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2009 AMC 12B Problems/Problem 16

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Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

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 $CD$ ?

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

Solution

Solution 1

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

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

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 through $D$ 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 $\mathrm{(B)}$ .

See also

2009 AMC 12B (Problems • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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