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

### Solution 2

Let $E$ be the intersection of $\overline {BC}$ and the line through $D$ parallel to $\overline {AB}.$ By construction $BE = AD$ 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 $\mathrm{(B)}$.

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