2004 AMC 10B Problem 20

Problem 20 of 25HarderAlgebra Geometry

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}?

Show solution

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.

More practice

Concepts: similarity · parallel lines · ratio and proportion

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.