AMC10 2014 A

AMC10 2014 A · Q22

AMC10 2014 A · Q22. It mainly tests Angle chasing, Triangles (properties).

In rectangle $ABCD$, $AB = 20$ and $BC = 10$. Let $E$ be a point on $CD$ such that $\angle CBE = 15^\circ$. What is $AE$?

在矩形 $ABCD$ 中，$AB = 20$，$BC = 10$。设 $E$ 为 $CD$ 上的点，使得 $\angle CBE = 15^\circ$。$AE$ 等于多少？

(A) $\frac{20\sqrt{3}}{3}$ $\frac{20\sqrt{3}}{3}$

(B) $10\sqrt{3}$ $10\sqrt{3}$

(D) $11\sqrt{3}$ $11\sqrt{3}$

(E) 20 20

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $E'$ be the point on $\overline{CD}$ such that $AE' = AB = 2AD$. Then $\triangle ADE'$ is a $30-60-90^\circ$ triangle, so $\angle DAE' = 60^\circ$. Hence $\angle BAE' = 30^\circ$. Also, $AE' = AB$ implies that $\angle E'BA = \angle BE'A = 75^\circ$, and then $\angle CBE' = 15^\circ$. Thus it follows that $E'$ and $E$ are the same point. Therefore, $AE = AE' = AB = 20$.

答案（E）：令 $E'$ 为 $\overline{CD}$ 上的一点，使得 $AE' = AB = 2AD$。则 $\triangle ADE'$ 是一个 $30-60-90^\circ$ 三角形，所以 $\angle DAE' = 60^\circ$。因此 $\angle BAE' = 30^\circ$。另外，由 $AE' = AB$ 可得 $\angle E'BA = \angle BE'A = 75^\circ$，进而 $\angle CBE' = 15^\circ$。因此可知 $E'$ 与 $E$ 是同一点。所以 $AE = AE' = AB = 20$。

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