{"ok":true,"question":{"id":22127,"question_uuid":"0d50d4d0-d2ff-47f8-9ff0-50c066a3df86","job_id":3,"source_file":"人教版/七下/2026江苏七年级期末数学试卷+答案65.pdf","source_page_start":2,"source_page_end":2,"source_bbox":"[31.3, 454.88, 246.82, 564.76]","subject_code":"math","grade_level":"SENIOR","tree_code":"KT_SENIOR_MATH_STANDARD","question_type":"subjective","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"如图，是直线OAB上一点，COD = 90o,? 平分 ．\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OE}{(1)} \\\\frac{BOC}{BOD}\"}, {\"text\": \"若 =，求 的度数；\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{60^{o}}{20^{o}} \\\\frac{COE}{DOE}\"}, {\"text\": \"若 =，求 的度数；\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{(2)}{(3)} \\\\frac{AOC}{DOE}\"}, {\"text\": \"若 =，求AOC的度数． (用含的式子表示)\", \"type\": \"text\"}, {\"kind\": \"embedded_image\", \"type\": \"figure\", \"width\": 739, \"height\": 505, \"sha256\": \"d270e94ba2be08d450599c972b48265e62c4b877b03c0e8f5766c8317971f3af\", \"caption\": \"\", \"asset_key\": \"math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0d50d4d0-d2ff-47f8-9ff0-50c066a3df86/d270e94ba2be.png\", \"public_url\": \"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0d50d4d0-d2ff-47f8-9ff0-50c066a3df86/d270e94ba2be.png\", \"source_bbox\": [30.0, 579.9049682617188, 179.10000610351562, 672.6799926757812], \"source_page\": 2}, {\"html\": \"\\n \\n \\n \\n \\n \\n \\n \\n \\n
【典例 1】\\n如图，O是直线AB上一点，COD =90 o,?\\nOE平分BOC．\\n(1)若BOD =60o，求COE的度数；\\n(2)若AOC = 20o，求DOE的度数；\\n(3)若DOE =，求AOC的度数．\\n(用含的式子表示)【典例 2】\\n下列说法中，正确的有\\n①若a与c相交，b与c相交，则a与b 相交；\\n②若aPb ,bPc，则aPc；\\n③过直线外一点有且只有一条直线与已\\n知直线平行；\\n④在同一平面内，不重合的两条直线的\\n位置关系有平行、相交、垂直三种．\\nA．①②③ B．②③④\\nC．①②④ D．②③
\", \"type\": \"table\", \"source_bbox\": [24.55, 422.15, 570.89, 685.42], \"source_page\": 2}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"(1) COD = 、 = 60,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{90}{BOD} \\\\frac{BOD}{COD}\"}, {\"text\": \" =  +  = 60+ 90 =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"150 \\\\frac{BOC}{平分}\"}, {\"text\": \" OE BOC COE =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{2}\"}, {\"text\": \" =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{2}\"}, {\"text\": \" = 75;\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{BOC}{90} \\\\frac{150}{AOC}\"}, {\"text\": \"(2)  = 、 = 20,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{COD}{180}\"}, {\"text\": \"BOD =  −COD −AOC = −90 − = 70,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{180}{BOC} \\\\frac{20}{BOD}\"}, {\"text\": \" =  + COD = + = 160,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{70}{OE}\\\\frac{90}{平分}\"}, {\"text\": \" ，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{BOC}{BOC}\"}, {\"text\": \" =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{2}\"}, {\"text\": \" =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{2}\"}, {\"text\": \" =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"80 \\\\frac{COE}{DOE} \\\\frac{160}{COE}\"}, {\"text\": \" =  − = 90− 80 =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"10 \\\\frac{COD}{90}\"}, {\"text\": \"(3) COD = 、 =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{DOE}{DOE}\"}, {\"text\": \" =  − = 90 −\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{COE}{平分} \\\\frac{COD}{BOC}\"}, {\"text\": \" OE ， BOC = 2COE = 2 (90 −) =  − 2,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{180}{AOC}\"}, {\"text\": \" =  −BOC\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{180}{(180} = 180− − 2) =\"}, {\"text\": \" 2.\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [31.3, 454.88, 246.82, 564.76], \"file\": \"人教版/七下/2026江苏七年级期末数学试卷+答案65.pdf\", \"regions\": [{\"bbox\": [31.3, 454.88, 246.82, 564.76], \"page\": 2}], \"section\": \"example\", \"page_end\": 2, \"page_start\": 2, \"section_label\": \"跟我学 关键母题，常考题型各个击破！\", \"answer_regions\": [{\"bbox\": [29.815757751464844, 142.31668090820312, 254.6657562255859, 523.6532592773438], \"page\": 3}], \"answer_original\": \"\", \"solution_original\": \"(1) COD = 、 = 60, \\\\frac{90}{BOD} \\\\frac{BOD}{COD}  =  +  = 60+ 90 = 150 \\\\frac{BOC}{平分}  OE BOC COE = \\\\frac{1}{2}  = \\\\frac{1}{2}  = 75; \\\\frac{BOC}{90} \\\\frac{150}{AOC} (2)  = 、 = 20, \\\\frac{COD}{180} BOD =  −COD −AOC = −90 − = 70, \\\\frac{180}{BOC} \\\\frac{20}{BOD}  =  + COD = + = 160, \\\\frac{70}{OE}\\\\frac{90}{平分}  ， \\\\frac{BOC}{BOC}  = \\\\frac{1}{2}  = \\\\frac{1}{2}  = 80 \\\\frac{COE}{DOE} \\\\frac{160}{COE}  =  − = 90− 80 = 10 \\\\frac{COD}{90} (3) COD = 、 = \\\\frac{DOE}{DOE}  =  − = 90 − \\\\frac{COE}{平分} \\\\frac{COD}{BOC}  OE ， BOC = 2COE = 2 (90 −) =  − 2, \\\\frac{180}{AOC}  =  −BOC \\\\frac{180}{(180} = 180− − 2) =  2.\"}, \"version\": 1, \"grade_level\": \"SENIOR\", \"number_label\": \"1\", \"subject_code\": \"math\", \"question_type\": \"subjective\", \"content_sha256\": \"b7657c697606473205fe1bb8833847c5ae5c7bbdb1875835512b4ab8960c15e2\"}","body_html":"
如图，是直线OAB上一点，COD = 90o,? 平分 ．\\(\\frac{OE}{(1)} \\frac{BOC}{BOD}\\)若 =，求 的度数；\\(\\frac{60^{o}}{20^{o}} \\frac{COE}{DOE}\\)若 =，求 的度数；\\(\\frac{(2)}{(3)} \\frac{AOC}{DOE}\\)若 =，求AOC的度数． (用含的式子表示)\"\"\n \n \n \n \n \n \n \n \n
【典例 1】\n如图，O是直线AB上一点，COD =90 o,?\nOE平分BOC．\n(1)若BOD =60o，求COE的度数；\n(2)若AOC = 20o，求DOE的度数；\n(3)若DOE =，求AOC的度数．\n(用含的式子表示)【典例 2】\n下列说法中，正确的有\n①若a与c相交，b与c相交，则a与b 相交；\n②若aPb ,bPc，则aPc；\n③过直线外一点有且只有一条直线与已\n知直线平行；\n④在同一平面内，不重合的两条直线的\n位置关系有平行、相交、垂直三种．\nA．①②③ B．②③④\nC．①②④ D．②③
","answer_html":"","ways_html":"(1) COD = 、 = 60,\\(\\frac{90}{BOD} \\frac{BOD}{COD}\\) =  +  = 60+ 90 =\\(150 \\frac{BOC}{平分}\\) OE BOC COE =\\(\\frac{1}{2}\\) =\\(\\frac{1}{2}\\) = 75;\\(\\frac{BOC}{90} \\frac{150}{AOC}\\)(2)  = 、 = 20,\\(\\frac{COD}{180}\\)BOD =  −COD −AOC = −90 − = 70,\\(\\frac{180}{BOC} \\frac{20}{BOD}\\) =  + COD = + = 160,\\(\\frac{70}{OE}\\frac{90}{平分}\\) ，\\(\\frac{BOC}{BOC}\\) =\\(\\frac{1}{2}\\) =\\(\\frac{1}{2}\\) =\\(80 \\frac{COE}{DOE} \\frac{160}{COE}\\) =  − = 90− 80 =\\(10 \\frac{COD}{90}\\)(3) COD = 、 =\\(\\frac{DOE}{DOE}\\) =  − = 90 −\\(\\frac{COE}{平分} \\frac{COD}{BOC}\\) OE ， BOC = 2COE = 2 (90 −) =  − 2,\\(\\frac{180}{AOC}\\) =  −BOCfrac180(180 = 180− − 2) = 2.","explain_html":null,"difficulty":2,"score":null,"year":null,"parse_status":"PARSED","qa_status":"QUARANTINED","retry_count":0,"final_score":0.917,"error_message":"QA failed: FORMULA,SOLVER,CRITIC,ANSWER","published_question_id":null,"content_sha256":"b7657c697606473205fe1bb8833847c5ae5c7bbdb1875835512b4ab8960c15e2","created_at":"2026-09-25T05:39:22","updated_at":"2026-09-25T05:53:42","answer_missing":true,"formula_fallback":true,"_number":"1","_section":"example","_section_label":"跟我学 关键母题，常考题型各个击破！"},"assets":[{"id":4817,"asset_uuid":"3d0c00ef56fc4c0bbb0dfbcf073ca608","job_id":3,"import_question_id":22127,"asset_type":"embedded_image","source_file":"2026江苏七年级期末数学试卷+答案65.pdf","source_page":2,"source_bbox":"[30.0, 579.9, 179.1, 672.68]","sha256":"d270e94ba2be08d450599c972b48265e62c4b877b03c0e8f5766c8317971f3af","local_path":"/www/wwwroot/tikupdfpng.mtwlkj.net_5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0d50d4d0-d2ff-47f8-9ff0-50c066a3df86/d270e94ba2be.png","asset_key":"math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0d50d4d0-d2ff-47f8-9ff0-50c066a3df86/d270e94ba2be.png","public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0d50d4d0-d2ff-47f8-9ff0-50c066a3df86/d270e94ba2be.png","width":739,"height":505,"qa_status":"PENDING","created_at":"2026-09-25T05:43:26","safe_public_url":"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/0d50d4d0-d2ff-47f8-9ff0-50c066a3df86/d270e94ba2be.png"}],"knowledge":[],"audits":[{"id":438141,"import_question_id":22127,"audit_type":"STRUCTURE","auditor_name":"structure-rule","status":"PASS","score":0.9,"input_snapshot":"如图，是直线OAB上一点，COD = 90o,? 平分 ． \\frac{OE}{(1)} \\frac{BOC}{BOD} 若 =，求 的度数； \\frac{60^{o}}{20^{o}} \\frac{COE}{DOE} 若 =，求 的度数； \\frac{(2)}{(3)} \\frac{AOC}{DOE} 若 =，求AOC的度数． (用含的式子表示) [图] 【典例 1】 如图，O是直线AB上一点，COD =90 o,? OE平分BOC． (1)若BOD =60o，求COE的度数； (2)若AOC = 20o，求DOE的度数； (3)若DOE =，求AOC的度数． (用含的式子表示) 【典例 2】 下列说法中，正确的有 ①若a与c相交，b与c相交，则a与b 相交； ②若aPb ,bPc，则aPc； ③过直线外一点有且只有一条直线与已 知直线平行； ④在同一平面内，不重合的两条直线的 位置关系有平行、相交、垂直三种． A．①②③ B．②③④ C．①②④ D．②③","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-25T05:48:04"},{"id":438142,"import_question_id":22127,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"FAIL","score":1.0,"input_snapshot":"\\frac{OE}{(1)} \\frac{BOC}{BOD}; \\frac{60^{o}}{20^{o}} \\frac{COE}{DOE}; \\frac{(2)}{(3)} \\frac{AOC}{DOE}; \\frac{90}{BOD} \\frac{BOD}{COD}; 150 \\frac{BOC}{平分}","result_json":"{\"risky\": 20, \"issues\": [\"非法 LaTeX #20: unbalanced delimiters\"], \"checked\": 20, \"invalid\": 1, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [{\"index\": 19, \"latex\": \"\\\\frac{180}{(180} = 180− − 2) =\", \"issues\": [\"unbalanced delimiters\"]}], \"visually_verified\": 0}","created_at":"2026-09-25T05:48:04"},{"id":438143,"import_question_id":22127,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"如图，是直线OAB上一点，COD = 90o,? 