Difference between revisions of "2009 AMC 12B Problems/Problem 16"

Revision as of 20:07, 16 February 2024

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 $CD$ ?

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

Solutions

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 $BCE$ 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

2009 AMC 12B (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 24: / Line 24: @@
 === Solution 2 ===
-Let <math>E</math> be the intersection of <math>\overline {BC}</math> and the line through <math>D</math> parallel to <math>\overline {AB}.</math>  By construction <math>BE = AD</math> and <math>\angle BDE = 23^{\circ}</math>; it follows that <math>DE</math> is the bisector of the angle <math>BDC</math>. So by the [[Angle Bisector Theorem]] we get
-<cmath>CD = \frac {CD}{BD} = \frac {EC}{BE} = \frac {BC - BE}{BE} = \frac {BC}{AD} -1 = \frac 95 - 1 = \boxed{\frac 45}.</cmath>
-The answer is <math>\mathrm{(B)}</math>.
 == See also ==
 {{AMC12 box|year=2009|ab=B|num-b=15|num-a=17}}
 {{MAA Notice}}

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