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

By the angle bisector theorem, $AM/AB=AC/BC$ and $CL/LB=AC/AB$ , so by rearranging the given equation and noting $AB=c$ and $BC=a$ , $k=c/a\rightarrow\boxed{\textbf{E}}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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1959 AHSME Problems/Problem 28

Problem 28

Solution

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