{"ok":true,"question":{"id":21095,"question_uuid":"985b7154-0234-45a1-85a2-70dce6a19297","job_id":4,"source_file":"A中考/2026江苏中考期末数学试卷+答案830.pdf","source_page_start":2,"source_page_end":2,"source_bbox":"[31.3, 333.58, 204.82, 618.08]","subject_code":"math","grade_level":"JUNIOR","tree_code":"KT_JUNIOR_MATH_STANDARD","question_type":"comprehensive","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"如图． = ， 与\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{APO}{M} \\\\frac{BPO}{OM} \\\\frac{PA}{OP} \\\\frac{e}{e} \\\\frac{O}{O}\"}, {\"text\": \"相切于点 、连接， 与 相交于点，过点作CD ⊥ 垂 C\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"C \\\\frac{OM}{PD}\"}, {\"text\": \"足为，交e O于点，连接 交 E D OM于点． F\", \"type\": \"text\"}, {\"kind\": \"embedded_image\", \"type\": \"figure\", \"width\": 693, \"height\": 546, \"sha256\": \"85355fc8f94f3cc7f45713beeb0ab66bcfaad0f81c00c295d7d134ec4e135745\", \"caption\": \"\", \"asset_key\": \"math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/985b7154-0234-45a1-85a2-70dce6a19297/85355fc8f94f.png\", \"public_url\": \"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/985b7154-0234-45a1-85a2-70dce6a19297/85355fc8f94f.png\", \"source_bbox\": [30.0, 436.969970703125, 168.0, 539.719970703125], \"source_page\": 2}, {\"type\": \"subquestion\", \"index\": 1, \"blocks\": [{\"text\": \"求证：PB是e O的切线．\", \"type\": \"text\"}]}, {\"type\": \"subquestion\", \"index\": 2, \"blocks\": [{\"text\": \"当PC = 6，PM =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{5}{4}\"}, {\"text\": \"CD时求线, 段MF的长．\", \"type\": \"text\"}]}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"小问详解过点作1:OON ⊥ PB于点，N (1) Q四边形ABCD是矩形,  =  = 90, = CD, A\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"C \\\\frac{AB}{RtQCD}\"}, {\"text\": \"在\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"V \\\\frac{RtABP}{CQ}\"}, {\"text\": \"和 V 中, AP =  =  = 90 A C  = CD\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{AB}{V}\"}, {\"text\": \"  QCD(SAS),\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{ABP}{BP}\\\\frac{V}{DQ}\"}, {\"text\": \" =. (2)设AP =, = 5 + a．\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{aAD}{PBQD}\"}, {\"text\": \"当四边形 是菱形时， PB = = 5，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{PD}{RtABP}\"}, {\"text\": \"在 V 中， 根据勾股定理得\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"AP^{2}\"}, {\"text\": \"+\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"AB^{2}\"}, {\"text\": \"=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"PB^{2}\"}, {\"text\": \"， 即\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"a^{2}\"}, {\"text\": \"+ =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"5^{2}\"}, {\"text\": \"，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{4^{2}}{a}\"}, {\"text\": \"可得： =，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{3}{3}\"}, {\"text\": \"所以 D= + 5 = 8． A 2026-06-22  与e 相切于点M，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{PA}{OM} \\\\frac{O}{PA}\"}, {\"text\": \" ⊥, APO = ,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{Q}{Q} \\\\frac{BPO}{APB}\"}, {\"text\": \"是 