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 $CD$ ?

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

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 $CD = \boxed{\frac 45}.$

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.

Page

Toolbox

Search

2009 AMC 12B Problems/Problem 16

Contents

Problem

Solutions

Solution 1

See also