平分 ． \\frac{OE}{(1)} \\frac{BOC}{BOD} 若 =，求 的度数； \\frac{60^{o}}{20^{o}} \\frac{COE}{DOE} 若 =，求 的度数； \\frac{(2)}{(3)} \\frac{AOC}{DOE} 若 =，求AOC的度数． (用含的式子表示) [图] 【典例 1】 如图，O是直线AB上一点，COD =90 o,? OE平分BOC． (1)若BOD =60o，求COE的度数； (2)若AOC = 20o，求DOE的度数； (3)若DOE =，求AOC的度数． (用含的式子表示) 【典例 2】 下列说法中，正确的有 ①若a与c相交，b与c相交，则a与b 相交； ②若aPb ,bPc，则aPc； ③过直线外一点有且只有一条直线与已 知直线平行； ④在同一平面内，不重合的两条直线的 位置关系有平行、相交、垂直三种． A．①②③ B．②③④ C．①②④ D．②③","result_json":"{\"picks\": [], \"tree_code\": \"KT_SENIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-25T05:48:04"},{"id":438165,"import_question_id":22127,"audit_type":"VISION","auditor_name":"vision-auditor","status":"PASS","score":0.95,"input_snapshot":"题干：\n如图，是直线OAB上一点，COD = 90o,? 平分 ．\\(\\frac{OE}{(1)} \\frac{BOC}{BOD}\\)若 =，求 的度数；\\(\\frac{60^{o}}{20^{o}} \\frac{COE}{DOE}\\)若 =，求 的度数；\\(\\frac{(2)}{(3)} \\frac{AOC}{DOE}\\)若 =，求AOC的度数． (用含的式子表示) 【典例 1】 如图，O是直线AB上一点，COD =90 o,? OE平分BOC． (1)若BOD =60o，求COE的度数； (2)若AOC = 20o，求DOE的度数； (3)若DOE =，求AOC的度数． (用含的式子表示) 【典例 2】 下列说法中，正确的有 ①若a与c相交，b与c相交，则a与b 相交； ②若aPb ,bPc，则aPc； ③过直线外一点有且只有一条直线与已 知直线平行； ④在同一平面内，不重合的两条直线的 位置关系有平行、相交、垂直三种． A．①②③ B．②③④ C．①②④ D．②③\n答案：\n\n解析：\n(1) COD = 、 = 60,\\(\\frac{90}{BOD} \\frac{BOD}{COD}\\) =  +  = 60+ 90 =\\(150 \\frac{BOC}{平分}\\) OE BOC COE =\\(\\frac{1}{2}\\) =\\(\\frac{1}{2}\\) = 75;\\(\\frac{BOC}{90} \\frac{150}{AOC}\\)(2)  = 、 = 20,\\(\\frac{COD}{180}\\)BOD =  −COD −AOC = −90 − = 70,\\(\\frac{180}{BOC} \\frac{20}{BOD}\\) =  + COD = + = 160,\\(\\frac{70}{OE}\\frac{90}{平分}\\) ，\\(\\frac{BOC}{BOC}\\) =\\(\\frac{1}{2}\\) =\\(\\frac{1}{2}\\) =\\(80 \\frac{COE}{DOE} \\frac{160}{COE}\\) =  − = 90− 80 =\\(10 \\frac{COD}{90}\\)(3) COD = 、 =\\(\\frac{DOE}{DOE}\\) =  − = 90 −\\(\\frac{COE}{平分} \\frac{COD}{BOC}\\) OE ， BOC = 2COE = 2 (90 −) =  − 2,\\(\\frac{180}{AOC}\\) =  −BOC frac180(180 = 180− − 2) =  2.","result_json":"{\"ok\": true, \"hard\": false, \"issues\": [], \"confidence\": 0.95, \"suggestions\": [\"题干中‘? 平分’疑似排版残留符号，建议核实原卷是否应为‘OE平分∠BOC’，当前已根据上下文和答案逻辑修正，但若原卷确为问号则需标注差异。\", \"题干第(3)问‘求∠AOC的度数’后括号注明‘用含α的式子表示’，但答案给出的是‘2α’，而解析中最终写的是‘= α 2’（即2α），表述顺序略有不同，建议统一书写规范。\", \"来源答案解析中步骤(3)最后一步‘∠AOC = 180° − (180° − 2α) = 2α’，但题干要求‘用含α的式子表示’，答案正确，但可补充说明‘即∠AOC = 2α’以增强清晰度。\"]}","created_at":"2026-09-25T05:49:10"},{"id":438178,"import_question_id":22127,"audit_type":"SOLVER","auditor_name":"blind-solver","status":"WARN","score":0.4,"input_snapshot":"如图，是直线OAB上一点，COD = 90o,? 平分 ． \\frac{OE}{(1)} \\frac{BOC}{BOD} 若 =，求 的度数； \\frac{60^{o}}{20^{o}} \\frac{COE}{DOE} 若 =，求 的度数； \\frac{(2)}{(3)} \\frac{AOC}{DOE} 若 =，求AOC的度数． (用含的式子表示) [图] 【典例 1】 如图，O是直线AB上一点，COD =90 o,? OE平分BOC． (1)若BOD =60o，求COE的度数； (2)若AOC = 20o，求DOE的度数； (3)若DOE =，求AOC的度数． (用含的式子表示) 【典例 2】 下列说法中，正确的有 ①若a与c相交，b与c相交，则a与b 相交； ②若aPb ,bPc，则aPc； ③过直线外一点有且只有一条直线与已 知直线平行； ④在同一平面内，不重合的两条直线的 位置关系有平行、相交、垂直三种． A．①②③ B．②③④ C．①②④ D．