{"ok":true,"question":{"id":21097,"question_uuid":"f1c9b27e-f189-418c-919a-01059409ac40","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":5,"source_page_end":5,"source_bbox":"[31.56, 172.69, 373.99, 247.84]","subject_code":"math","grade_level":"JUNIOR","tree_code":"KT_JUNIOR_MATH_STANDARD","question_type":"comprehensive","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"【小试1】 如图,△ABC内接于⊙O, AB是直径,⊙O的切线PC交BA的延长 线于点P, OF ∥BC交于点E, 交PC于点F, 连接AF;\", \"type\": \"text\"}, {\"kind\": \"embedded_image\", \"type\": \"figure\", \"width\": 708, \"height\": 310, \"sha256\": \"5da0c12aeb2f755e6e42f4120b53e33a15e8ecfdc9aec1b1558dab762a5a91c3\", \"caption\": \"\", \"asset_key\": \"math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f1c9b27e-f189-418c-919a-01059409ac40/5da0c12aeb2f.png\", \"public_url\": \"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f1c9b27e-f189-418c-919a-01059409ac40/5da0c12aeb2f.png\", \"source_bbox\": [31.5, 248.614013671875, 173.1999969482422, 310.010009765625], \"source_page\": 5}, {\"type\": \"subquestion\", \"index\": 1, \"blocks\": [{\"text\": \"判断AF与⊙O的位置关系并说明理由.\", \"type\": \"text\"}]}, {\"type\": \"subquestion\", \"index\": 2, \"blocks\": [{\"text\": \"若⊙O的半径为4, AF = 3, 求AC的长．\", \"type\": \"text\"}]}, {\"type\": \"answer\", \"blocks\": [{\"text\": \"(1)AF 是 ⊙O 的切线; (2) _{5}.^{24}\", \"type\": \"text\"}]}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"(1)AF是⊙O的切线, 连接OC, 因为OF ∥BC, 所以∠1 = ∠2, ∠B = ∠3,∵OC = OB, ∴∠B = ∠1, ∠3 = ∠2, 在 OA = OC △OAF 和 △OCF 中, {∠3 = ∠2，∴△OAF ≅△OCF, ∴∠OAF = ∠OCF, ∵PC 是 ⊙O 的切线, ∴ OF = OF ∠OCF=90°, ∠OAF = 90°, ∴FA ⊥OA, ∴AF是⊙O的切线; (2)∵⊙O的半径为4, AF = 3, ∠OAF = 90°, ∴OF =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\sqrt{AF^{2}}\"}, {\"text\": \"+\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"OA^{2}\"}, {\"text\": \"=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\sqrt{3^{2}}\"}, {\"text\": \"+\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"4^{2}\"}, {\"text\": \"= 5,∵AF, CF是⊙O的切线 1 1 , ∴AF = CF, 因为OA = OC, 所以OF ⊥AC,∴AC = 2AE, ∵OF ⊥AC, 所以S△OAF= _{2}AF\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\times\"}, {\"text\": \"OA = _{2}OF ⋅AE, 所 以\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3 \\\\times 4 = 5 \\\\times\"}, {\"text\": \"AE, AE = _{5}, AC = 2AE = _{5}.^{25} 12\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [31.56, 172.69, 373.99, 247.84], \"file\": \"A中考/2026江苏中考期末数学试卷+答案830.pdf\", \"regions\": [{\"bbox\": [31.56, 172.69, 373.99, 247.84], \"page\": 5}], \"section\": \"quiz\", \"page_end\": 5, \"page_start\": 5, \"section_label\": \"小试牛刀   ★ 独立完成，见证实力！\", \"answer_regions\": [{\"bbox\": [30.0, 173.62, 560.16, 367.75], \"page\": 10}]}, \"version\": 1, \"grade_level\": \"JUNIOR\", \"number_label\": \"1\", \"subject_code\": \"math\", \"question_type\": \"comprehensive\", \"content_sha256\": \"70e23e15e1810f80cd794b360fede7023ab747efe322639738acecbdbdc54b59\"}","body_html":"<div>【小试1】 如图,△ABC内接于⊙O, AB是直径,⊙O的切线PC交BA的延长 线于点P, OF ∥BC交于点E, 交PC于点F, 连接AF;<img class=\"question-figure\" src=\"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f1c9b27e-f189-418c-919a-01059409ac40/5da0c12aeb2f.png\" style=\"width:188.9px!important;max-width:100%;height:auto\" alt=\"\" title=\"点击查看原图\"><br>（1）判断AF与⊙O的位置关系并说明理由.<br>（2）若⊙O的半径为4, AF = 3, 求AC的长．</div>","answer_html":"(1)AF 是 ⊙O 的切线; (2) _{5}.