AMC 8 1995 Problem 13

AMC8 1995 年第 13 题

三角形★★★☆☆
In the figure, ∠A\angle A , ∠B\angle B , and ∠C\angle C are right angles. If ∠AEB=40∘\angle AEB = 40^\circ and ∠BED=∠BDE\angle BED = \angle BDE , then ∠CDE=\angle CDE =
在图中，∠A\angle A、∠B\angle B 和 ∠C\angle C 是直角。如果 ∠AEB=40∘\angle AEB = 40^\circ 且 ∠BED=∠BDE\angle BED = \angle BDE，则 ∠CDE=\angle CDE =
AMC 8 1995 Problem 13 figure
  1. A.75∘75^\circ
  2. B.80∘80^\circ
  3. C.85∘85^\circ
  4. D.90∘90^\circ
  5. E.95∘95^\circ

正确答案 · Correct Answer

E

解析 · Solution

Solution 1

Because ∠BED=∠BDE\angle BED=\angle BDE, ∠B=90∘\angle B=90^\circ, and △BED\triangle BED is a triangle, we get: ∠B+∠BED+∠BDE=180\angle B + \angle BED +\angle BDE=180 90+∠BED+∠BED=18090 +\angle BED +\angle BED=180 2∠BED=902\angle BED=90 ∠BED=∠BDE=45∘\angle BED=\angle BDE=45^\circ
So ∠AED=∠AEB+∠BED=40+45=85∘\angle AED=\angle AEB +\angle BED=40 +45=85^\circ Since ACDE is a quadrilateral, the sum of its angles is 360. Therefore: ∠A+∠C+∠CDE+∠AED=360\angle A +\angle C +\angle CDE +\angle AED=360 90+90+∠CDE+85=36090 +90 +\angle CDE +85=360 ∠CDE=95∘(E)\angle CDE=95^\circ \text{(E)}

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