In rectangle $ABCD$, we have $AB=8$, $BC=9$, $H$ is on $BC$ with $BH=6$, $E$ is on $AD$ with $DE=4$, line $EC$ intersects line $AH$ at $G$, and $F$ is on line $AD$ with $GF \perp AF$. Find the length of $GF$. $\mathrm{(A) \ } 16\qquad$ $ \mathrm{(B) \ } 20\qquad$ $ \mathrm{(C) \ } 24\qquad$ $ \mathrm{(D) \ } 28\qquad$ $ \mathrm{(E) \ } 30$
Solution
Let $FG=x$ and $DF=y$, where $F$ lies on line $AD$ beyond $D$.
$CD\parallel GF$ (both perpendicular to $AD$), so $\triangle EDC\sim \triangle EFG$ and $ED:EF=CD:FG$, i.e. $4:(4+y)=8:x$.
$AD\parallel BC$ gives $\angle GAF=\angle AHB$, and $\angle B=\angle F=90^\circ$, so $\triangle HBA\sim\triangle AFG$ and $AB:GF=BH:FA$, i.e. $8:x=6:(9+y)$.
Comparing, $4:(4+y)=6:(9+y)$, so $36+4y=24+6y$, $y=6$, and $x=\frac{8(4+6)}4=20$.
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 $GF$.
$\mathrm{(A) \ } 16\qquad$ $ \mathrm{(B) \ } 20\qquad$ $ \mathrm{(C) \ } 24\qquad$ $ \mathrm{(D) \ } 28\qquad$ $ \mathrm{(E) \ } 30$ 
