Question

如下图所示，点 $E$ 位于直线 $CD$ 所确定的、与点 $A$ 相反的半平面内，且 $\angle CDE = 110^\circ$ 。点 $F$ 在 $\overline{AD}$ 上，满足 $DE=DF$ ，并且 $ABCD$ 是正方形。 $\angle AFE$ 的度数是多少？

As shown in the figure below, point $E$ lies in the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ.$ Point $F$ lies on $\overline{AD}$ so that $DE=DF,$ and $ABCD$ is a square. What is the degree measure of $\angle AFE?$

$160$

$164$

$166$

$170$

$174$

答案：D

知识点：导角等腰三角形正方形（几何）

难度评级：960

解答：

因为 $\angle ADC = 90^{\circ}$ 所以又因为 $\triangle FDE$ 是等腰三角形，所以最后 $\angle FDE = 360^{\circ} - 90^{\circ} - 110^{\circ}$ $= 160^{\circ}.$ $\angle EFD = \dfrac{180^{\circ} - 160^{\circ}}{2} = 10^{\circ}.$ $\angle AFE = 180^{\circ} - 10^{\circ} = 170^{\circ}.$

所以正确答案是 D。

Since $\angle ADC = 90^{\circ},$ we get that $\angle FDE = 360^{\circ} - 90^{\circ} - 110^{\circ}$ $= 160^{\circ}.$ Also since $\triangle FDE$ is isosceles, we get that $\angle EFD = \dfrac{180^{\circ} - 160^{\circ}}{2} = 10^{\circ}.$ Finally, we get that $\angle AFE = 180^{\circ} - 10^{\circ} = 170^{\circ}.$

Thus, D is the correct answer.

2021 AMC 10A Fall 第 7 题

答案：D

难度评级：960

解答：

其他年份的第 7 题