{"ok":true,"question":{"id":21120,"question_uuid":"f4f0ece3-40c6-4b63-a802-7058ba0908f8","job_id":4,"source_file":"A中考/2026江苏中考期末数学试卷+答案835.pdf","source_page_start":1,"source_page_end":1,"source_bbox":"[45.0, 124.66, 437.59, 270.91]","subject_code":"math","grade_level":"SENIOR","tree_code":"KT_SENIOR_MATH_STANDARD","question_type":"fill_blank","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"如图，点\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"O\"}, {\"text\": \"是\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\Delta ABC\"}, {\"text\": \"的内切圆圆心，若\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\angle BAC = 80^{\\\\circ}\"}, {\"text\": \"，则\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\angle BOC\"}, {\"text\": \"度数等于( ).\", \"type\": \"text\"}, {\"kind\": \"embedded_image\", \"type\": \"figure\", \"width\": 708, \"height\": 416, \"sha256\": \"1a04227800d8781ebde549e46e7d1e42c169f252cb0e593befb45e920920fecc\", \"caption\": \"\", \"asset_key\": \"math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f4f0ece3-40c6-4b63-a802-7058ba0908f8/1a04227800d8.png\", \"public_url\": \"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f4f0ece3-40c6-4b63-a802-7058ba0908f8/1a04227800d8.png\", \"source_bbox\": [45.0, 152.9840087890625, 186.6999969482422, 240.58001708984375], \"source_page\": 1}, {\"type\": \"option\", \"label\": \"A\", \"blocks\": [{\"text\": \"100∘ (B)110∘ (C)120∘ (D)130∘\", \"type\": \"text\"}], \"source_bbox\": [45.0, 257.17, 304.01, 270.91], \"source_page\": 1}, {\"type\": \"answer\", \"blocks\": [{\"text\": \"2.(1)见解析; (2)3√2. (1)连接OC, ∵AB是⊙O的直径, ∴∠ACB = 90∘, ∴∠A + ∠ABC = 90∘,∵OB = OC, ∴∠ABC = ∠OCB, ∵ ∠BCD = ∠A, ∴∠BCD + ∠OCB = 90∘,即∠OCD = 90∘, ∴OC ⊥CD. ∵OC为⊙O的半径, ∴CD是⊙O的切线. (2)∵ 点 B 是 AD 的中点, ∴BD = AB = 2OC, ∵OB = OC, ∴OD = OB +BD= 3OC, ∴ OC\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"1 _{OD}\"}, {\"text\": \"= _{3}, ∵BE ⊥AD, ∴ ∠DBE = 90∘, 又∵∠OCD = 90∘,∴sinD = BE OC\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"1 _{DE}\"}, {\"text\": \"= _{OD}= _{3}, ∴DE = 3BE = 9, 在Rt △DBE中,BD=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\sqrt{DE^{2}}\"}, {\"text\": \"−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"BE^{2}\"}, {\"text\": \"=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\sqrt{9^{2}}\"}, {\"text\": \"−\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3^{2}\"}, {\"text\": \"= 6√2. ∴OC = 3√2, 即⊙O半径为3√2． 3.