AMC 8 1996 Problem 24

AMC8 1996 年第 24 题

三角形★★★☆☆
The measure of angle ABCABC is 50∘50^\circ , AD‾\overline{AD} bisects angle BACBAC , and DC‾\overline{DC} bisects angle BCABCA . The measure of angle ADCADC is
∠ABC\angle ABC 的度数为 50∘50^\circ，AD‾\overline{AD} 平分 ∠BAC\angle BAC，DC‾\overline{DC} 平分 ∠BCA\angle BCA。∠ADC\angle ADC 的度数是
AMC 8 1996 Problem 24 figure
  1. A.90∘90^\circ
  2. B.100∘100^\circ
  3. C.115∘115^\circ
  4. D.122.5∘122.5^\circ
  5. E.125∘125^\circ

正确答案 · Correct Answer

C

解析 · Solution

Solution 1

Let ∠CAD=∠BAD=x\angle CAD = \angle BAD = x, and let ∠ACD=∠BCD=y\angle ACD = \angle BCD = y
From △ABC\triangle ABC, we know that 50 + 2x + 2y = 180, leading to x + y = 65.
From △ADC\triangle ADC, we know that x+y+∠D=180x + y + \angle D = 180. Plugging in x + y = 65, we get ∠D=180−65=115\angle D = 180 - 65 = 115, which is answer CC.

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