AMC10 2019 A

AMC10 2019 A · Q13

AMC10 2019 A · Q13. It mainly tests Triangles (properties), Circle theorems.

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 $\angle BFC$?

设 $\triangle ABC$ 是等腰三角形，$BC = AC$，且 $\angle ACB = 40^\circ$。构造以 $\overline{BC}$ 为直径的圆，让 $D$ 和 $E$ 分别为该圆与边 $\overline{AC}$ 和 $\overline{AB}$ 的其他交点。让 $F$ 为四边形 $BCDE$ 对角线的交点。$\angle BFC$ 的度数是多少？

(A) 90 90

(B) 100 100

(D) 110 110

(E) 120 120

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Because $BC=AC$ and $\angle ACB=40^\circ$, it follows that $\angle BAC=\angle ABC=70^\circ$. Because $\angle BAC=\frac12(\widehat{BC}-\widehat{DE})$ and $\widehat{BC}=180^\circ$, it follows that $\widehat{DE}=40^\circ$. Then \[ \angle BFC=\frac12(\widehat{BC}+\widehat{DE})=\frac12(180^\circ+40^\circ)=110^\circ. \] Because $D$ and $E$ lie on the circle with diameter $\overline{BC}$, both $\angle BDC$ and $\angle BEC$ are right angles, so $\angle ADF$ and $\angle AEF$ are also right angles. Therefore in quadrilateral $AEFD$ \[ \angle DFE=180^\circ-\angle DAE=180^\circ-\frac12(180^\circ-\angle ACB)=110^\circ, \] and $\angle BFC$ has the same measure.

答案（D）：因为 $BC=AC$ 且 $\angle ACB=40^\circ$，所以 $\angle BAC=\angle ABC=70^\circ$。又因为 $\angle BAC=\frac12(\widehat{BC}-\widehat{DE})$ 且 $\widehat{BC}=180^\circ$，可得 $\widehat{DE}=40^\circ$。于是 \[ \angle BFC=\frac12(\widehat{BC}+\widehat{DE})=\frac12(180^\circ+40^\circ)=110^\circ. \] 因为 $D$ 和 $E$ 在以 $\overline{BC}$ 为直径的圆上，$\angle BDC$ 与 $\angle BEC$ 都是直角，所以 $\angle ADF$ 和 $\angle AEF$ 也都是直角。因此在四边形 $AEFD$ 中 \[ \angle DFE=180^\circ-\angle DAE=180^\circ-\frac12(180^\circ-\angle ACB)=110^\circ, \] 并且 $\angle BFC$ 的度数相同。

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