(1)见解析; (2)1. (1)证明: 连接OA, 过O作OM ⊥AC于点M, ∵AB = AC且O为BC中点,所以AO平分∠BAC, ∵⊙ O与AB相切于点D, ∴OD ⊥AB, ∵OM ⊥AC,所以OM = OD = r, 所以AC是⊙O的切线. (2)过点O 作 ON ⊥GH, FN =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"1 _{2}\"}, {\"text\": \"GF, ∴OM = NH, ∵OD ⊥AB, GH ⊥AC, 所以∠GOF=∠DOB = 90° −∠B, ∠GFO = ∠GFH = 90° −∠C, ∵AB = AC, 所以∠B= ∠C, ∠GOF = ∠GFO, GO = GF, ∵OF = OG, 所以△OGF为等边三 角形, ∴∠GOF = ∠DOE = 60°, ∵OD = OE,所以 △ODE 为等边三角形, ∴OD = DE = r = 2,所以 FN =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"1 _{2}\"}, {\"text\": \"GF =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"1 _{2}\"}, {\"text\": \"DE = 1, NH = OM =r= 2, ∴FH = NH −FN = 1. 4.C 因为⊙O是\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\Delta{}ABC\"}, {\"text\": \"的内切圆, 所以OB, OC分别平分∠ABC, ∠ACB, 因为= 46∘, ∠ACB = 84∘, 所以∠OBC = 1\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"1 _{2}\"}, {\"text\": \"∠ABC = 23∘, ∠OCB = _{2}\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\times\"}, {\"text\": \"∠ACB= 42∘, ∴∠BOC = 180∘−∠OBC −∠OCB = 115∘. 故选: C. 5.(1)见解析; (2)2√3． (1)连接OD,⊙O是△ABC的外接圆, AB是⊙O的直径, ∴∠ACB = 90∘,OA= OD, ∴∠OAD = ∠ODA, ∵DE ∥ BC, ∴∠AED = ∠ACB = 90∘,∵∠BAC的平分线交⊙O于点D, ∴∠EAD = ∠BAD, ∴∠EAD =∠ADO,∴OD ∥ AE, ∴∠ODE = 180∘−∠AED = 90∘, ∵OD为⊙O的半径, ∴DE是⊙O的切线; (2)设OD交BC于点F, ∵∠BAC = 60∘, ∠ACB = 90∘,∴∠ABC = 30∘, ∵CE ∥DF, DE ∥CF, ∠E = 90∘, ∴ 四边形CEDF为矩形,∴∠DFC = 90∘, DF = CE = √3, ∴∠OFB = 90∘, 设⊙O的半径为r, 则:OB= OD = r, OF = OD −DF = r −√3, ∵∠OFB = 90∘, ∠ABC = 30∘, ∴OB =2OF,∴r = 2(r −√3), ∴r = 2√3, ∴⊙ O的半径为2√3． 6.(1)见解析; (2)3√5. (1)证明: 如图连接OC, OA = OC, 所以∠AOC = 2∠ABC = 90°, ∠OAC=∠OCA = 45°, ∵AD是⊙O的切线, ∴ ∠OAD = 90°, ∴∠OAC = ∠EAC=45°, ∵CE = AE, ∴∠ECA = ∠EAC = 45°, 所以∠AEC = 90°, 所以四边形 OAEC为正方形, ∴∠ECO = 90°, ∴EC ⊥OC, ∴CE是⊙O的切线; (2)因为四边形OAEC为正方形, 所以EC ∥AF, 所以△ECD ∼△AFD,EC:AF= DE: AD, 因为DE = 2AE = 4, 所以AF = 3, OF = 1, CF = √5, CD=2√5, DF = CF + CD = 3√5. 7.20∘ 连接OC, 由圆周角定理得, ∠COD = 2∠A = 70∘, 因为CD为⊙O的切线,所以OC ⊥CD, ∴∠D = 90∘−∠COD = 20∘. 8.(1)见解析; (2)①2; ②0或4． (1)∵正六边形ABCDEF内接于⊙O, ∴AB = BC = CD = DE = EF =FA,∠A = ∠ABC = ∠C = ∠D = ∠DEF = ∠F. ∵点P, Q同时分别从A,\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [45.0, 124.66, 437.59, 270.91], \"file\": \"A中考/2026江苏中考期末数学试卷+答案835.pdf\", \"regions\": [{\"bbox\": [45.0, 124.66, 437.59, 270.91], \"page\": 1}], \"section\": \"generic\", \"page_end\": 1, \"page_start\": 1, \"section_label\": \"\", \"answer_regions\": [{\"bbox\": [30.0, 93.87199401855467, 581.7239990234375, 725.6640014648438], \"page\": 8}, {\"bbox\": [30.0, 262.1419982910156, 580.6439819335938, 750.3839721679688], \"page\": 9}, {\"bbox\": [30.0, 80.31199645996094, 579.56396484375, 