1959 AHSME 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}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.