As shown in the figure, the side length of square $A B C D$ is 4, and the side length of square $C E F G$ is 12. Points $D$, $C$, and $E$ are collinear. Connect $D F$, and draw $G I / / D F$ intersecting the extension of $D A$ at point $I$. Draw $I H \perp D F$ intersecting $D F$ at point $H$. Then the length of $I H$ is $\qquad$.
Geometry, Pythagorean Theorem, Similar Triangles.
(Method 1)
Draw $G J \perp D F$, with the foot of the perpendicular at $J$.
In Rt $\triangle D E F$, $\angle E=90^{\circ}, E F=12, D E=4+12=16$; by the Pythagorean Theorem, we get $D F=20$;
Since $\angle E=90^{\circ}$;
Therefore, $\angle E D F+\angle E F D=90^{\circ}$;
Since $\angle E F G=\angle J F G+\angle E F D=90^{\circ}$;
Therefore, $\angle E D F=\angle J F G$;
In $\triangle D E F$ and $\triangle F J G$,
$$
\left\{\begin{array}{c}
\angle D E F=\angle F J G=90^{\circ} \\
\angle E D F=\angle J F G
\end{array} ;\right.
$$

Therefore, $\triangle D E F \sim \triangle F J G$;
Therefore, $\frac{F E}{G J}=\frac{D F}{F G}$;
Therefore, $\frac{12}{G J}=\frac{20}{12}$;
Solving, we get $G J=\frac{36}{5}$;
Since $G I / / D F, I H \perp D F, G J \perp D F$;
Therefore, $I H=G J$;
Therefore, $I H=\frac{36}{5}$.
(Method 2)
Extend $D I$ and $F G$ to intersect at point $K$.
In Rt $\triangle K D F$, $\angle K=90^{\circ}, K D=12, K F=4+12=16$; by the Pythagorean Theorem, we get $D F=20$;
Since $G I / / D F$;
Therefore, $\triangle K D F \sim \triangle K I G$;
(Pyramid Model)
Therefore, $\frac{K F}{K G}=\frac{K D}{K I}=\frac{D F}{I G}$;
Therefore, $\frac{16}{4}=\frac{12}{K I}=\frac{20}{I G}$;
Solving, we get $K I=3, \quad I G=5$;
$S_{\triangle K I G}=K G \times K I \div 2=4 \times 3 \div 2=6$;
$S_{\triangle K D F}=K F \times K D \div 2=16 \times 12 \div 2=96$;
$S_{\text {trapezoid } D F G I}=S_{\triangle K D F}-S_{\triangle K I G}=96-6=90$;
$S_{\text {trapezoid } D F G I}=(I G+D F) \times I H \div 2$;
$(5+20) \times I H \div 2=90$
Solving, we get $I H=\frac{36}{5}$.