749.4240112304688], \"page\": 10}, {\"bbox\": [30.0, 79.54291534423828, 580.8359985351562, 612.1240234375], \"page\": 11}]}, \"version\": 1, \"grade_level\": \"SENIOR\", \"number_label\": \"1\", \"subject_code\": \"math\", \"question_type\": \"fill_blank\", \"content_sha256\": \"28e3b477bfffedd3fcdd03a785209690940adaabb98fa5f2b175b2dbf18043ec\"}","body_html":"<div>如图，点\\(O\\)是\\(\\Delta ABC\\)的内切圆圆心，若\\(\\angle BAC = 80^{\\circ}\\)，则\\(\\angle BOC\\)度数等于( ).<img class=\"question-figure\" src=\"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/f4f0ece3-40c6-4b63-a802-7058ba0908f8/1a04227800d8.png\" style=\"width:188.9px!important;max-width:100%;height:auto\" alt=\"\" title=\"点击查看原图\"><br>A. 100∘ (B)110∘ (C)120∘ (D)130∘</div>","answer_html":"2.(1)见解析; (2)3√2. (1)连接OC, ∵AB是⊙O的直径, ∴∠ACB = 90∘, ∴∠A + ∠ABC = 90∘,∵OB = OC, ∴∠ABC = ∠OCB, ∵ ∠BCD = ∠A, ∴∠BCD + ∠OCB = 90∘,即∠OCD = 90∘, ∴OC ⊥CD. ∵OC为⊙O的半径, ∴CD是⊙O的切线. (2)∵ 点 B 是 AD 的中点, ∴BD = AB = 2OC, ∵OB = OC, ∴OD = OB +BD= 3OC, ∴ OC\\(1 _{OD}\\)= _{3}, ∵BE ⊥AD, ∴ ∠DBE = 90∘, 又∵∠OCD = 90∘,∴sinD = BE OC\\(1 _{DE}\\)= _{OD}= _{3}, ∴DE = 3BE = 9, 在Rt △DBE中,BD=\\(\\sqrt{DE^{2}}\\)−\\(BE^{2}\\)=\\(\\sqrt{9^{2}}\\)−\\(3^{2}\\)= 6√2. ∴OC = 3√2, 即⊙O半径为3√2． 3.(1)见解析; (2)1. (1)证明: 连接OA, 过O作OM ⊥AC于点M, ∵AB = AC且O为BC中点,所以AO平分∠BAC, ∵⊙ O与AB相切于点D, ∴OD ⊥AB, ∵OM ⊥AC,所以OM = OD = r, 所以AC是⊙O的切线. (2)过点O 作 ON ⊥GH, FN =\\(1 _{2}\\)GF, ∴OM = NH, ∵OD ⊥AB, GH ⊥AC, 所以∠GOF=∠DOB = 90° −∠B, ∠GFO = ∠GFH = 90° −∠C, ∵AB = AC, 所以∠B= ∠C, ∠GOF = ∠GFO, GO = GF, ∵OF = OG, 所以△OGF为等边三 角形, ∴∠GOF = ∠DOE = 60°, ∵OD = OE,所以 △ODE 为等边三角形, ∴OD = DE = r = 2,所以 FN =\\(1 _{2}\\)GF =\\(1 _{2}\\)DE = 1, NH = OM =r= 2, ∴FH = NH −FN = 1. 4.C 因为⊙O是\\(\\Delta{}ABC\\)的内切圆, 所以OB, OC分别平分∠ABC, ∠ACB, 因为= 46∘, ∠ACB = 84∘, 所以∠OBC = 1\\(1 _{2}\\)∠ABC = 23∘, ∠OCB = _{2}\\(\\times\\)∠ACB= 42∘, ∴∠BOC = 180∘−∠OBC −∠OCB = 115∘. 故选: C. 5.(1)见解析; (2)2√3． (1)连接OD,⊙O是△ABC的外接圆, AB是⊙O的直径, ∴∠ACB = 90∘,OA= OD, ∴∠OAD = ∠ODA, ∵DE ∥ BC, ∴∠AED = ∠ACB = 90∘,∵∠BAC的平分线交⊙O于点D, ∴∠EAD = ∠BAD, ∴∠EAD =∠ADO,∴OD ∥ AE, ∴∠ODE = 180∘−∠AED = 90∘, ∵OD为⊙O的半径, ∴DE是⊙O的切线; (2)设OD交BC于点F, ∵∠BAC = 60∘, ∠ACB = 90∘,∴∠ABC = 30∘, ∵CE ∥DF, DE ∥CF, ∠E = 90∘, ∴ 四边形CEDF为矩形,∴∠DFC = 90∘, DF = CE = √3, ∴∠OFB = 90∘, 设⊙O的半径为r, 则:OB= OD = r, OF = OD −DF = r −√3, ∵∠OFB = 90∘, ∠ABC = 30∘, ∴OB =2OF,∴r = 2(r −√3), ∴r = 2√3, ∴⊙ O的半径为2√3． 6.(1)见解析; (2)3√5. (1)证明: 如图连接OC, OA = OC, 所以∠AOC = 2∠ABC = 90°, ∠OAC=∠OCA = 45°, ∵AD是⊙O的切线, ∴ ∠OAD = 90°, ∴∠OAC = ∠EAC=45°, ∵CE = AE, ∴∠ECA = ∠EAC = 45°, 所以∠AEC = 90°, 所以四边形 OAEC为正方形, ∴∠ECO = 90°, ∴EC ⊥OC, ∴CE是⊙O的切线; (2)因为四边形OAEC为正方形, 所以EC ∥AF, 所以△ECD ∼△AFD,EC:AF= DE: AD, 因为DE = 2AE = 4, 所以AF = 3, OF = 1, CF = √5, CD=2√5, DF = CF + CD = 3√5. 