AMC10 2022 B

AMC10 2022 B · Q20

AMC10 2022 B · Q20. It mainly tests Angle chasing, Triangles (properties).

Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$. Let $E$ be the midpoint of $\overline{CD}$, and let $F$ be the point on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$. What is the degree measure of $\angle BFC$?

设 $ABCD$ 是一个菱形，$\angle ADC = 46^\circ$。令 $E$ 为 $\overline{CD}$ 的中点，$F$ 为 $\overline{BE}$ 上的点，使得 $\overline{AF}$ 与 $\overline{BE}$ 垂直。求 $\angle BFC$ 的度数。

(A) 110 110

(B) 111 111

(D) 113 113

(E) 114 114

Answer

Correct choice: (D)

正确答案：（D）

Solution

Extend segments $\overline{AD}$ and $\overline{BE}$ until they meet at point $G$. Because $\overline{AB} \parallel \overline{ED}$, we have $\angle ABG = \angle DEG$ and $\angle GDE = \angle GAB$, so $\triangle ABG \sim \triangle DEG$ by AA. Because $ABCD$ is a rhombus, $AB = CD = 2DE$, so $AG = 2GD$, meaning that $D$ is a midpoint of segment $\overline{AG}$. Now, $\overline{AF} \perp \overline{BE}$, so $\triangle GFA$ is right and median $FD = AD$. So now, because $ABCD$ is a rhombus, $FD = AD = CD$. This means that there exists a circle from $D$ with radius $AD$ that passes through $F$, $A$, and $C$. AG is a diameter of this circle because $\angle AFG=90^\circ$. This means that $\angle GFC = \angle GAC = \frac{1}{2} \angle GDC$, so $\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ$, which means that $\angle BFC = \boxed{\textbf{(D)} \ 113}$

延长线段 $\overline{AD}$ 和 $\overline{BE}$ 直到相交于点 $G$。因为 $\overline{AB} \parallel \overline{ED}$，我们有 $\angle ABG = \angle DEG$ 且 $\angle GDE = \angle GAB$，所以由 AA 相似，$\triangle ABG \sim \triangle DEG$。因为 $ABCD$ 是菱形，$AB = CD = 2DE$，所以 $AG = 2GD$，意味着 $D$ 是线段 $\overline{AG}$ 的中点。现在，$\overline{AF} \perp \overline{BE}$，所以 $\triangle GFA$ 是直角且中位线 $FD = AD$。所以现在，因为 $ABCD$ 是菱形，$FD = AD = CD$。这意味着存在以 $D$ 为圆心、半径 $AD$ 的圆经过 $F$、$A$ 和 $C$。 $AG$ 是此圆的直径因为 $\angle AFG=90^\circ$。这意味着 $\angle GFC = \angle GAC = \frac{1}{2} \angle GDC$，所以 $\angle GFC = \frac{1}{2}(180^\circ - 46^\circ)=67^\circ$，从而 $\angle BFC = \boxed{\textbf{(D)} \ 113}$

Topics