{"ok":true,"question":{"id":21008,"question_uuid":"1507a145-7703-4126-8834-4e078659e7b2","job_id":4,"source_file":"A中考/2026江苏中考期末数学试卷+答案820.pdf","source_page_start":5,"source_page_end":5,"source_bbox":"[31.56, 663.21, 557.04, 796.15]","subject_code":"math","grade_level":"JUNIOR","tree_code":"KT_JUNIOR_MATH_STANDARD","question_type":"fill_blank","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"【小试1】 如图所示,在\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\Delta{}ABC\"}, {\"text\": \"中, D 是边 BC 上一点,以 AD 为直径的 ⊙O交 AC于E, 连接DE. 若⊙O与BC相切, ∠ADE = 55∘, 则∠C的度数为____. 2026-06-22\", \"type\": \"text\"}], \"source\": {\"bbox\": [31.56, 663.21, 557.04, 796.15], \"file\": \"A中考/2026江苏中考期末数学试卷+答案820.pdf\", \"regions\": [{\"bbox\": [31.56, 663.21, 557.04, 796.15], \"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\": [], \"answer_original\": \"\", \"solution_original\": \"\"}, \"version\": 1, \"grade_level\": \"JUNIOR\", \"number_label\": \"1\", \"subject_code\": \"math\", \"question_type\": \"fill_blank\", \"content_sha256\": \"86928986b188212ea3e32764e20320a2bebaf3c5798ccd0d141899a6bb1749b7\"}","body_html":"<div>【小试1】 如图所示,在\\(\\Delta{}ABC\\)中, D 是边 BC 上一点,以 AD 为直径的 ⊙O交 AC于E, 连接DE. 若⊙O与BC相切, ∠ADE = 55∘, 则∠C的度数为____. 2026-06-22</div>","answer_html":"","ways_html":"","explain_html":null,"difficulty":2,"score":null,"year":null,"parse_status":"PARSED","qa_status":"QUARANTINED","retry_count":0,"final_score":0.925,"error_message":"QA failed: COHERENCE,ANSWER","published_question_id":null,"content_sha256":"86928986b188212ea3e32764e20320a2bebaf3c5798ccd0d141899a6bb1749b7","created_at":"2026-09-24T21:48:30","updated_at":"2026-09-24T21:58:03","answer_missing":true,"formula_fallback":false,"_number":"1","_section":"quiz","_section_label":"小试牛刀   ★ 独立完成，见证实力！"},"assets":[],"knowledge":[],"audits":[{"id":413105,"import_question_id":21008,"audit_type":"STRUCTURE","auditor_name":"structure-rule","status":"PASS","score":0.9,"input_snapshot":"【小试1】 如图所示,在 \\Delta{}ABC 中, D 是边 BC 上一点,以 AD 为直径的 ⊙O交 AC于E, 连接DE. 若⊙O与BC相切, ∠ADE = 55∘, 则∠C的度数为____. 2026-06-22","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-24T21:52:22"},{"id":413106,"import_question_id":21008,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"PASS","score":1.0,"input_snapshot":"\\Delta{}ABC","result_json":"{\"risky\": 0, \"issues\": [], \"checked\": 1, \"invalid\": 0, \"visual_check\": \"not_run_no_formula_bbox\", \"invalid_details\": [], \"visually_verified\": 0}","created_at":"2026-09-24T21:52:22"},{"id":413107,"import_question_id":21008,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"【小试1】 如图所示,在 \\Delta{}ABC 中, D 是边 BC 上一点,以 AD 为直径的 ⊙O交 AC于E, 连接DE. 若⊙O与BC相切, ∠ADE = 55∘, 则∠C的度数为____. 2026-06-22","result_json":"{\"picks\": [], \"tree_code\": \"KT_JUNIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-24T21:52:22"},{"id":413121,"import_question_id":21008,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":0.9,"input_snapshot":"题干：\n【小试1】 如图所示,在\\(\\Delta{}ABC\\)中, D 是边 BC 上一点,以 AD 为直径的 ⊙O交 AC于E, 连接DE. 若⊙O与BC相切, ∠ADE = 55∘, 则∠C的度数为____. 2026-06-22\n答案：\n\n解析：\n","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 0.9, \"suggestions\": [\"建议补充图形，以便更直观理解题意\"]}","created_at":"2026-09-24T21:52:38"},{"id":413211,"import_question_id":21008,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":0.9,"input_snapshot":"【小试1】 如图所示,在 \\Delta{}ABC 中, D 是边 BC 上一点,以 AD 为直径的 ⊙O交 AC于E, 连接DE. 若⊙O与BC相切, ∠ADE = 55∘, 则∠C的度数为____. 2026-06-22","result_json":"{\"basis\": \"连接OE。因为AD是直径，所以∠AED=90°，则∠CDE=90°。又⊙O与BC相切于D，故OD⊥BC，即AD⊥BC，所以∠ADC=90°，从而∠CDE=∠ADC=90°，说明E在DC上，即D、E、C共线且DE⊥AC。在Rt△DEC中，∠C=90°−∠CDE的余角？重新梳理：由AD为直径得∠AED=90°，故DE⊥AC。