7.20∘ 连接OC, 由圆周角定理得, ∠COD = 2∠A = 70∘, 因为CD为⊙O的切线,所以OC ⊥CD, ∴∠D = 90∘−∠COD = 20∘. 8.(1)见解析; (2)①2; ②0或4． (1)∵正六边形ABCDEF内接于⊙O, ∴AB = BC = CD = DE = EF =FA,∠A = ∠ABC = ∠C = ∠D = ∠DEF = ∠F. ∵点P, Q同时分别从A,","ways_html":"","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=8091): Read timed out. 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[图] A. 100∘ (B)110∘ (C)120∘ (D)130∘","result_json":"{\"ok\": true, \"rule\": \"clean-override\", \"issues\": [], \"confidence\": 0.9}","created_at":"2026-09-24T23:04:57"},{"id":415672,"import_question_id":21120,"audit_type":"FORMULA","auditor_name":"formula-auditor","status":"PASS","score":1.0,"input_snapshot":"O; \\Delta ABC; \\angle BAC = 80^{\\circ}; \\angle BOC; 1 _{OD}","result_json":"{\"risky\": 12, \"issues\": [], \"checked\": 16, \"invalid\": 0, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [], \"visually_verified\": 0}","created_at":"2026-09-24T23:04:57"},{"id":415673,"import_question_id":21120,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"如图，点 O 是 \\Delta ABC 的内切圆圆心，若 \\angle BAC = 80^{\\circ} ，则 \\angle BOC 度数等于( ). [图] A. 100∘ (B)110∘ (C)120∘ (D)130∘","result_json":"{\"picks\": [], \"tree_code\": \"KT_SENIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-24T23:04:57"},{"id":416059,"import_question_id":21120,"audit_type":"VISION","auditor_name":"vision-auditor","status":"WARN","score":0.2,"input_snapshot":"题干：\n如图，点\\(O\\)是\\(\\Delta ABC\\)的内切圆圆心，若\\(\\angle BAC = 80^{\\circ}\\)，则\\(\\angle BOC\\)度数等于( ). A. 100∘ (B)110∘ (C)120∘ (D)130∘\n答案：\n2.(1)见解析; (2)3√2. (1)连接OC, ∵AB是⊙O的直径, ∴∠ACB = 90∘, ∴∠A + ∠ABC = 90∘,∵OB = OC, ∴∠ABC = ∠OCB, ∵ ∠BCD = ∠A, ∴∠BCD + ∠OCB = 90∘,即∠OCD = 90∘, ∴OC ⊥CD. ∵OC为⊙O的半径, ∴CD是⊙O的切线. (2)∵ 点 B 是 AD 的中点, ∴BD = AB = 2OC, ∵OB = OC, ∴OD = OB +BD= 3OC, ∴ OC\\(1 _{OD}\\)= _{3}, ∵BE ⊥AD, ∴ ∠DBE = 90∘, 又∵∠OCD = 90∘,∴sinD = BE OC\\(1 _{DE}\\)= _{OD}= _{3}, ∴DE = 3BE = 9, 在Rt △DBE中,BD=\\(\\sqrt{DE^{2}}\\)−\\(BE^{2}\\)=\\(\\sqrt{9^{2}}\\)−\\(3^{2}\\)= 6√2. ∴OC = 3√2, 即⊙O半径为3√2． 3.(1)见解析; (2)1. (1)证明: 连接OA, 过O作OM ⊥AC于点M, ∵AB = AC且O为BC中点,所以AO平分∠BAC, ∵⊙ O与AB相切于点D, ∴OD ⊥AB, ∵OM ⊥AC,所以OM = OD = r, 所以AC是⊙O的切线. (2)过点O 作 ON ⊥GH, FN =\\(1 _{2}\\)GF, ∴OM = NH, ∵OD ⊥AB, GH ⊥AC, 所以∠GOF=∠DOB = 90° −∠B, ∠GFO = ∠GFH = 90° −∠","result_json":"{\"note\": \"视觉比对不可用\", \"error\": \"RuntimeError\", \"unavailable\": true}","created_at":"2026-09-24T23:08:57"}]}