19. As shown in Figure 4, in rectangle $A B C D$, $A B=6, B C$ $=3$, point $E$ is on side $B C$, point $F$ is on side $C E$, $B E=$ $E F=F C$. Connecting $A E$ and $A F$, they intersect $B D$ at points $P$ and $Q$ respectively. If $B P: P Q: Q D=r: s: t\left(r \backslash s, t \in \mathbf{Z}_{+},(r, s\right.$, $t)=1)$, then $r+s+t=(\quad)$.
(A) 7
(B) 9
(C) 12
(D) 15
(E) 20
19. E.
$$
\begin{array}{l}
\text { Given } A D / / B C \Rightarrow \triangle A P D \backsim \triangle E P B \\
\Rightarrow \frac{r}{s+t}=\frac{B E}{A D}=\frac{1}{3} .
\end{array}
$$

Similarly, $\frac{r+s}{t}=\frac{B F}{A D}=\frac{2}{3}$.
$$
\begin{array}{l}
\text { Then }\left\{\begin{array}{l}
s+t=3 r, \\
3 r+3 s=2 t
\end{array} \Rightarrow t=4 s \Rightarrow 5 s=3 r\right. \\
\Rightarrow r: s: t=5: 3: 12 \\
\Rightarrow r+s+t=5+3+12=20 .
\end{array}
$$