{"ok":true,"question":{"id":21051,"question_uuid":"6f57568f-4776-401b-bf39-de55246f99b5","job_id":4,"source_file":"A中考/2026江苏中考期末数学试卷+答案825.pdf","source_page_start":2,"source_page_end":2,"source_bbox":"[31.31, 334.48, 456.83, 492.6]","subject_code":"math","grade_level":"JUNIOR","tree_code":"KT_JUNIOR_MATH_STANDARD","question_type":"comprehensive","canonical_json":"{\"year\": null, \"blocks\": [{\"text\": \"如图， 是  O 的弦,过点 作直线\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{AB}{以为顶点作}\"}, {\"text\": \"B 如图①,在矩形ABCD中，AB = 4, EF, O  = ,分\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{AOC}{CD}\\\\frac{90}{CB} =\"}, {\"text\": \"，点是 边上一动点，连接\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{BC}{AE} \\\\frac{6}{DE} \\\\frac{BC}{ABE}\"}, {\"text\": \"E 别交,,,若 = CD．\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{EFAB}{试判断直线}\\\\frac{于点}{EF}\"}, {\"text\": \"、，作 的外接，交AD\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{O}{FG}F \\\\frac{DE}{DFG}\"}, {\"text\": \"G 于点，交 于点，连接 ． （）若 = 60，则 = ____ ：\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{2} \\\\frac{AED}{DFG}\"}, {\"text\": \"（）当CE的长为 ____时， 为等腰 三角形； （）如图②当3, O与CD相切时求,CE的长．\", \"type\": \"text\"}, {\"kind\": \"embedded_image\", \"type\": \"figure\", \"width\": 837, \"height\": 578, \"sha256\": \"754a4453792493b34f5d9862f96d9a3c25833a3a06cc41b1d785984d1f4f54d8\", \"caption\": \"\", \"asset_key\": \"math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/6f57568f-4776-401b-bf39-de55246f99b5/754a44537924.png\", \"public_url\": \"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/6f57568f-4776-401b-bf39-de55246f99b5/754a44537924.png\", \"source_bbox\": [30.0, 487.03997802734375, 202.6000061035156, 597.5299682617188], \"source_page\": 2}, {\"kind\": \"embedded_image\", \"type\": \"figure\", \"width\": 664, \"height\": 490, \"sha256\": \"07f76ceab20917730fe2dcbe0ad79d63c11769873b97540ddc46fb1b4855aa49\", \"caption\": \"\", \"asset_key\": \"math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/6f57568f-4776-401b-bf39-de55246f99b5/07f76ceab209.png\", \"public_url\": \"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/6f57568f-4776-401b-bf39-de55246f99b5/07f76ceab209.png\", \"source_bbox\": [450.2999877929687, 333.09197998046875, 669.699951171875, 422.3399658203125], \"source_page\": 2}, {\"type\": \"subquestion\", \"index\": 1, \"blocks\": [{\"text\": \"与 O 的位置关系, 并说明理由；\", \"type\": \"text\"}]}, {\"type\": \"subquestion\", \"index\": 2, \"blocks\": [{\"text\": \"若\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\bigcirc O\"}, {\"text\": \"的半径为3,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\tan\\\\angle OAD=\\\\frac{1}{3}\"}, {\"text\": \",求\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"BC\"}, {\"text\": \"的长．