In $\triangle ABC$, sides $a,b$ and $c$ are opposite $\angle{A},\angle{B}$ and $\angle{C}$ respectively. $AD$ bisects $\angle{A}$ and meets $BC$ at $D$. 
Then if $x = \overline{CD}$ and $y = \overline{BD}$ the correct proportion is: 
$\textbf{(A)}\ \frac {x}{a} = \frac {a}{b + c} \qquad \textbf{(B)}\ \frac {x}{b} = \frac {a}{a + c} \qquad \textbf{(C)}\ \frac{y}{c}=\frac{c}{b+c}\\ \textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}\qquad \textbf{(E)}\ \frac{x}{y}=\frac{c}{b}$
By the [Angle Bisector Theorem](https://artofproblemsolving.com/wiki/index.php/Angle_Bisector_Theorem), $\frac{x}{b}=\frac{y}{c}$. Because of this, $\frac{x}{b}$ must equal $\frac{ky}{kc}$, where $k$ is some real number. Therefore, $\frac{x+y}{b+c}=\frac{y+ky}{c+kc}$. Factoring $\frac{k+1}{k+1}$ out, we get $\frac{y}{c}=\frac{x+y}{b+c}$. However, $x+y=a$, so the answer is $\textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}$.