Trapezoid ABCD has AD \parallel BC, BD = 1, \angle DBA = 23°, and \angle BDC = 46°. The ratio BC : AD is 9 : 5. What is CD?

2009 AMC 12B Problem 16 — Solution

2009 AMC 12B Problem 16

Below is the professionally curated solution for Problem 16 of the 2009 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 12B solutions, or check the answer key.

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Concepts:angle bisector theorem trapezoid parallel lines

Difficulty rating: 1800

16.

Trapezoid $ABCD$ has $AD \parallel BC,$ $BD = 1,$ $\angle DBA = 23^\circ,$ and $\angle BDC = 46^\circ.$ The ratio $BC : AD$ is $9 : 5.$ What is $CD?$

$\dfrac{7}{9}$

$\dfrac{4}{5}$

$\dfrac{13}{15}$

$\dfrac{8}{9}$

$\dfrac{14}{15}$

Solution:

Draw the line through $D$ parallel to $AB,$ meeting $BC$ at $E,$ so $ABED$ is a parallelogram with $BE = AD.$ Then $\angle BDE = \angle DBA = 23^\circ,$ and since $\angle BDC = 46^\circ,$ segment $DE$ bisects $\angle BDC.$

By the angle bisector theorem in $\triangle BDC,$ $\dfrac{EC}{BE} = \dfrac{DC}{DB},$ so $CD = DB \cdot \dfrac{BC - AD}{AD} = 1 \cdot \left(\dfrac{9}{5} - 1\right) = \dfrac{4}{5}.$

Thus, the correct answer is B.

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