②③","result_json":"{\"note\": \"盲解+评判合并调用未返回\"}","created_at":"2026-09-25T05:52:41"},{"id":438179,"import_question_id":22127,"audit_type":"CRITIC","auditor_name":"critic","status":"WARN","score":0.5,"input_snapshot":"如图，是直线OAB上一点，COD = 90o,? 平分 ． \\frac{OE}{(1)} \\frac{BOC}{BOD} 若 =，求 的度数； \\frac{60^{o}}{20^{o}} \\frac{COE}{DOE} 若 =，求 的度数； \\frac{(2)}{(3)} \\frac{AOC}{DOE} 若 =，求AOC的度数． (用含的式子表示) [图] 【典例 1】 如图，O是直线AB上一点，COD =90 o,? OE平分BOC． (1)若BOD =60o，求COE的度数； (2)若AOC = 20o，求DOE的度数； (3)若DOE =，求AOC的度数． (用含的式子表示) 【典例 2】 下列说法中，正确的有 ①若a与c相交，b与c相交，则a与b 相交； ②若aPb ,bPc，则aPc； ③过直线外一点有且只有一条直线与已 知直线平行； ④在同一平面内，不重合的两条直线的 位置关系有平行、相交、垂直三种． A．①②③ B．②③④ C．①②④ D．②③","result_json":"{\"note\": \"Critic 未返回结果\"}","created_at":"2026-09-25T05:52:41"},{"id":438180,"import_question_id":22127,"audit_type":"SYMBOLIC","auditor_name":"symbolic-auditor","status":"WARN","score":0.5,"input_snapshot":"答案: || 盲解:","result_json":"{\"checks\": [], \"detail\": {\"reason\": \"未提取到来源答案\"}, \"status\": \"WARN\", \"applicable\": true}","created_at":"2026-09-25T05:52:41"},{"id":438539,"import_question_id":22127,"audit_type":"COHERENCE","auditor_name":"coherence-auditor","status":"PASS","score":0.9,"input_snapshot":"答案: || 解析:(1) COD = 、 = 60,\\(\\frac{90}{BOD} \\frac{BOD}{COD}\\) =  +  = 60+ 90 =\\(150 \\frac{BOC}{平分}\\) OE BOC COE =\\(\\frac{1}{2}\\) =\\(\\frac{1}{2}\\) = 75;\\(\\frac{BOC}{90} \\frac{150}{AOC}\\)(2) ","result_json":"{\"ok\": true, \"which\": [], \"issues\": [], \"confidence\": 0.9, \"equation_checks\": [{\"valid\": true, \"reason\": \"Correct application of angle addition.\", \"equation\": \"∠BOC = ∠BOD + ∠COD = 60° + 90° = 150°\"}, {\"valid\": true, \"reason\": \"Correct use of angle bisector property.\", \"equation\": \"∠COE = 1/2 ∠BOC = 1/2 × 150° = 75°\"}, {\"valid\": true, \"reason\": \"Correct calculation using straight line angle sum.\", \"equation\": \"∠BOD = 180° - ∠COD - ∠AOC = 180° - 90° - 20° = 70°\"}, {\"valid\": true, \"reason\": \"Correct application of angle addition.\", \"equation\": \"∠BOC = ∠BOD + ∠COD = 70° + 90° = 160°\"}, {\"valid\": true, \"reason\": \"Correct use of angle bisector property.\", \"equation\": \"∠COE = 1/2 ∠BOC = 1/2 × 160° = 80°\"}, {\"valid\": true, \"reason\": \"Correct subtraction to find ∠DOE.\", \"equation\": \"∠DOE = ∠COD - ∠COE = 90° - 80° = 10°\"}, {\"valid\": true, \"reason\": \"Correct expression for ∠COE in terms of α.\", \"equation\": \"∠COE = 90° - α\"}, {\"valid\": true, \"reason\": \"Correct use of angle bisector property.\", \"equation\": \"∠BOC = 2∠COE = 2(90° - α) = 180° - 2α\"}, {\"valid\": true, \"reason\": \"Correct calculation using straight line angle sum.\", \"equation\": \"∠AOC = 180° - ∠BOC = 180° - (180° - 2α) = 2α\"}]}","created_at":"2026-09-25T05:53:42"},{"id":438540,"import_question_id":22127,"audit_type":"QUALITY_GATE","auditor_name":"quality-gate","status":"FAIL","score":0.917,"input_snapshot":"","result_json":"{\"warnings\": [\"SOLVER\"], \"graded_gate\": true, \"hard_failed\": [\"FORMULA\", \"CRITIC\", \"ANSWER\"]}","created_at":"2026-09-25T05:53:42"}]}