$10 \cdot 84$ Given that $O A$ and $O B$ are two different lines, take any point on $O A$ and draw a perpendicular to $O B$, then from this foot of the perpendicular draw a perpendicular to $O A$, and from the second foot of the perpendicular draw a perpendicular to $O B$, and so on to infinity. If the lengths of the first two perpendicular segments are $a, b(a \neq b)$, then the sum of these perpendicular segments, as the number of perpendiculars increases indefinitely, approaches a limit. This limit value is
(A) $\frac{b}{a-b}$.
(B) $\frac{a}{a-b}$.
(C) $\frac{a b}{a-b}$.
(D) $\frac{b^{2}}{a-b}$.
(E) $\frac{a^{2}}{a-b}$.
(16th American High School Mathematics Examination, 1965)
[Solution] As shown in the figure, it is easy to prove $\triangle P P_{1} P_{2} \backsim \triangle P_{1} P_{2} P_{3} \backsim \triangle P_{2} P_{3} P_{4} \backsim \cdots$, then and $\square$
$$
\begin{array}{l}
\frac{a}{b}=\frac{b}{P_{2} P_{3}}=\frac{P_{2} P_{3}}{P_{3} P_{4}}=\cdots, \\
P_{2} P_{3}=\frac{b^{2}}{a}, \quad P_{3} P_{4}=\frac{b^{3}}{a^{2}}, \cdots, \\
P_{n} P_{n+1}=\frac{b^{n}}{a^{n-1}} .
\end{array}
$$

Thus, the sum of the lengths of the segments is
$$
\begin{aligned}
S & =a+b+\frac{b^{2}}{a}+\frac{b^{3}}{a^{2}}+\cdots=a\left[1+\frac{b}{a}+\left(\frac{b}{a}\right)^{2}+\cdots\right] \\
& =a \cdot \frac{1}{1-\frac{b}{a}}=\frac{a^{2}}{a-b} .
\end{aligned}
$$

Therefore, the answer is $(E)$.