AMC10 2004 B

AMC10 2004 B · Q24

AMC10 2004 B · Q24. It mainly tests Angle chasing, Circle theorems.

In $\triangle ABC$ we have $AB = 7$, $AC = 8$, and $BC = 9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects $\angle BAC$. What is the value of $AD/CD$?

在 $\triangle ABC$ 中，$AB = 7$，$AC = 8$，$BC = 9$。点 $D$ 在三角形的外接圆上，使得 $AD$ 平分 $\angle BAC$。$AD/CD$ 的值是多少？

(A) \frac{9}{8} \frac{9}{8}

(B) \frac{5}{3} \frac{5}{3}

(D) \frac{17}{7} \frac{17}{7}

(E) \frac{5}{2} \frac{5}{2}

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Suppose that $AD$ and $BC$ intersect at $E$. Since $\angle ADC$ and $\angle ABC$ cut the same arc of the circumscribed circle, the Inscribed Angle Theorem implies that \[ \angle ABC=\angle ADC. \] Also, $\angle EAB=\angle CAD$, so $\triangle ABE$ is similar to $\triangle ADC$, and \[ \frac{AD}{CD}=\frac{AB}{BE}. \] By the Angle Bisector Theorem, \[ \frac{BE}{EC}=\frac{AB}{AC}, \] so \[ BE=\frac{AB}{AC}\cdot EC=\frac{AB}{AC}(BC-BE) \quad\text{and}\quad BE=\frac{AB\cdot BC}{AB+AC}. \] Hence \[ \frac{AD}{CD}=\frac{AB}{BE}=\frac{AB+AC}{BC}=\frac{7+8}{9}=\frac{5}{3}. \]

（B）设 $AD$ 与 $BC$ 交于 $E$。由于 $\angle ADC$ 与 $\angle ABC$ 截同一外接圆弧，由圆周角定理可得 \[ \angle ABC=\angle ADC. \] 又有 $\angle EAB=\angle CAD$，因此 $\triangle ABE \sim \triangle ADC$，从而 \[ \frac{AD}{CD}=\frac{AB}{BE}. \] 由角平分线定理， \[ \frac{BE}{EC}=\frac{AB}{AC}, \] 所以 \[ BE=\frac{AB}{AC}\cdot EC=\frac{AB}{AC}(BC-BE) \quad\text{并且}\quad BE=\frac{AB\cdot BC}{AB+AC}. \] 因此 \[ \frac{AD}{CD}=\frac{AB}{BE}=\frac{AB+AC}{BC}=\frac{7+8}{9}=\frac{5}{3}. \]

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