AMC10 2004 B

AMC10 2004 B · Q20

AMC10 2004 B · Q20. It mainly tests Similarity, Ratios in geometry.

In $\triangle ABC$ points $D$ and $E$ lie on $BC$ and $AC$, respectively. If $AD$ and $BE$ intersect at $T$ so that $AT/DT = 3$ and $BT/ET = 4$, what is $CD/BD$?

在$\triangle ABC$中，点$D$和$E$分别在$BC$和$AC$上。若$AD$和$BE$相交于$T$，使得$AT/DT = 3$且$BT/ET = 4$，则$CD/BD$是多少？

(A) \frac{1}{8} \frac{1}{8}

(B) \frac{2}{9} \frac{2}{9}

(D) \frac{4}{11} \frac{4}{11}

(E) \frac{5}{12} \frac{5}{12}

Answer

Correct choice: (D)

正确答案：（D）

Solution

20. (D) Let $F$ be a point on $\overline{AC}$ such that $\overline{DF}$ is parallel to $\overline{BE}$. Let $BT=4x$ and $ET=x$. Because $\triangle ATE$ and $\triangle ADF$ are similar, we have $$\frac{DF}{x}=\frac{AD}{AT}=\frac{4}{3}, \quad \text{and} \quad DF=\frac{4x}{3}.$$ Also, $\triangle BEC$ and $\triangle DFC$ are similar, so $$\frac{CD}{BC}=\frac{DF}{BE}=\frac{4x/3}{5x}=\frac{4}{15}.$$ Thus $$\frac{CD}{BD}=\frac{CD/BC}{1-(CD/BC)}=\frac{4/15}{1-4/15}=\frac{4}{11}.$$

20.（D）设点 $F$ 在 $\overline{AC}$ 上，且 $\overline{DF}\parallel\overline{BE}$。设 $BT=4x$，$ET=x$。因为 $\triangle ATE$ 与 $\triangle ADF$ 相似，所以 $$\frac{DF}{x}=\frac{AD}{AT}=\frac{4}{3}, \quad \text{因此} \quad DF=\frac{4x}{3}。$$ 另外，$\triangle BEC$ 与 $\triangle DFC$ 相似，所以 $$\frac{CD}{BC}=\frac{DF}{BE}=\frac{4x/3}{5x}=\frac{4}{15}。$$ 因此 $$\frac{CD}{BD}=\frac{CD/BC}{1-(CD/BC)}=\frac{4/15}{1-4/15}=\frac{4}{11}。$$

Topics