Question

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

$\dfrac{1}{8}$

$\dfrac{2}{9}$

$\dfrac{3}{10}$

$\dfrac{4}{11}$

$\dfrac{5}{12}$

Answer: D

Concepts:similarity parallel lines ratio and proportion

Difficulty rating: 1840

Small Hint:

Draw the line through $D$ parallel to $\overline{BE}$ meeting $\overline{AC}$ at a point $F$

Big Hint:

Two pairs of similar triangles give $\frac{CD}{BC},$ which converts to $\frac{CD}{BD}$

Solution:

Let $F$ be on $\overline{AC}$ with $DF \parallel BE,$ and write $ET = x,$ $BT = 4x.$

From $\triangle ATE \sim \triangle ADF,$ $\dfrac{DF}{x} = \dfrac{AD}{AT} = \dfrac{4}{3},$ so $DF = \dfrac{4x}{3}.$

From $\triangle BEC \sim \triangle DFC,$ $\dfrac{CD}{BC} = \dfrac{DF}{BE} = \dfrac{\frac{4x}{3}}{5x} = \dfrac{4}{15}.$

Therefore $\begin{aligned} \dfrac{CD}{BD} &= \dfrac{\frac{CD}{BC}}{1 - \frac{CD}{BC}} \\ &= \dfrac{\frac{4}{15}}{\frac{11}{15}} = \dfrac{4}{11}. \end{aligned}$

Thus, the correct answer is D.

2004 AMC 10B Problem 20

Answer: D

Difficulty rating: 1840

Small Hint:

Big Hint:

Solution:

Problem 20 in Other Years