f2bc43eeaf1074cc38/media/image1.png)
![](./Images/868c80b8ff2547adb2d6b6c62c9b4c41/media/image1.png)
height="1.91044072615923in"}\
①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。
②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)。
③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\|。
求证：在三角形ABC中，P为其内部任意一点。请证明：∠BPC \> ∠A。
证明过程：
![](./Images/868c80b8ff2547adb2d6b6c62c9b4c41/media/image2.png)
延长BP交AC于D
∵∠BPC是△PCD的一个外角，∠PDC是△BAD的一个外角
∴∠BPC=∠PCD+∠PDC，∠PDC=∠DBA+∠A
∴∠BPC=∠PCD+∠DBA+∠A
∴∠BPC＞∠A

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