AMC12 2015 B

AMC12 2015 B · Q13

AMC12 2015 B · Q13. It mainly tests Angle chasing, Triangles (properties).

Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC = 70^\circ$, $\angle ADB = 40^\circ$, $AD = 4$, and $BC = 6$. What is $AC$?

四边形 $ABCD$ 内接于一个圆，$\angle BAC = 70^\circ$，$\angle ADB = 40^\circ$，$AD = 4$，$BC = 6$。$AC$ 等于多少？

(A) 3 + \sqrt{5} 3 + \sqrt{5}

(B) 6 6

(D) 8 - \sqrt{2} 8 - \sqrt{2}

(E) 7 7

Answer

Correct choice: (B)

正确答案：（B）

Solution

Because $\angle BAC$ and $\angle BDC$ intercept the same arc, $\angle BDC = 70^\circ$. Then $\angle ADC = 110^\circ$ and $\angle ABC = 180^\circ - \angle ADC = 70^\circ$. Thus $\triangle ABC$ is isosceles, and therefore $AC = BC = 6$.

因为 $\angle BAC$ 和 $\angle BDC$ 截取相同的圆弧，所以 $\angle BDC = 70^\circ$。则 $\angle ADC = 110^\circ$，$\angle ABC = 180^\circ - \angle ADC = 70^\circ$。因此 $\triangle ABC$ 是等腰三角形，故 $AC = BC = 6$。

Topics