One. (35 points) The lines $l_{1}$ and $l_{2}$ are tangent to a circle at points $A$ and $B$, respectively. On lines $l_{1}$ and $l_{2}$, take 1993 points $A_{1}, A_{2}, \cdots, A_{1993}$ and $B_{1}, B_{2}, \cdots, B_{1993}$, such that $A A_{i}=(i+1) B B_{i} (i=1, 2, \cdots, 1993)$. If the extension of $A_{i} B_{i}$ intersects the extension of $A B$ at point $M_{i} (i=1, 2, \cdots, 1993)$, find:
$$
\frac{A_{1} B_{1}}{A_{1} M_{1}} \cdot \frac{A_{2} B_{2}}{A_{2} M_{2}} \cdot \cdots \cdot \frac{A_{1993} B_{1993}}{A_{1993} M_{1993}}=?
$$
For a fixed $i$,
consider the point $A_{i}$ on $l_{1}$ and the point $B_{i}$ on $l_{2}$. At this time, $A A_{i} = (i+1) B B_{i}$ and the extension of $A_{i} B_{i}$ intersects the extension of $A B$ at point $M_{i}$ (as shown in the figure). Draw a line through point $B_{i}$ parallel to $A A_{i}$, intersecting $A B$ at point $B_{i}^{\prime}$. Clearly, $\triangle B_{i} B_{i}^{\prime} B$ is an isosceles triangle, so $B B_{i} = B_{i}^{\prime} B_{i}$. Since $B_{i}^{\prime} B_{i} \parallel A A_{i}$, we have
$$
\triangle A A_{i} M_{i} \sim \triangle B_{i}^{\prime} B_{i} M_{i}.
$$

Thus, $\frac{A A_{i}}{B^{\prime} B_{i}} = \frac{A_{i} M_{i}}{B_{i} M_{i}}$, which implies $\frac{A A_{i}}{A_{i} M_{i}} = \frac{B B_{i}}{B_{i} M_{i}}$. 
From (1), we have $\frac{A A_{i}}{A_{i} B_{i} + B_{i} M_{i}} = \frac{B B_{i}}{B_{i} M_{i}}$, solving this gives
$$
B_{i} M_{i} = \frac{B B_{i} \cdot A_{i} B_{i}}{A A_{i} - B B_{i}}.
$$

Substituting $A A_{i} = (i+1) B B_{i}$ into the above equation, we get
$$
B_{i} M_{i} = \frac{A_{1} S_{i}}{i},
$$

which means $A_{i} B_{i} = i D_{i} M_{i}$. Therefore, $\frac{A_{i} B_{i}}{A_{i} M_{i}} = \frac{A_{i} B_{i}}{A_{i} B_{i} + B_{i} M_{i}}$
$$
\begin{array}{l}
= \frac{i B_{i} M_{i}}{i B_{i} M_{i} + B_{i} M_{i}} \\
= \frac{i}{i+1}.
\end{array}
$$

Thus, $\frac{A_{1} B_{1}}{A_{1} M_{1}} \cdot \frac{A_{2} B_{2}}{A_{2} M_{2}} \cdots \cdots \frac{A_{1993} B_{1993}}{A_{1993} M_{1993}}$
$$
\begin{array}{l}
= \frac{1}{2} \cdot \frac{2}{3} \cdot \cdots \cdot \frac{1993}{1993+1} \\
= \frac{1}{1993+1} = \frac{1}{1994}.
\end{array}
$$