# 1953 AHSME Problems/Problem 28

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## Problem 28

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:

$\textbf{(A)}\ \frac {x}{a} = \frac {a}{b + c} \qquad \textbf{(B)}\ \frac {x}{b} = \frac {a}{a + c} \qquad \textbf{(C)}\ \frac{y}{c}=\frac{c}{b+c}\\ \textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}\qquad \textbf{(E)}\ \frac{x}{y}=\frac{c}{b}$

## Solution

$[asy]pair A,B,C,D; A = (5,6); B = (9,0); C = (0,0); D = (B+C)/2; draw(D--A--B--C--A); label("A",A,N); label("B",B,SW); label("C",C,SE); label("D",D,S); label("b",(A+C)/2,NW); label("c",(B+A)/2,NE); label("x",(C+D)/2,S); label("y",(D+B)/2,S);[/asy]$ By the 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 $\textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}$.