# 1953 AHSME Problems/Problem 28

## 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 $\textbf{(D)}\ \frac{y}{c}=\frac{a}{b+c}$.

## See Also

 1953 AHSC (Problems • Answer Key • Resources) Preceded byProblem 27 Followed byProblem 29 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username
Login to AoPS