AMC10 2019 B

AMC10 2019 B · Q16

AMC10 2019 B · Q16. It mainly tests Angle chasing, Triangles (properties).

In $\triangle ABC$ with a right angle at $C$, point $D$ lies in the interior of $\overline{AB}$ and point $E$ lies in the interior of $\overline{BC}$ so that $AC = CD$, $DE = EB$, and the ratio $AC:DE = 4:3$. What is the ratio $AD:DB$?

在 $\triangle ABC$ 中，$角 C$ 为直角，点 $D$ 位于 $\overline{AB}$ 的内部，点 $E$ 位于 $\overline{BC}$ 的内部，使得 $AC = CD$，$DE = EB$，且比例 $AC:DE = 4:3$。$AD:DB$ 的比值为多少？

(A) $2:3$ $2:3$

(B) $2:\sqrt{5}$ $2:\sqrt{5}$

(D) $3:\sqrt{5}$ $3:\sqrt{5}$

(E) $3:2$ $3:2$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let $\overline{AC}=\overline{DC}=4x$ and $\overline{DE}=\overline{BE}=3x$. Because $\angle A\cong\angle ADC$, $\angle B\cong\angle EDB$, and $\angle A$ and $\angle B$ are complementary, it follows that $\angle CDE$ is a right angle. Thus $\overline{CE}=5x$. Let $F$ and $G$ lie on $\overline{AB}$ so that $\overline{CF}$ and $\overline{EG}$ are perpendicular to $\overline{AB}$. Then it follows that $$ \frac{3}{8} = \frac{\overline{BE}}{\overline{BC}} = \frac{\overline{BG}}{\overline{BF}} = \frac{\frac{1}{2}\overline{BD}}{\overline{BD}+\frac{1}{2}\overline{AD}} = \frac{\overline{BD}}{2\overline{BD}+\overline{AD}}, $$ so $8\overline{BD}=6\overline{BD}+3\overline{AD}$. It follows that $AD:DB=2:3$.

答案（A）：设 $\overline{AC}=\overline{DC}=4x$，且 $\overline{DE}=\overline{BE}=3x$。因为 $\angle A\cong\angle ADC$，$\angle B\cong\angle EDB$，并且 $\angle A$ 与 $\angle B$ 互为余角，所以可得 $\angle CDE$ 为直角。因此 $\overline{CE}=5x$。令 $F$ 与 $G$ 在 $\overline{AB}$ 上，使得 $\overline{CF}$ 和 $\overline{EG}$ 均垂直于 $\overline{AB}$。则有 $$ \frac{3}{8} = \frac{\overline{BE}}{\overline{BC}} = \frac{\overline{BG}}{\overline{BF}} = \frac{\frac{1}{2}\overline{BD}}{\overline{BD}+\frac{1}{2}\overline{AD}} = \frac{\overline{BD}}{2\overline{BD}+\overline{AD}}, $$ 所以 $8\overline{BD}=6\overline{BD}+3\overline{AD}$。从而 $AD:DB=2:3$。

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