的平分线，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{PO}{ON}\"}, {\"text\": \" =,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OM}{e}\"}, {\"text\": \"Q OM为 的半径,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{O}{O}\"}, {\"text\": \" 为e 的半径,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{ON}{ON}\"}, {\"text\": \"Q ⊥ PB,PB是 e O的切线; 小问详解2 : Q CD ⊥ OM，OM为半径， CE = DE =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{2}\"}, {\"text\": \"CD, Q PM =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{5}{4}\"}, {\"text\": \"CD,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{CD}{PM} = \\\\frac{4}{5}\"}, {\"text\": \"\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{CE}{PM} = 2\"}, {\"text\": \", 5 Q  = 90, = 90,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OMP}{PM} \\\\frac{OEC}{OMP}\"}, {\"text\": \"CD∥,V ∽VOEC, \", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{CE}{PM} = \\\\frac{OC}{OP}\"}, {\"text\": \"Q PC = 6,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{2}{5} = \\\\frac{OC}{OC}, + 6\"}, {\"text\": \" = 4,OC = OM = 4,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OC}{R}\"}, {\"text\": \"在 tVM P中，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{O}{\\\\sqrt{OP^{2}−OM^{2}}}\"}, {\"text\": \"PM = = （\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\sqrt{6+4）^{2}−4^{2}} = 2 \\\\sqrt{21}\"}, {\"text\": \"， Q  = ,MFP = EFD,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{FMP}{MFP} \\\\frac{FED}{EFD}\"}, {\"text\": \"MP V ∽V,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{MF}{EF} =\"}, {\"text\": \"E D 设MF = x， 则EF = 4 − x −\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{8}{5} = \\\\frac{12}{5}\"}, {\"text\": \"− x，  x = 221 ,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{12}{5}\"}, {\"text\": \"− x\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"4 \\\\sqrt{21} 5\"}, {\"text\": \"解得x =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{12}{7}\"}, {\"text\": \"，M = 12. F 7 2026-06-22\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [31.3, 333.58, 204.82, 618.08], \"file\": \"A中考/2026江苏中考期末数学试卷+答案830.pdf\", \"regions\": [{\"bbox\": [31.3, 333.58, 204.82, 618.08], \"page\": 2}], \"section\": \"example\", \"page_end\": 2, \"page_start\": 2, \"section_label\": \"跟我学        中考必考考点，常考题型各个击破！\", \"answer_regions\": [{\"bbox\": [31.30277442932129, 665.0416259765625, 223.91000366210935, 679.5272827148438], \"page\": 2}, {\"bbox\": [320.13665771484375, 122.60568237304688, 557.0399780273438, 796.1519775390625], \"page\": 3}, {\"bbox\": [29.986812591552734, 121.23283386230467, 557.0399780273438, 796.1519775390625], \"page\": 4}]}, \"version\": 1, \"grade_level\": \"JUNIOR\", \"number_label\": \"1\", \"subject_code\": \"math\", \"question_type\": \"comprehensive\", \"content_sha256\": \"064a6bde4f8762cf9fdb00abd366b529133b849513e0b010d1dc819b0b3d7338\"}","body_html":"<div>如图． = ， 与\\(\\frac{APO}{M} \\frac{BPO}{OM} \\frac{PA}{OP} \\frac{e}{e} \\frac{O}{O}\\)相切于点 、连接， 与 相交于点，过点作CD ⊥ 垂 C\\(C \\frac{OM}{PD}\\)足为，交e O于点，连接 交 E D OM于点． F<img class=\"question-figure\" src=\"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/985b7154-0234-45a1-85a2-70dce6a19297/85355fc8f94f.png\" style=\"width:184.0px!important;max-width:100%;height:auto\" alt=\"\" title=\"点击查看原图\"><br>（1）求证：PB是e O的切线．