AMC10 2010 A

AMC10 2010 A · Q14

AMC10 2010 A · Q14. It mainly tests Angle chasing, Triangles (properties).

Triangle ABC has AB = 2 · AC. Let D and E be on AB and BC, respectively, such that ∠BAE = ∠ACD. Let F be the intersection of segments AE and CD, and suppose that △CFE is equilateral. What is ∠ACB?

三角形 ABC 有 AB = 2 · AC。D 和 E 分别在 AB 和 BC 上，使得 ∠BAE = ∠ACD。F 是线段 AE 和 CD 的交点，且假设 △CFE 是等边三角形。∠ACB 是多少度？

(A) 60° 60°

(B) 75° 75°

(D) 105° 105°

(E) 120° 120°

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $\alpha=\angle BAE=\angle ACD=\angle ACF$. Because $\triangle CFE$ is equilateral, it follows that $\angle CFA=120^\circ$ and then $$ \angle FAC=180^\circ-120^\circ-\angle ACF=60^\circ-\alpha. $$ Therefore $$ \angle BAC=\angle BAE+\angle FAC=\alpha+(60^\circ-\alpha)=60^\circ. $$ Because $AB=2\cdot AC$, it follows that $\triangle BAC$ is a $30-60-90^\circ$ triangle, and thus $\angle ACB=90^\circ$.

答案（C）：设 $\alpha=\angle BAE=\angle ACD=\angle ACF$。因为 $\triangle CFE$ 是等边三角形，所以 $\angle CFA=120^\circ$，因此 $$ \angle FAC=180^\circ-120^\circ-\angle ACF=60^\circ-\alpha。 $$ 因此 $$ \angle BAC=\angle BAE+\angle FAC=\alpha+(60^\circ-\alpha)=60^\circ。 $$ 因为 $AB=2\cdot AC$，可知 $\triangle BAC$ 是一个 $30-60-90^\circ$ 三角形，从而 $\angle ACB=90^\circ$。

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