One. (20 points) As shown in Figure 4, in $\triangle ABC$, $\angle C = 90^{\circ}, AC = BC, O$ is a point inside $\triangle ABC$, and the distances from point $O$ to the sides of $\triangle ABC$ are all equal to 1. $\triangle ABC$ is rotated $45^{\circ}$ clockwise around point $O$ to get $\triangle A_{1} B_{1} C_{1}$, and the common part of the two triangles is the polygon $KLMNPQ$.
(1) Prove that $\triangle AKL, \triangle BMN, \triangle CPQ$ are all isosceles right triangles;
(2) Find the area of the common part of $\triangle ABC$ and $\triangle A_{1} B_{1} C_{1}$.
(1) Connect $O C, O C_{1}$, intersecting $Q P, P N$ at points $D, E$ respectively.
According to the problem, $\angle C O C_{1}=45^{\circ}$.
Since the distances from point $O$ to $A C, B C$ are both equal to 1, we know that $O C$ is the angle bisector of $\angle A C B$.
Since $\angle A C B=90^{\circ}$, therefore,
$\angle O C E=\angle O C Q=45^{\circ}$.
Similarly, $\angle O C_{1} D=\angle O C_{1} N=45^{\circ}$.
Thus, $\angle O E C=\angle O D C_{1}=90^{\circ}$.
Therefore, $\angle C Q P=\angle C P Q=45^{\circ}$.
Hence, $\triangle C P Q$ is an isosceles right triangle.
Since $\angle B N M=\angle C_{1} N P=\angle C Q P=45^{\circ}$, then $\triangle B M N$ is an isosceles right triangle.
Similarly, $\triangle A K L$ is also an isosceles right triangle.
(2) In the right triangles $\triangle O D C_{1}$ and $\triangle O E C$, given $O D=O E=1, \angle C O C_{1}=45^{\circ}$, we get $O C=\sqrt{2}$.
Thus, $C D=\sqrt{2}-1$.
Then $P Q=N P=2(\sqrt{2}-1)=2 \sqrt{2}-2$, $C Q=C P=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}$.
Therefore, $S_{\triangle C P Q}=\frac{1}{2}(2-\sqrt{2})^{2}=3-2 \sqrt{2}$.
Extend $C O$ to intersect $A B$ at point $H$.
Since $C O$ bisects $\angle A C B$ and $A C=B C$, we have $C H \perp A B$.
Also, $C H=C O+O H=\sqrt{2}+1$, $A C=B C=\sqrt{2}(\sqrt{2}+1)=2+\sqrt{2}$, hence $S_{\triangle A B C}=\frac{1}{2}(2+\sqrt{2})^{2}=3+2 \sqrt{2}$.
Since $A_{1} Q=B N$
$$
=(2+\sqrt{2})-(2 \sqrt{2}-2)-(2-\sqrt{2})=2 \text {, }
$$

Therefore, $K Q=M N=\frac{2}{\sqrt{2}}=\sqrt{2}$.
Thus, $S_{\triangle B M N}=\frac{1}{2}(\sqrt{2})^{2}=1$.
Since $A K=(2+\sqrt{2})-(2-\sqrt{2})-\sqrt{2}=\sqrt{2}$, we get $S_{\triangle A K L}=\frac{1}{2}(\sqrt{2})^{2}=1$.

Therefore, $S_{\text {hexagon KLMNPQ }}$
$$
\begin{array}{l}
=S_{\triangle A B C}-S_{\triangle C P Q}-S_{\triangle B M N}-S_{\triangle M K L} \\
=(3+2 \sqrt{2})-(3-2 \sqrt{2})-1-1 \\
=4 \sqrt{2}-2 .
\end{array}
$$