<br>（2）当PC = 6，PM =\\(\\frac{5}{4}\\)CD时求线, 段MF的长．</div>","answer_html":"","ways_html":"小问详解过点作1:OON ⊥ PB于点，N (1) Q四边形ABCD是矩形,  =  = 90, = CD, A\\(C \\frac{AB}{RtQCD}\\)在\\(V \\frac{RtABP}{CQ}\\)和 V 中, AP =  =  = 90 A C  = CD\\(\\frac{AB}{V}\\)  QCD(SAS),\\(\\frac{ABP}{BP}\\frac{V}{DQ}\\) =. (2)设AP =, = 5 + a．\\(\\frac{aAD}{PBQD}\\)当四边形 是菱形时， PB = = 5，\\(\\frac{PD}{RtABP}\\)在 V 中， 根据勾股定理得\\(AP^{2}\\)+\\(AB^{2}\\)=\\(PB^{2}\\)， 即\\(a^{2}\\)+ =\\(5^{2}\\)，\\(\\frac{4^{2}}{a}\\)可得： =，\\(\\frac{3}{3}\\)所以 D= + 5 = 8． A 2026-06-22  与e 相切于点M，\\(\\frac{PA}{OM} \\frac{O}{PA}\\) ⊥, APO = ,\\(\\frac{Q}{Q} \\frac{BPO}{APB}\\)是 的平分线，\\(\\frac{PO}{ON}\\) =,\\(\\frac{OM}{e}\\)Q OM为 的半径,\\(\\frac{O}{O}\\) 为e 的半径,\\(\\frac{ON}{ON}\\)Q ⊥ PB,PB是 e O的切线; 小问详解2 : Q CD ⊥ OM，OM为半径， CE = DE =\\(\\frac{1}{2}\\)CD, Q PM =\\(\\frac{5}{4}\\)CD,\\(\\frac{CD}{PM} = \\frac{4}{5}\\)\\(\\frac{CE}{PM} = 2\\), 5 Q  = 90, = 90,\\(\\frac{OMP}{PM} \\frac{OEC}{OMP}\\)CD∥,V ∽VOEC, \\(\\frac{CE}{PM} = \\frac{OC}{OP}\\)Q PC = 6,\\(\\frac{2}{5} = \\frac{OC}{OC}, + 6\\) = 4,OC = OM = 4,\\(\\frac{OC}{R}\\)在 tVM P中，\\(\\frac{O}{\\sqrt{OP^{2}−OM^{2}}}\\)PM = = （\\(\\sqrt{6+4）^{2}−4^{2}} = 2 \\sqrt{21}\\)， Q  = ,MFP = EFD,\\(\\frac{FMP}{MFP} \\frac{FED}{EFD}\\)MP V ∽V,\\(\\frac{MF}{EF} =\\)E D 设MF = x， 则EF = 4 − x −\\(\\frac{8}{5} = \\frac{12}{5}\\)− x，  x = 221 ,\\(\\frac{12}{5}\\)− x\\(4 \\sqrt{21} 5\\)解得x =\\(\\frac{12}{7}\\)，M = 12. F 7 2026-06-22","explain_html":null,"difficulty":2,"score":null,"year":null,"parse_status":"PARSED","qa_status":"QUARANTINED","retry_count":0,"final_score":null,"error_message":"处理异常: LLM call failed after retries: HTTPConnectionPool(host='127.0.0.1', port=8092): Read timed out. (read timeout=120) [endpoints: http://127.0.0.1:8091/v1 -> http://127.0.0.1:8092/v1]","published_question_id":null,"content_sha256":"064a6bde4f8762cf9fdb00abd366b529133b849513e0b010d1dc819b0b3d7338","created_at":"2026-09-24T22:35:08","updated_at":"2026-09-24T22:50:27","answer_missing":true,"formula_fallback":false,"_number":"1","_section":"example","_section_label":"跟我学        中考必考考点，常考题型各个击破！"},"assets":[{"id":4601,"asset_uuid":"5c9eb7bf1810401bb6ed1c6aa15c66cc","job_id":4,"import_question_id":21095,"asset_type":"embedded_image","source_file":"2026江苏中考期末数学试卷+答案830.pdf","source_page":2,"source_bbox":"[30.0, 436.97, 168.0, 539.72]","sha256":"85355fc8f94f3cc7f45713beeb0ab66bcfaad0f81c00c295d7d134ec4e135745","local_path":"/www/wwwroot/tikupdfpng.mtwlkj.net_5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/985b7154-0234-45a1-85a2-70dce6a19297/85355fc8f94f.png","asset_key":"math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/985b7154-0234-45a1-85a2-70dce6a19297/85355fc8f94f.png","public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/985b7154-0234-45a1-85a2-70dce6a19297/85355fc8f94f.png","width":693,"height":546,"qa_status":"PENDING","created_at":"2026-09-24T22:35:08","safe_public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/985b7154-0234-45a1-85a2-70dce6a19297/85355fc8f94f.png"}],"knowledge":[],"audits":[{"id":415171,"import_question_id":21095,"audit_type":"STRUCTURE","auditor_name":"structure-rule","status":"PASS","score":0.9,"input_snapshot":"如图． = ， 