$17 \cdot 86$ In a right $\triangle ABC$, the hypotenuse $AB=5$, the right-angle side $AC=3$. The angle bisector of $\angle A$ intersects the opposite side at $A_{1}$. Then, on the hypotenuse, $PQ=A_{1}B$, and on the right-angle side, $PR=A_{1}C$, a second right $\triangle PQR$ is constructed. If the angle bisector of $\angle P$ intersects the opposite side at $P_{1}$, then the length of $PP_{1}$ is
(A) $\frac{3 \sqrt{6}}{4}$.
(B) $\frac{3 \sqrt{5}}{4}$.
(C) $\frac{3 \sqrt{3}}{4}$.
(D) $\frac{3 \sqrt{2}}{4}$.
(E) $\frac{15 \sqrt{2}}{16}$.
(18th American High School Mathematics Examination, 1967)
[Solution] Let the side $BC$ of the right triangle $\triangle ABC$ be divided into segments $A_1B$ and $A_1C$ by the angle bisector $AA_1$ of $\angle A$. By the angle bisector property, we have
$$
\frac{A_1B}{A_1C} = \frac{A_1B}{BC - A_1B} = \frac{AB}{AC},
$$

Given $BC = \sqrt{5^2 - 3^2} = 4$, then
$$
\begin{array}{l}
\frac{A_1B}{4 - A_1B} = \frac{5}{3}, \\
A_1B = \frac{5}{2} = PQ,
\end{array}
$$

and $A_1C = \frac{3}{2} = PR$.

In the right triangle $\triangle PQR$, $RQ = \sqrt{PQ^2 - PR^2} = 2$.
Clearly, $\triangle PQR \sim \triangle ABC$ with a similarity ratio of 1:2.
Thus, $PP_1 = \frac{1}{2} AA_1 = \frac{1}{2} \sqrt{AC^2 + A_1C^2} = \frac{3\sqrt{5}}{4}$.
Therefore, the answer is $(B)$.