2004 AMC 10B Problem 20

Attempt Problem 20 of the 2004 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

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.