\", \"type\": \"text\"}]}, {\"type\": \"answer\", \"blocks\": [{\"text\": \"；60   = 90,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{AOC}{ADO}\"}, {\"text\": \"(2)连接EF，如图①所示：  +  = 90, 四边形FGEA是 O的内接四边形，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OAD}{CDB}\"}, {\"text\": \" = ， 又 ADO = ，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{DGF}{GDF} \\\\frac{DAE}{ADE}\"}, {\"text\": \"又 = ，DFG  DEA,  =  = CBD,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{ADO}{CBD} \\\\frac{CDB}{OAD}\"}, {\"text\": \"当DEA为等腰三角形时，  +  = 90,  OA = OB, = , DFG为等腰三角形，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OAD}{OBD} \\\\frac{OBD}{CBO}\"}, {\"text\": \" 四边形ABCD是矩形，AB = 4, BC = 6，  +  =  = 90,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{CBD}{OB}\"}, {\"text\": \"CD = AB = 4, AD = = 6, 即 ⊥ BC,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{BC}{BCD}\"}, {\"text\": \" = ABC =  =  = 90,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{BAD}{O} \\\\frac{ADC}{ABE} \\\\frac{OB}{EF}\"}, {\"text\": \" 为半径，  是 的外接圆， = 90，  与 O相切\", \"type\": \"text\"}]}, {\"type\": \"solution\", \"blocks\": [{\"text\": \"【讲解】 2026-06-22 (1)EF与 O相切；理由如下： (1)\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{四边形}{AED}\\\\frac{AEGF}{DFG}\"}, {\"text\": \"是 O的内接四边形，  =  = 60， 如图，连接OB，  CB = CD,CDB = CBD, 故答案为：；60   = 90,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{AOC}{ADO}\"}, {\"text\": \"(2)连接EF，如图①所示：  +  = 90, 四边形FGEA是 O的内接四边形，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OAD}{CDB}\"}, {\"text\": \" = ， 又 ADO = ，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{DGF}{GDF} \\\\frac{DAE}{ADE}\"}, {\"text\": \"又 = ，DFG  DEA,  =  = CBD,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{ADO}{CBD} \\\\frac{CDB}{OAD}\"}, {\"text\": \"当DEA为等腰三角形时，  +  = 90,  OA = OB, = , DFG为等腰三角形，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OAD}{OBD} \\\\frac{OBD}{CBO}\"}, {\"text\": \" 四边形ABCD是矩形，AB = 4, BC = 6，  +  =  = 90,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{CBD}{OB}\"}, {\"text\": \"CD = AB = 4, AD = = 6, 即 ⊥ BC,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{BC}{BCD}\"}, {\"text\": \" = ABC =  =  = 90,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{BAD}{O} \\\\frac{ADC}{ABE} \\\\frac{OB}{EF}\"}, {\"text\": \" 为半径，  是 的外接圆， = 90，  与 O相切；\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{ABE}{O}\"}, {\"text\": \" AE是  的直径, (2)如(1)图， = 90，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{CBO}{3}\"}, {\"text\": \"AFE = 90,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{O}{OA}\"}, {\"text\": \" 的半径为，  = OB = 3,  = 180− = 180 − 90 = 90，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{DFE}{CDF} \\\\frac{AFE}{DFE}\"}, {\"text\": \" AOC = 90,tan OAD =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{3}\"}, {\"text\": \" = DCE =  = 90, 四边形 是矩形，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{DCEF}{CEEF}\"}, {\"text\": \"  =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{DO}{AO} = \\\\frac{1}{3},OD = 1\"}, {\"text\": \"DF =, = CD = 4,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{tan}{BC} \\\\frac{OAD}{CD} \\\\frac{AED}{AE}\"}, {\"text\": \"若 为等腰三角形，分三种情况： 设 = = x, = DE时，  = 90,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{AFE}{DF}\"}, {\"text\": \"①当 CO = + DO = x +1，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{CD}{RtBOC}\"}, {\"text\": \"⊥  AF = =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{1}{2}\"}, {\"text\": \"AD = 3,  在  中，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{EF}{CE} \\\\frac{AD}{DF}CO^{2}\"}, {\"text\": \"= +\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"OB^{2}\"}, {\"text\": \"，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"CB^{2}_{2}\"}, {\"text\": \" = = 3; （x +1） =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"3^{2}\"}, {\"text\": \"+\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"x^{2}\"}, {\"text\": \"， = AD = 6时，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{AE}{RtABE}\"}, {\"text\": \"②当 解得：x = 4, 在  中，由勾股定理得： BC = 4. = =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\sqrt{6^{2}−4^{2}} = 2 \\\\sqrt{5}\"}, {\"text\": \"，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{BE}{CE} \\\\frac{\\\\sqrt{AE^{2}−AB^{2}}}{BC}\"}, {\"text\": \" = − = 6 −\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2 \\\\sqrt{5}\"}, {\"text\": \"；\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{CE}{DA}\"}, {\"text\": \"③当DE = = 6时，在RtDCE中， 由勾股定理得： CE =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\sqrt{DE^{2}−CD^{2}} = \\\\sqrt{6^{2}−4^{2}} = 2 \\\\sqrt{5}\"}, {\"text\": \"； 综上所述，当BE的长为或36 −\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2 \\\\sqrt{5}\"}, {\"text\": \"或\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2 \\\\sqrt{5}\"}, {\"text\": \"时， DFG为等腰三角形， 故答案为：或36 −\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2 \\\\sqrt{5}\"}, {\"text\": \"或2 5; 2026-06-22 (3)过作OOH ⊥ CD于点， H 如图②所示： 则OH∥AD∥CE，\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{四边形}{ABC}\\\\frac{ABCD}{90}\"}, {\"text\": \" 是矩形，  = ，  为 \", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{AE}{OA}\"}, {\"text\": \"O的直径，  =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OE}{是梯形}\"}, {\"text\": \"\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OH}{OH}\"}, {\"text\": \"ADCE的中位线，  = 1 (+)\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"2\\\\frac{AD}{AD} \\\\frac{CE}{CE}\"}, {\"text\": \"2 = +,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OH}{O}\"}, {\"text\": \"相切， 为切点,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{CD}{OA}\"}, {\"text\": \"H  与  =\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{OH}{AE}\"}, {\"text\": \",  = 2OH = AD + CE = 6 + CE, 在Rt ABE中由勾股定理得, : + =,\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{AB^{2}}{4^{2}} \\\\frac{BE^{2}}{6}AE^{2}_{2}\"}, {\"text\": \"即 +（− CE）（=6 + CE）2, 解得 : CE=\", \"type\": \"text\"}, {\"type\": \"math_inline\", \"latex\": \"\\\\frac{2}{3}\"}, {\"text\": \". 2026-06-22\", \"type\": \"text\"}]}], \"source\": {\"bbox\": [31.31, 334.48, 456.83, 492.6], \"file\": \"A中考/2026江苏中考期末数学试卷+答案825.pdf\", \"regions\": [{\"bbox\": [31.31, 334.48, 456.83, 492.6], \"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\": [232.1300048828125, 613.11962890625, 557.0399780273438, 796.1519775390625], \"page\": 2}, {\"bbox\": [29.816051483154297, 123.30118560791016, 557.0399780273438, 796.1519775390625], \"page\": 3}, {\"bbox\": [231.9165496826172, 251.11669921875, 557.0399780273438, 796.1519775390625], \"page\": 4}]}, \"version\": 1, \"grade_level\": \"JUNIOR\", \"number_label\": \"2\", \"subject_code\": \"math\", \"question_type\": \"comprehensive\", \"content_sha256\": \"fe47fee64eea02ca3bac128082a9e55da201fe201d4c4acf2e3241c809ce8f69\"}","body_html":"<div>如图， 是  O 的弦,过点 作直线\\(\\frac{AB}{以为顶点作}\\)B 如图①,在矩形ABCD中，AB = 4, EF, O  = ,分\\(\\frac{AOC}{CD}\\frac{90}{CB} =\\)，点是 边上一动点，连接\\(\\frac{BC}{AE} \\frac{6}{DE} \\frac{BC}{ABE}\\)E 别交,,,若 = CD．