与 \\frac{APO}{M} \\frac{BPO}{OM} \\frac{PA}{OP} \\frac{e}{e} \\frac{O}{O} 相切于点 、连接， 与 相交于点，过点作CD ⊥ 垂 C C \\frac{OM}{PD} 足为，交e O于点，连接 交 E D OM于点． F [图] (1) 求证：PB是e O的切线． (2) 当PC = 6，PM = \\frac{5}{4} CD时求线, 段MF的长．","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-24T22:46:27"},{"id":415172,"import_question_id":21095,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"PASS","score":1.0,"input_snapshot":"\\frac{APO}{M} \\frac{BPO}{OM} \\frac{PA}{OP} \\frac{e}{e} \\frac{O}{O}; C \\frac{OM}{PD}; \\frac{5}{4}; C \\frac{AB}{RtQCD}; V \\frac{RtABP}{CQ}","result_json":"{\"risky\": 38, \"issues\": [], \"checked\": 38, \"invalid\": 0, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [], \"visually_verified\": 0}","created_at":"2026-09-24T22:46:27"},{"id":415173,"import_question_id":21095,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"如图． = ， 与 \\frac{APO}{M} \\frac{BPO}{OM} \\frac{PA}{OP} \\frac{e}{e} \\frac{O}{O} 相切于点 、连接， 与 相交于点，过点作CD ⊥ 垂 C C \\frac{OM}{PD} 足为，交e O于点，连接 交 E D OM于点． F [图] (1) 求证：PB是e O的切线． (2) 当PC = 6，PM = \\frac{5}{4} CD时求线, 段MF的长．","result_json":"{\"picks\": [], \"tree_code\": \"KT_JUNIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-24T22:46:27"},{"id":415193,"import_question_id":21095,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":1.0,"input_snapshot":"题干：\n如图． = ， 与\\(\\frac{APO}{M} \\frac{BPO}{OM} \\frac{PA}{OP} \\frac{e}{e} \\frac{O}{O}\\)相切于点 、连接， 与 相交于点，过点作CD ⊥ 垂 C\\(C \\frac{OM}{PD}\\)足为，交e O于点，连接 交 E D OM于点． F （1）求证：PB是e O的切线． （2）当PC = 6，PM =\\(\\frac{5}{4}\\)CD时求线, 段MF的长．\n答案：\n\n解析：\n小问详解过点作1:OON ⊥ PB于点，N (1) Q四边形ABCD是矩形,  =  = 90, = CD, A\\(C \\frac{AB}{RtQCD}\\)在\\(V \\frac{RtABP}{CQ}\\)和 V 中, AP =  =  = 90 A C  = CD\\(\\frac{AB}{V}\\)  QCD(SAS),\\(\\frac{ABP}{BP}\\frac{V}{DQ}\\) =. (2)设AP =, = 5 + a．\\(\\frac{aAD}{PBQD}\\)当四边形 是菱形时， PB = = 5，\\(\\frac{PD}{RtABP}\\)在 V 中， 根据勾股定理得\\(AP^{2}\\)+\\(AB^{2}\\)=\\(PB^{2}\\)， 即\\(a^{2}\\)+ =\\(5^{2}\\)，\\(\\frac{4^{2}}{a}\\)可得： =，\\(\\frac{3}{3}\\)所以 D= + 5 = 8． A 2026-06-22  与e 相切于点M，\\(\\frac{PA}{OM} \\frac{O}{PA}\\) ⊥, APO = ,\\(\\frac{Q}{Q} \\frac{BPO}{APB}\\)是 的平分线，\\(\\frac{PO}{ON}\\) =,\\(\\frac{OM}{e}\\)Q OM为 的半径,\\(\\frac{O}{O}\\) 为e 的半径,\\(\\frac{ON}{ON}\\)Q ⊥ PB,PB是 e O的切线; 小问详解2 : Q CD ⊥ OM，OM为半径， CE = DE =\\(\\frac{1}{2}\\)CD, Q PM =\\(\\frac{5}{4}\\)CD,\\(\\frac{CD}{PM} = \\frac{4}{5}\\)\\(\\frac{CE}{PM} = 2\\), 5 Q  = 90, = 90,\\(\\frac{OMP}{PM} \\frac{OEC}{OMP}\\)CD∥,V ∽VOEC, \\(\\frac{CE}{PM} = \\frac{OC}{OP}\\)Q PC = 6,\\(\\frac{2}{5} = \\frac{OC}{OC}, + 6\\) = 4,OC = OM = 4,\\(\\frac{OC}{R}\\)在 tVM P中，\\(\\frac{O}{\\sqrt{O","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 1.0, \"suggestions\": []}","created_at":"2026-09-24T22:50:17"}]}