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∘

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)
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 01:06:16 | PUBLISH | INFO | 已发布 t_question.f_id=11321 qkey=4c1ce269185c83ddcaca17e1bb9812be |
| 2026-09-26 01:06:16 | QA | INFO | QA: PASS (score=0.912) |