# 1959 AHSME Problems/Problem 28

## Problem 28

In triangle $ABC$, $AL$ bisects angle $A$, and $CM$ bisects angle $C$. Points $L$ and $M$ are on $BC$ and $AB$, respectively. The sides of $\triangle ABC$ are $a$, $b$, and $c$. Then $\frac{AM}{MB}=k\frac{CL}{LB}$ where $k$ is: $\textbf{(A)}\ 1 \qquad\textbf{(B)}\ \frac{bc}{a^2}\qquad\textbf{(C)}\ \frac{a^2}{bc}\qquad\textbf{(D)}\ \frac{c}{b}\qquad\textbf{(E)}\ \frac{c}{a}$

## Solution

$[asy] import geometry; point A = (0,0); point B = (3,4); point C = (8,0); point L,M; // Triangle ABC draw(A--B--C--A); dot(A); label("A",A,SW); dot(B); label("B",B,NW); dot(C); label("C",C,SE); // Angle Bisectors pair[] l = intersectionpoints(bisector(line(A,B),line(A,C)),B--C); L = l[0]; dot(L); label("L",L,NE); draw(A--L); markscalefactor = 0.15; draw(anglemark(L,A,B)); markscalefactor = 0.17; draw(anglemark(C,A,L)); markscalefactor = 0.18; draw(anglemark(L,A,B)); markscalefactor = 0.2; draw(anglemark(C,A,L)); pair[] m = intersectionpoints(bisector(line(B,C), line(A,C)), A--B); M = m[0]; dot(M); label("M",M,NW); draw(C--M); markscalefactor = 0.15; draw(anglemark(M,C,A)); markscalefactor = 0.17; draw(anglemark(B,C,M)); // Length Labels label("a", midpoint(B--C), (5,5)); label("b", midpoint(A--C), S); label("c", midpoint(A--B), (-5,5)); [/asy]$

By the Angle Bisector Theorem, $\frac{AM}{AB}=\frac{AC}{BC}$ and $\frac{CL}{LB}=\frac{AC}{AB}$, so by rearranging the given equation and noting $AB=c$ and $BC=a$, $k=\frac{c}{a}\rightarrow\boxed{\textbf{(E)}}$.