AHSME 1967 Problem 21

In right triangle $ABC$ the hypotenuse $\overline{AB}=5$ and leg $\overline{AC}=3$. The bisector of angle $A$ meets the opposite side in $A_1$. A second right triangle $PQR$ is then constructed with hypotenuse $\overline{PQ}=A_1B$ and leg $\overline{PR}=A_1C$. If the bisector of angle $P$ meets the opposite side in $P_1$, the length of $PP_1$ is: $\textbf{(A)}\ \frac{3\sqrt{6}}{4} \textbf{(B)}\ \frac{3\sqrt{5}}{4} \textbf{(C)}\ \frac{3\sqrt{3}}{4} \textbf{(D)}\ \frac{3\sqrt{2}}{2} \textbf{(E)}\ \frac{15\sqrt{2}}{16}$