又⊙O与BC相切于D，则OD⊥BC，而O为AD中点，故AD⊥BC，即∠ADC=90°。因此在△ADC中，∠DAC+∠C=90°。又∠ADE=55°，而∠ADE与∠C互余（因为DE⊥AC，AD⊥BC），故∠C=90°−55°=35°。\", \"answer\": \"35°\", \"confidence\": 0.9, \"handout_chars\": 0}","created_at":"2026-09-24T21:55:11"},{"id":413212,"import_question_id":21008,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":1.0,"input_snapshot":"来源答案: || 盲解:35°","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 1.0}","created_at":"2026-09-24T21:55:11"},{"id":413213,"import_question_id":21008,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"WARN","score":0.5,"input_snapshot":"答案: || 盲解:35°","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"未提取到来源答案\"}, \"status\": \"WARN\", \"applicable\": true}","created_at":"2026-09-24T21:55:11"},{"id":413229,"import_question_id":21008,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.9,"input_snapshot":"答案: || 解析:","result_json":"{\"ok\": false, \"which\": [\"solution\"], \"issues\": [\"解析为空，缺少必要的几何推导步骤；答案未给出，仅标注‘参考：盲解答案为 35°’，不符合完整解答要求。\"], \"confidence\": 0.9, \"equation_checks\": []}","created_at":"2026-09-24T21:55:35"},{"id":413230,"import_question_id":21008,"audit_type":"QUALITY_GATE","auditor_name":"quality-gate","status":"FAIL","score":0.925,"input_snapshot":"","result_json":"{\"warnings\": [], \"graded_gate\": true, \"hard_failed\": [\"COHERENCE\", \"ANSWER\"]}","created_at":"2026-09-24T21:55:35"},{"id":413284,"import_question_id":21008,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"PASS","score":0.9,"input_snapshot":"【小试1】 如图所示,在 \\Delta{}ABC 中, D 是边 BC 上一点,以 AD 为直径的 ⊙O交 AC于E, 连接DE. 若⊙O与BC相切, ∠ADE = 55∘, 则∠C的度数为____. 2026-06-22","result_json":"{\"basis\": \"连接OE。因为AD是⊙O的直径，所以∠AED=90°，即DE⊥AC。又因为⊙O与BC相切于D，所以AD⊥BC，即∠ADC=90°。在Rt△CDE中，∠C+∠CDE=90°。因为OA=OE，所以∠OAE=∠OEA。又因为∠ADE=55°，所以∠OEA=90°-55°=35°，从而∠C=35°。\", \"answer\": \"35°\", \"confidence\": 0.9, \"handout_chars\": 0}","created_at":"2026-09-24T21:57:04"},{"id":413285,"import_question_id":21008,"audit_type":"CRITIC","auditor_name":"critic","status":"PASS","score":0.9,"input_snapshot":"来源答案:35° || 盲解:35°","result_json":"{\"ok\": true, \"issues\": [], \"merged\": true, \"confidence\": 0.9}","created_at":"2026-09-24T21:57:04"},{"id":413316,"import_question_id":21008,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"FAIL","score":0.85,"input_snapshot":"答案:35° || 解析:连接OD. 因为⊙O与BC相切于D, 所以OD⊥BC, 即∠ODB=90°. 因为OA=OD（半径）, 所以∠OAD=∠ODA. 在△AOD中, ∠AOD=180°-2∠ODA. 又因为∠ADE=55°, 且∠ADE与∠ODA互余（因OD⊥BC, DE为弦）, 故∠ODA=90°-55°=35°. 所以∠C=∠ODA=35°（同弧所对圆周角相等）.","result_json":"{\"ok\": false, \"which\": [\"solution\"], \"issues\": [\"解析中‘∠C=∠ODA（同弧所对圆周角相等）’的推理不成立，缺乏几何依据。\", \"∠C = ∠ODA: ∠C 与 ∠ODA 不是同弧所对的圆周角。∠ODA 是弦切角（或三角形内角），而 ∠C 是圆周角，它们所对的弧不同，不能直接相等。正确推导应为：∠C = ∠ADE - ∠EDC 或通过其他几何关系求解，但原解析未给出严谨步骤。\"], \"confidence\": 0.85, \"equation_checks\": [{\"valid\": false, \"reason\": \"∠C 与 ∠ODA 不是同弧所对的圆周角。∠ODA 是弦切角（或三角形内角），而 ∠C 是圆周角，它们所对的弧不同，不能直接相等。正确推导应为：∠C = ∠ADE - ∠EDC 或通过其他几何关系求解，但原解析未给出严谨步骤。\", \"equation\": \"∠C = ∠ODA\"}]}","created_at":"2026-09-24T21:58:03"},{"id":413317,"import_question_id":21008,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"PASS","score":1.0,"input_snapshot":"答案:35° || 盲解:35°","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"盲解与来源答案一致\"}, \"status\": \"PASS\", \"applicable\": true}","created_at":"2026-09-24T21:58:03"},{"id":413318,"import_question_id":21008,"audit_type":"AUTO_REPAIR_ATTEMPT","auditor_name":"answer-repair","status":"FAIL","score":0.9,"input_snapshot":null,"result_json":"{\"critic\": true, \"symbolic\": true, \"coherence\": false, \"consensus\": \"two_solvers+symbolic+critic+coherence\", \"confidence\": 0.9, 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