\\(\\frac{EFAB}{试判断直线}\\frac{于点}{EF}\\)、，作 的外接，交AD\\(\\frac{O}{FG}F \\frac{DE}{DFG}\\)G 于点，交 于点，连接 ． （）若 = 60，则 = ____ ：\\(\\frac{1}{2} \\frac{AED}{DFG}\\)（）当CE的长为 ____时， 为等腰 三角形； （）如图②当3, O与CD相切时求,CE的长．<img class=\"question-figure\" src=\"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/6f57568f-4776-401b-bf39-de55246f99b5/754a44537924.png\" style=\"width:230.1px!important;max-width:100%;height:auto\" alt=\"\" title=\"点击查看原图\"><img class=\"question-figure\" src=\"http://tikupdfpng.mtwlkj.net:5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/6f57568f-4776-401b-bf39-de55246f99b5/07f76ceab209.png\" style=\"width:292.5px!important;max-width:100%;height:auto\" alt=\"\" title=\"点击查看原图\"><br>（1）与 O 的位置关系, 并说明理由；<br>（2）若\\(\\bigcirc O\\)的半径为3,\\(\\tan\\angle OAD=\\frac{1}{3}\\),求\\(BC\\)的长．</div>","answer_html":"；60   = 90,\\(\\frac{AOC}{ADO}\\)(2)连接EF，如图①所示：  +  = 90, 四边形FGEA是 O的内接四边形，\\(\\frac{OAD}{CDB}\\) = ， 又 ADO = ，\\(\\frac{DGF}{GDF} \\frac{DAE}{ADE}\\)又 = ，DFG  DEA,  =  = CBD,\\(\\frac{ADO}{CBD} \\frac{CDB}{OAD}\\)当DEA为等腰三角形时，  +  = 90,  OA = OB, = , DFG为等腰三角形，\\(\\frac{OAD}{OBD} \\frac{OBD}{CBO}\\) 四边形ABCD是矩形，AB = 4, BC = 6，  +  =  = 90,\\(\\frac{CBD}{OB}\\)CD = AB = 4, AD = = 6, 即 ⊥ BC,\\(\\frac{BC}{BCD}\\) = ABC =  =  = 90,\\(\\frac{BAD}{O} \\frac{ADC}{ABE} \\frac{OB}{EF}\\) 为半径，  是 的外接圆， = 90，  与 O相切","ways_html":"【讲解】 2026-06-22 (1)EF与 O相切；理由如下： (1)\\(\\frac{四边形}{AED}\\frac{AEGF}{DFG}\\)是 O的内接四边形，  =  = 60， 如图，连接OB，  CB = CD,CDB = CBD, 故答案为：；60   = 90,\\(\\frac{AOC}{ADO}\\)(2)连接EF，如图①所示：  +  = 90, 四边形FGEA是 O的内接四边形，\\(\\frac{OAD}{CDB}\\) = ， 又 ADO = ，\\(\\frac{DGF}{GDF} \\frac{DAE}{ADE}\\)又 = ，DFG  DEA,  =  = CBD,\\(\\frac{ADO}{CBD} \\frac{CDB}{OAD}\\)当DEA为等腰三角形时，  +  = 90,  OA = OB, = , DFG为等腰三角形，\\(\\frac{OAD}{OBD} \\frac{OBD}{CBO}\\) 四边形ABCD是矩形，AB = 4, BC = 6，  +  =  = 90,\\(\\frac{CBD}{OB}\\)CD = AB = 4, AD = = 6, 即 ⊥ BC,\\(\\frac{BC}{BCD}\\) = ABC =  =  = 90,\\(\\frac{BAD}{O} \\frac{ADC}{ABE} \\frac{OB}{EF}\\) 为半径，  是 的外接圆， = 90，  与 O相切；\\(\\frac{ABE}{O}\\) AE是  的直径, (2)如(1)图， = 90，\\(\\frac{CBO}{3}\\)AFE = 90,\\(\\frac{O}{OA}\\) 的半径为，  = OB = 3,  = 180− = 180 − 90 = 90，\\(\\frac{DFE}{CDF} \\frac{AFE}{DFE}\\) AOC = 90,tan OAD =\\(\\frac{1}{3}\\) = DCE =  = 90, 四边形 是矩形，\\(\\frac{DCEF}{CEEF}\\)  =\\(\\frac{DO}{AO} = \\frac{1}{3},OD = 1\\)DF =, = CD = 4,\\(\\frac{tan}{BC} \\frac{OAD}{CD} \\frac{AED}{AE}\\)若 为等腰三角形，分三种情况： 设 = = x, = DE时，  = 90,\\(\\frac{AFE}{DF}\\)①当 CO = + DO = x +1，\\(\\frac{CD}{RtBOC}\\)⊥  AF = =\\(\\frac{1}{2}\\)AD = 3,  在  中，\\(\\frac{EF}{CE} \\frac{AD}{DF}CO^{2}\\)= +\\(OB^{2}\\)，\\(CB^{2}_{2}\\) = = 3; （x +1） =\\(3^{2}\\)+\\(x^{2}\\)， = AD = 6时，\\(\\frac{AE}{RtABE}\\)②当 解得：x = 4, 在  中，由勾股定理得： BC = 