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题目 #24475

UUID
a766bf27-b9a1-4f2e-852b-84e3e5449825
来源
人教版/九上/2026江苏九年级期中数学试卷+答案1210.pdf 第 3-4 页,bbox [31.56, 619.77, 398.49, 203.2]
题型 / 学科
single_choice / math / SENIOR
知识树
KT_SENIOR_MATH_STANDARD
难度
难度 2 / 5 简单 · 单一知识点
QA 状态
PUBLISHED 重试 0 次,得分 0.912
最终题 ID
11321

最终题干渲染(f_body 预览)

[小试2] 如图,在⊙O内接四边形ABCD中,∠BCD = 60°, AB ∥ CD,点E为CD上任意一点,则∠AEC的度数为( ).
A. 30∘
B. 60∘
C. 90∘
D. 120∘

答案 / 解析

答案(f_answer)

B

解析(f_ways)

因为AB\(\parallel\)CD, \therefore\(\angle\)BCD +\(\angle\)ABC = 180^{\(\circ\)}, \because\(\angle\)BCD = 60^{\(\circ\)}, \therefore\(\angle\)ABC = 180^{\(\circ\)} - 60^{\(\circ\)} = 120^{\(\circ\)}, \because 四边形ABCE为\odot O内接四边形, \therefore\(\angle\)AEC +\(\angle\)ABC = 180^{\(\circ\)}, 所以\(\angle\)AEC = 60^{\(\circ\)},故选:B.

原图与资产(1)

预览类型页source bboxhashURL本地路径
embedded_image4 [30.0, 221.45, 71.19, 257.4] 0eafc29bb8e9b69b…
打开
/www/wwwroot/tikupdfpng.mtwlkj.net_5129/qbank/math/2026/09/2b06312b-7e3f-494e-ae3e-21d10207fcf7/74c99a02-5fc3-4b2c-9302-c2ef5eabb364/0eafc29bb8e9.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 0.950
SOLVER blind-solver PASS 0.800
CRITIC critic PASS 0.900
SYMBOLIC symbolic-auditor PASS 1.000
COHERENCE coherence-auditor PASS 0.900
QUALITY_GATE quality-gate PASS 0.912

知识点(0)

无

正式题库记录(t_question.f_id=11321)

f_qkey
4c1ce269185c83ddcaca17e1bb9812be
f_style / 难度
1 / 2
入库题干
[小试2] 如图,在⊙O内接四边形ABCD中,∠BCD = 60°, AB ∥ CD,点E为CD上任意一点,则∠AEC的度数为( ).
A. 30∘
B. 60∘
C. 90∘
D. 120∘
入库答案
B
入库解析
因为AB\(\parallel\)CD, \therefore\(\angle\)BCD +\(\angle\)ABC = 180^{\(\circ\)}, \because\(\angle\)BCD = 60^{\(\circ\)}, \therefore\(\angle\)ABC = 180^{\(\circ\)} - 60^{\(\circ\)} = 120^{\(\circ\)}, \because 四边形ABCE为\odot O内接四边形, \therefore\(\angle\)AEC +\(\angle\)ABC = 180^{\(\circ\)}, 所以\(\angle\)AEC = 60^{\(\circ\)},故选:B.

事件

时间阶段级别消息
2026-09-26 01:06:16PUBLISHINFO已发布 t_question.f_id=11321 qkey=4c1ce269185c83ddcaca17e1bb9812be
2026-09-26 01:06:16QAINFOQA: PASS (score=0.912)