$17 \cdot 46$ In $\triangle A B C$, $A L$ bisects $\angle A, C M$ bisects $\angle C$, and points $L$ and $M$ are on $B C$ and $A B$ respectively. If the three sides of $\triangle A B C$ are $a, b, c$, then $\frac{A M}{M B}=k \frac{C L}{L B}$, where $k$ is
(A) 1.
(B) $\frac{b c}{a^{2}}$.
(C) $\frac{a^{2}}{b c}$.
(D) $\frac{c}{b}$.
(E) $\frac{\mathcal{C}}{a}$.
(10th American High School Mathematics Examination, 1959)
［Solution］By the Angle Bisector Theorem of a triangle, we have
$$
\frac{A M}{M B}=\frac{A C}{B C}, \quad \frac{C L}{L B}=\frac{A C}{A B} \text {. }
$$

Dividing the two equations above, we get
$$
\begin{array}{c}
\frac{A M}{M B} \div \frac{C L}{L B}=\frac{A C}{B C} \div \frac{A C}{A B}=\frac{A B}{B C}=k . \\
k=\frac{c}{a} .
\end{array}
$$

Therefore, the answer is $(E)$.