4. = =\\(\\sqrt{6^{2}−4^{2}} = 2 \\sqrt{5}\\)，\\(\\frac{BE}{CE} \\frac{\\sqrt{AE^{2}−AB^{2}}}{BC}\\) = − = 6 −\\(2 \\sqrt{5}\\)；\\(\\frac{CE}{DA}\\)③当DE = = 6时，在RtDCE中， 由勾股定理得： CE =\\(\\sqrt{DE^{2}−CD^{2}} = \\sqrt{6^{2}−4^{2}} = 2 \\sqrt{5}\\)； 综上所述，当BE的长为或36 −\\(2 \\sqrt{5}\\)或\\(2 \\sqrt{5}\\)时， DFG为等腰三角形， 故答案为：或36 −\\(2 \\sqrt{5}\\)或2 5; 2026-06-22 (3)过作OOH ⊥ CD于点， H 如图②所示： 则OH∥AD∥CE，\\(\\frac{四边形}{ABC}\\frac{ABCD}{90}\\) 是矩形，  = ，  为 \\(\\frac{AE}{OA}\\)O的直径，  =\\(\\frac{OE}{是梯形}\\)\\(\\frac{OH}{OH}\\)ADCE的中位线，  = 1 (+)\\(2\\frac{AD}{AD} \\frac{CE}{CE}\\)2 = +,\\(\\frac{OH}{O}\\)相切， 为切点,\\(\\frac{CD}{OA}\\)H  与  =\\(\\frac{OH}{AE}\\),  = 2OH = AD + CE = 6 + CE, 在Rt ABE中由勾股定理得, : + =,\\(\\frac{AB^{2}}{4^{2}} \\frac{BE^{2}}{6}AE^{2}_{2}\\)即 +（− CE）（=6 + CE）2, 解得 : CE=\\(\\frac{2}{3}\\). 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. 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\\frac{AOC}{CD}\\frac{90}{CB} =; \\frac{BC}{AE} \\frac{6}{DE} \\frac{BC}{ABE}; \\frac{EFAB}{试判断直线}\\frac{于点}{EF}; \\frac{O}{FG}F \\frac{DE}{DFG}","result_json":"{\"risky\": 59, \"issues\": [], \"checked\": 61, \"invalid\": 0, \"visual_check\": \"delegated_to_question_vision\", \"invalid_details\": [], \"visually_verified\": 0}","created_at":"2026-09-24T22:30:03"},{"id":414620,"import_question_id":21051,"audit_type":"KNOWLEDGE","auditor_name":"knowledge-auditor","status":"WARN","score":0.0,"input_snapshot":"如图， 是  O 的弦,过点 作直线 \\frac{AB}{以为顶点作} B 如图①,在矩形ABCD中，AB = 4, EF, O  = ,分 \\frac{AOC}{CD}\\frac{90}{CB} = ，点是 边上一动点，连接 \\frac{BC}{AE} \\frac{6}{DE} \\frac{BC}{ABE} E 别交,,,若 = CD． \\frac{EFAB}{试判断直线}\\frac{于点}{EF} 、，作 的外接，交AD \\frac{O}{FG}F \\frac{DE}{DFG} G 于点，交 于点，连接 ． （）若 = 60，则 = ____ ： \\frac{1}{2} \\frac{AED}{DFG} （）当CE的长为 ____时， 为等腰 三角形； （）如图②当3, O与CD相切时求,CE的长． [图] [图] (1) 与 O 的位置关系, 并说明理由； (2) 若 \\bigcirc O 的半径为3, \\tan\\angle OAD=\\frac{1}{3} ,求 BC 的长．","result_json":"{\"picks\": [], \"tree_code\": \"KT_JUNIOR_MATH_STANDARD\", \"candidates\": 0, \"classifier_version\": null}","created_at":"2026-09-24T22:30:03"},{"id":414650,"import_question_id":21051,"audit_type":"VISION","auditor_name":"vision-auditor","status":"WARN","score":0.2,"input_snapshot":"题干：\n如图， 是  O 的弦,过点 作直线\\(\\frac{AB}{以为顶点作}\\)B 如图①,在矩形ABCD中，AB = 4, EF, O  = ,分\\(\\frac{AOC}{CD}\\frac{90}{CB} =\\)，点是 边上一动点，连接\\(\\frac{BC}{AE} \\frac{6}{DE} \\frac{BC}{ABE}\\)E 别交,,,若 = CD．\\(\\frac{EFAB}{试判断直线}\\frac{于点}{EF}\\)、，作 的外接，交AD\\(\\frac{O}{FG}F \\frac{DE}{DFG}\\)G 于点，交 于点，连接 ． （）若 = 60，则 = ____ ：\\(\\frac{1}{2} \\frac{AED}{DFG}\\)（）当CE的长为 ____时， 为等腰 三角形； （）如图②当3, O与CD相切时求,CE的长． （1）与 O 的位置关系, 并说明理由； （2）若\\(\\bigcirc O\\)的半径为3,\\(\\tan\\angle OAD=\\frac{1}{3}\\),求\\(BC\\)的长．\n答案：\n；60   = 90,\\(\\frac{AOC}{ADO}\\)(2)连接EF，如图①所示：  +  = 90, 四边形FGEA是 O的内接四边形，\\(\\frac{OAD}{CDB}\\) = ， 又 ADO = ，\\(\\frac{DGF}{GDF} \\frac{DAE}{ADE}\\)又 = ，DFG  DEA,  =  = CBD,\\(\\frac{ADO}{CBD} \\frac{CDB}{OAD}\\)当DEA为等腰三角形时，  +  = 90,  OA = OB, = , DFG为等腰三角形，\\(\\frac{OAD}{OBD} \\frac{OBD}{CBO}\\) 四边形AB","result_json":"{\"note\": \"视觉比对不可用\", \"error\": \"RuntimeError\", \"unavailable\": true}","created_at":"2026-09-24T22:34:04"}]}