Question

设 $\triangle ABC$ 是等腰三角形， $BC = AC$ ，且 $\angle ACB = 40^{\circ}$ 。以 $\overline{BC}$ 为直径作圆，令 $D$ 和 $E$ 分别为该圆与边 $\overline{AC}$ 和 $\overline{AB}$ 的另一个交点。令 $F$ 为四边形 $BCDE$ 的两条对角线交点。求 $\angle BFC$ 的度数。

Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}.$ Construct the circle with diameter $\overline{BC},$ and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB},$ respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE.$ What is the degree measure of $\angle BFC ?$

$90$

$100$

$105$

$110$

$120$

答案：D

知识点：圆周角导角等腰三角形

难度评级：1420

解答：

因为 $\overline{BC}$ 是圆的直径， $\angle BDC$ 和 $\angle BEC$ 都是直角。

可知 $\angle ABC = 70^{\circ}$ ，这是因为 $\triangle ABC$ 是等腰三角形。

利用三角形内角和为 $\triangle BCE$ 得以及 $\triangle BCD$ $\begin{aligned} \angle ECB&=180^{\circ}-70^{\circ}-90^{\circ}\\ &=20^{\circ} \end{aligned}$ $\begin{aligned} \angle DBC&=180^{\circ}-40^{\circ}-90^{\circ}\\ &=50^{\circ}. \end{aligned}$

在 $\triangle BFC$ 中，所以正确答案是 D。 $F$ $BD$ $CE$ $50^{\circ}$ $20^{\circ}$ $\begin{aligned} \angle BFC&=180^{\circ}-50^{\circ}-20^{\circ}\\ &=110^{\circ}. \end{aligned}$

Since $\overline{BC}$ is the diameter of the circle, we get that $\angle BDC$ and $\angle BEC$ are right angles.

We know that $\angle ABC = 70^{\circ}$ from the fact that $\triangle ABC$ is isosceles.

In $\triangle BCE$ and $\triangle BCD$ , respectively, $\begin{aligned} \angle ECB&=180^{\circ}-70^{\circ}-90^{\circ}\\ &=20^{\circ} \end{aligned}$ and $\begin{aligned} \angle DBC&=180^{\circ}-40^{\circ}-90^{\circ}\\ &=50^{\circ}. \end{aligned}$

Because $F$ lies on $BD$ and $CE$ , the other two angles of $\triangle BFC$ are $50^{\circ}$ and $20^{\circ}$ . Hence $\begin{aligned} \angle BFC&=180^{\circ}-50^{\circ}-20^{\circ}\\ &=110^{\circ}. \end{aligned}$ Thus, D is the correct answer.

2019 AMC 10A 第 13 题

答案：D

难度评级：1420

解答：

其他年份的第 13 题