^{24}","ways_html":"(1)AF是⊙O的切线, 连接OC, 因为OF ∥BC, 所以∠1 = ∠2, ∠B = ∠3,∵OC = OB, ∴∠B = ∠1, ∠3 = ∠2, 在 OA = OC △OAF 和 △OCF 中, {∠3 = ∠2，∴△OAF ≅△OCF, ∴∠OAF = ∠OCF, ∵PC 是 ⊙O 的切线, ∴ OF = OF ∠OCF=90°, ∠OAF = 90°, ∴FA ⊥OA, ∴AF是⊙O的切线; (2)∵⊙O的半径为4, AF = 3, ∠OAF = 90°, ∴OF =\\(\\sqrt{AF^{2}}\\)+\\(OA^{2}\\)=\\(\\sqrt{3^{2}}\\)+\\(4^{2}\\)= 5,∵AF, CF是⊙O的切线 1 1 , ∴AF = CF, 因为OA = OC, 所以OF ⊥AC,∴AC = 2AE, ∵OF ⊥AC, 所以S△OAF= _{2}AF\\(\\times\\)OA = _{2}OF ⋅AE, 所 以\\(3 \\times 4 = 5 \\times\\)AE, AE = _{5}, AC = 2AE = _{5}.^{25} 12","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":"70e23e15e1810f80cd794b360fede7023ab747efe322639738acecbdbdc54b59","created_at":"2026-09-24T22:35:08","updated_at":"2026-09-24T22:50:07","answer_missing":false,"formula_fallback":false,"_number":"1","_section":"quiz","_section_label":"小试牛刀   ★ 独立完成，见证实力！"},"assets":[{"id":4604,"asset_uuid":"6df6b90c04d349ce894050d4c6be865c","job_id":4,"import_question_id":21097,"asset_type":"embedded_image","source_file":"2026江苏中考期末数学试卷+答案830.pdf","source_page":5,"source_bbox":"[31.5, 248.61, 173.2, 310.01]","sha256":"5da0c12aeb2f755e6e42f4120b53e33a15e8ecfdc9aec1b1558dab762a5a91c3","local_path":"/www/wwwroot/tikupdfpng.mtwlkj.net_5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f1c9b27e-f189-418c-919a-01059409ac40/5da0c12aeb2f.png","asset_key":"math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f1c9b27e-f189-418c-919a-01059409ac40/5da0c12aeb2f.png","public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f1c9b27e-f189-418c-919a-01059409ac40/5da0c12aeb2f.png","width":708,"height":310,"qa_status":"PENDING","created_at":"2026-09-24T22:36:48","safe_public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f1c9b27e-f189-418c-919a-01059409ac40/5da0c12aeb2f.png"}],"knowledge":[],"audits":[{"id":415168,"import_question_id":21097,"audit_type":"STRUCTURE","auditor_name":"structure-rule","status":"PASS","score":0.9,"input_snapshot":"【小试1】 如图,△ABC内接于⊙O, AB是直径,⊙O的切线PC交BA的延长 线于点P, OF ∥BC交于点E, 交PC于点F, 连接AF; [图] (1) 判断AF与⊙O的位置关系并说明理由. (2) 若⊙O的半径为4, AF = 3, 求AC的长．","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-24T22:46:07"},{"id":415169,"import_question_id":21097,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"PASS","score":1.0,"input_snapshot":"\\sqrt{AF^{2}}; OA^{2}; \\sqrt{3^{2}}; 4^{2}; \\times","result_json":"{\"risky\": 6, \"issues\": [], \"checked\": 6, \"invalid\": 0, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [], \"visually_verified\": 0}","created_at":"2026-09-24T22:46:07"},{"id":415170,"import_question_id":21097,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"【小试1】 如图,△ABC内接于⊙O, AB是直径,⊙O的切线PC交BA的延长 线于点P, OF ∥BC交于点E, 交PC于点F, 连接AF; [图] (1) 判断AF与⊙O的位置关系并说明理由. (2) 若⊙O的半径为4, AF = 3, 求AC的长．","result_json":"{\"picks\": [], \"tree_code\": \"KT_JUNIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-24T22:46:07"},{"id":415180,"import_question_id":21097,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":0.95,"input_snapshot":"题干：\n【小试1】 如图,△ABC内接于⊙O, AB是直径,⊙O的切线PC交BA的延长 线于点P, OF ∥BC交于点E, 交PC于点F, 连接AF; （1）判断AF与⊙O的位置关系并说明理由. （2）若⊙O的半径为4, AF = 3, 求AC的长．\n答案：\n(1)AF 是 ⊙O 的切线; (2) _{5}.^{24}\n解析：\n(1)AF是⊙O的切线, 连接OC, 因为OF ∥BC, 所以∠1 = ∠2, ∠B = ∠3,∵OC = OB, ∴∠B = ∠1, ∠3 = ∠2, 在 OA = OC △OAF 和 △OCF 中, {∠3 = ∠2，∴△OAF ≅△OCF, ∴∠OAF = ∠OCF, ∵PC 是 ⊙O 的切线, ∴ OF = OF ∠OCF=90°, ∠OAF = 90°, ∴FA ⊥OA, ∴AF是⊙O的切线; (2)∵⊙O的半径为4, AF = 3, ∠OAF = 90°, ∴OF =\\(\\sqrt{AF^{2}}\\)+\\(OA^{2}\\)=\\(\\sqrt{3^{2}}\\)+\\(4^{2}\\)= 5,∵AF, CF是⊙O的切线 1 1 , ∴AF = CF, 因为OA = OC, 所以OF ⊥AC,∴AC = 2AE, ∵OF ⊥AC, 所以S△OAF= _{2}AF\\(\\times\\)OA = _{2}OF ⋅AE, 所 以\\(3 \\times 4 = 5 \\times\\)AE, AE = _{5}, AC = 2AE = _{5}.^{25} 12","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 0.95, \"suggestions\": [\"解析中‘因为OA = OC, 所以OF ⊥ AC’的推导可补充说明：由AF、CF均为⊙O切线得OA=OC且OF平分∠AOC，故OF垂直平分AC\"]}","created_at":"2026-09-24T22:48:06"}]}