In rectangle $ABCD,$ $AB=12$ and $BC=8$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $BE=EF=FC$. Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$, respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s$ and $t$ is 1. What is $r+s+t$?

1. Let $A=(0,0), B=(12,0), C=(12,8), D=(0,8)$. Since $BE=EF=FC=\frac83$, we have $E=(12,\frac83)$ and $F=(12,\frac{16}3)$.
2. A cevian from $A$ to $(12,h)$ meets $BD$ at a point that is a fraction $\dfrac{h}{h+8}$ of the way from $B$ to $D$. Thus $BP/BD=\frac14$ and $BQ/BD=\frac25$.
3. Therefore $BP:PQ:QD=\frac14:(\frac25-\frac14):(1-\frac25)=5:3:12$.
4. The sum is $r+s+t=5+3+12=\boxed{20}$.