In rectangle $ABCD$, we have $AB=10$, $BC=14$, $H$ is on $BC$ with $BH=8$, $E$ is on $AD$ with $DE=6$, line $EC$ intersects line $AH$ at $G$, and $F$ is on line $AD$ with $GF \perp AF$. Find the length of $GF$.

Let $FG=x$ and $DF=y$, where $F$ lies on line $AD$ beyond $D$.
Since $CD\parallel GF$, $\triangle EDC\sim \triangle EFG$: $6:(6+y)=DC:FG=10:x$, so $x=\frac{10(6+y)}6=\frac{5(6+y)}3$.
Since $\angle AHB=\angle GAF$ ($AD\parallel BC$) and both triangles are right-angled, $\triangle HBA\sim\triangle AFG$: $AB:FG=BH:AF$, $10:x=8:(14+y)$, so $x=\frac{10(14+y)}8=\frac{5(14+y)}4$.
Setting equal: $\frac{5(6+y)}3=\frac{5(14+y)}4 \Rightarrow 4(6+y)=3(14+y)\Rightarrow 24+4y=42+3y\Rightarrow y=18$.
Then $x=\frac{5(6+18)}3=\frac{5(24)}3=\boxed{40}$.