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题目 #18669

UUID
a43662d5-cb04-4790-8486-a61a82fa8f0f
来源
A中考/2026江苏中考期末数学试卷+答案545.pdf 第 3 页,bbox [31.56, 337.24, 395.39, 412.15]
题型 / 学科
comprehensive / 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 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)

【小试2】 如图, 在▱ABCD中, 对角线AC, BD相交于点O, AE ⊥BD于点E,且DF ⊥ AC于点F, 且AE = DF.
(1)求证:四边形ABCD是矩形.
(2)若∠BAE: ∠EAD = 4: 5,求∠EAO的度数.

答案 / 解析

答案(f_answer)

(1)见解析;(2)10∘.

解析(f_ways)

(1)因为四边形 ABCD 是平行四边形, ∴OA = OC = 1\(1 _{2}\)AC, OB = OD = _{2}BD, ∵AE ⊥BD于点E, DF ⊥ AC于点F,∴∠AEO = ∠DFO = 90∘, 又∵∠AOE = ∠DOF, AE = DF,所以\(\Delta{}AEO\)≅\(\Delta{}DFO\)∴OA = OD, ∴ AC = BD, ∴四边形ABCD是矩形; (2)由(1)得:四边形ABCD是矩形,所以∠ABC = ∠BAD = 90∘, OA = OB, ∴∠OAB = ∠OBA, ∵ ∠BAE: ∠EAD等于4: 5, ∴∠BAE = 40∘, ∴∠OBA = ∠OAB = 90∘−40∘= 50∘, ∴∠EAO = ∠OAB −∠BAE等于 50∘−40∘= 10∘.

原图与资产(1)

预览类型页source bboxhashURL本地路径
embedded_image3 [31.5, 412.95, 173.2, 490.1] ac52e0cfb3325fb5…
打开
/www/wwwroot/tikupdfpng.mtwlkj.net_5129/qbank/math/2026/09/16de2e24-a616-456b-83c5-3364ab6bd928/a43662d5-cb04-4790-8486-a61a82fa8f0f/ac52e0cfb332.png

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 1.000

知识点(0)

无

事件

时间阶段级别消息
2026-09-24 05:42:07ERRORERROR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 05:38:07FORMULAWARN公式转换未确认,保留原文本: solution
2026-09-24 05:34:07FORMULAWARN公式转换未确认,保留原文本: sub_2
2026-09-24 05:32:44FORMULAWARN公式转换未确认,保留原文本: stem