2009 AMC 12B Problems/Problem 16

Problem

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

$\mathrm{(A)}\ \frac 79\qquad \mathrm{(B)}\ \frac 45\qquad \mathrm{(C)}\ \frac {13}{15}\qquad \mathrm{(D)}\ \frac 89\qquad \mathrm{(E)}\ \frac {14}{15}$

Solution

Solution 1

Extend $\overline AB$ and $\overline DC$ to meet at $E$ . Then

$\begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*}$

Thus $\triangle BDE$ is isosceles with $DE = BD$ . Because $\overline {AD} \parallel \overline {BC}$ , it follows that the triangles $BCD$ and $ADE$ are similar. Therefore $\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,$ so $CD = \boxed{\frac 45}.$

Solution 2

Let $E$ be the intersection of $\overline {BC}$ and the line parallel to $\overline {AB}.$ By constuction $BE = AD$ and $\angle BDE = 23^{\circ}$ ; it follows that $DE$ is the bisector of the angle $BDC$ . So by the Angle Bisector Theorem we get $CD = \frac {CD}{BD} = \frac {EC}{BE} = \frac {BC - BE}{BE} = \frac {BC}{AD} -1 = \frac 95 - 1 = \boxed{\frac 45}.$ The answer is $\mathrm{(B)}$ .

2009 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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2009 AMC 12B Problems/Problem 16

Contents

Problem

Solution

Solution 1

Solution 2

See also