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题目 #18667

UUID
c6988fb1-87d8-4953-a5b0-0a64ab43f8fc
来源
A中考/2026江苏中考期末数学试卷+答案545.pdf 第 2 页,bbox [314.0, 239.08, 554.75, 307.83]
题型 / 学科
subjective / math / JUNIOR
知识树
KT_JUNIOR_MATH_STANDARD
难度
难度 2 / 5 简单 · 单一知识点
QA 状态
存疑 / 待修复 缺少答案 公式已降级 重试 0 次,得分 -
最终题 ID
-
错误
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]

最终题干渲染(f_body 预览)

如图,四边形$ABCD中,对角线\(AC\),\(BD\)相交于点\(O\),\(AO=OC\),\(BO=OD\),且\(angle AOB=2angle OAD\). (1)求证:四边形$ABCD$是矩形; (2)若\(\angle AOB:\angle ODC=6:7\),求\(\angle ADO\)的度数.

答案 / 解析

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解析(f_ways)

(1) AO =, =,四边形ABCD是平行四\(\frac{OCBO}{AOB} \frac{OD}{DAO}\)边形,  =  +  = 2,\(\frac{ADO}{DO} \frac{OAD}{BD}\) = ADO, AO =, AC =,平行\(\frac{DAO}{四边形}\)ABCD是矩形; (2)四边形 是矩形,,BAD = 90,\(\frac{ABCD}{CDO} \frac{ABCD}{ODC}\) = , AOB :  = 6:7,\(\frac{ABO}{AOB}\) :  = 6:7,BAO : AOB : \(\frac{ABO}{ABO} \frac{ABO}{BAD} = 7\):6:7, =180 \(\frac{7}{6} = 63\),  7 + + 7 = 90,ADO = 90 − 63 = 27 .

原图与资产(1)

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审计记录

阶段审计器结果得分依据(简短)
STRUCTURE structure-rule PASS 0.900
FORMULA formula-auditor PASS 1.000
KNOWLEDGE knowledge-auditor WARN -
VISION vision-auditor PASS 0.950
SOLVER blind-solver WARN 0.400
CRITIC critic WARN 0.500
SYMBOLIC symbolic-auditor WARN 0.500

知识点(0)

无

事件

时间阶段级别消息
2026-09-24 05:40:10ERRORERROR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]
2026-09-24 05:32:37FORMULAWARN公式转换未确认,保留原文本: solution