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 $AT/DT = 3$ and $BT/ET = 4,$ what is $CD/BD?$

$\dfrac{1}{8}$

$\dfrac{2}{9}$

$\dfrac{3}{10}$

$\dfrac{4}{11}$

$\dfrac{5}{12}$

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

Therefore $\dfrac{CD}{BD} = \dfrac{CD/BC}{1 - CD/BC} = \dfrac{4/15}{11/15} = \dfrac{4}{11}.$

Thus, the correct answer is D.

2004 AMC 10B Problem 20

Solution:

Problem 20 in Other Years