Rectangle $ABCD$ has $AB=6$ and $BC=5$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$, and point $F$ lies on $\overline{CD}$ so that $DF=1$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\frac{PQ}{EF}$?

$AE=AB-EB=5$, $FC=CD-DF=5$. By $\triangle APE\sim\triangle CPF$, $PE:PF=5:5=1:1$, so $PE:EF=1:2$.

Extend $AG$ to meet $DC$ at $H$; by similar triangles, $CH:AB=CG:GB=1:4$, so $CH=\frac64=\frac32$.
$DH=DC+CH=6+\frac32=\frac{15}2$, so $FH=DH-DF=\frac{15}2-1=\frac{13}2$.
By $\triangle AQE\sim \triangle HQF$: $EQ:QF=AE:HF=5:\frac{13}2=10:13$, so $EQ:EF=10:23$.
$PQ:EF=\frac12-\frac{10}{23}=\frac{23-20}{46}=\boxed{\frac3{46}}$.