56. As shown in the figure, the side length of square $\mathrm{ABCD}$ is $10, \mathrm{AG}=\mathrm{CH}=8, \mathrm{BG}=\mathrm{DH}=6$, connect $\mathrm{GH}$, then the length of line segment $\mathrm{GH}$ is
Answer: $2 \sqrt{2}$
Solution: As shown in the figure, extend $\mathrm{BG}$ to intersect $\mathrm{CH}$ at point $\mathrm{E}$.
Since
$$
6^{2}+8^{2}=10^{2},
$$

$\triangle \mathrm{ABG}$ and $\triangle \mathrm{CDH}$ are both right triangles, and $\triangle \mathrm{ABG} \cong \triangle \mathrm{CDH}$,
Therefore,
$$
\angle 1=\angle 5, \angle 2=\angle 6 \text { . }
$$

Since
$$
\begin{array}{l}
\angle 2+\angle 3=90^{\circ}, \quad \angle 4+\angle 5=90^{\circ}, \\
\angle 1=\angle 3=\angle 5, \angle 2=\angle 4=\angle 6, \\
\angle 1=\angle 3, \quad \mathrm{AB}=\mathrm{BC}, \angle 2=\angle 4,
\end{array}
$$

Therefore,
in $\triangle \mathrm{ABG}$ and $\triangle \mathrm{BCE}$,
Therefore,
$$
\triangle \mathrm{ABG} \cong \triangle \mathrm{BCE},
$$

We can get
$$
\mathrm{BE}=\mathrm{AG}=8, \mathrm{CE}=\mathrm{BG}=6, \angle \mathrm{BEC}=\angle \mathrm{AGB}=90^{\circ} \text {, }
$$

Thus
$$
\mathrm{GE}=\mathrm{BE}-\mathrm{BG}=8-6=2 \text {, }
$$

Similarly
$$
\mathrm{HE}=\mathrm{CH}-\mathrm{CE}=8-6=2,
$$

In the right triangle $\mathrm{GHE}$, by the Pythagorean theorem, we have $\mathrm{GH}=\sqrt{\mathrm{GE}^{2}+\mathrm{HE}^{2}}=\sqrt{2^{2}+2^{2}}=2 \sqrt{2}$.