14. As shown in Figure 7, $\triangle ABC$ is an isosceles right triangle, $\angle C = 90^{\circ}, O$ 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: $\triangle AKL, \triangle BMN, \triangle CPQ$ are all isosceles right triangles;
(2) Find the area of the common part of $\triangle ABC$ and $\triangle A_{1} B_{1} C_{1}$.
14. (1) Connect $O C, O C_{1}$, intersecting $P Q, P N$ at points $D$ and $E$ respectively. According to the problem, $\angle C O C_{1}=45^{\circ}$. Since point $O$ is equidistant from the sides of $\triangle A B C$ and this distance is 1, $O C$ bisects $\angle A C B$.
Since $\angle A C B=90^{\circ}$, we have $\angle O C E=\angle O C Q=45^{\circ}$.
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, $\angle C Q P=\angle C P Q=\angle C_{1} P N=\angle C_{1} N P=45^{\circ}$.
Hence, $\triangle C P Q$ and $\triangle C_{1} N P$ are both isosceles right triangles.
Since $\angle B N M=\angle C_{1} N P=\angle A_{1} Q K=\angle C Q P$
$$
=\angle B=\angle A_{1}=45^{\circ},
$$

$\triangle B M N$ and $\triangle A_{1} K Q$ are both isosceles right triangles.
We have $\angle B_{1} M L=\angle B M N=\angle A K L=\angle A_{1} K Q=90^{\circ}$, $\angle B_{1}=\angle A=45^{\circ}$,
Thus, $\triangle B_{1} M L$ and $\triangle A K L$ are both isosceles right triangles.
(2) From (1), in right triangles $\triangle O D C_{1}$ and $\triangle O E C$, we have $O D=O E=1, \angle C O C_{1}=45^{\circ}$,
Therefore, $O C=O C_{1}=\sqrt{2}, C D=C_{1} E=\sqrt{2}-1$.
Thus, $P Q=P N=2 \times(\sqrt{2}-1)=2 \sqrt{2}-2$, $C Q=C P=C_{1} P=C_{1} N=\sqrt{2} \times(\sqrt{2}-1)=2-\sqrt{2}$.
Hence, $S_{\triangle C P Q}=\frac{1}{2} \times(2-\sqrt{2})^{2}=3-2 \sqrt{2}$.
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 $C H \perp A B$.
Therefore, $C H=C O+O H=\sqrt{2}+1$.
Thus, $A C=B C=A_{1} C_{1}=B_{1} C_{1}$
$$
=\sqrt{2} \times(\sqrt{2}+1)=2+\sqrt{2}.
$$

Therefore, $S_{\triangle A B C}=\frac{1}{2} \times(2+\sqrt{2})^{2}=3+2 \sqrt{2}$.
Since $A_{1} Q=B N=(2+\sqrt{2})-(2 \sqrt{2}-2)-(2-\sqrt{2})=2$,
We have $K Q=M N=\sqrt{2}$.
Thus, $S_{\triangle B M N}=\frac{1}{2} \times(\sqrt{2})^{2}=1$.
Since $A K=(2+\sqrt{2})-(2-\sqrt{2})-\sqrt{2}=\sqrt{2}$,
We have $S_{\triangle A K L}=\frac{1}{2} \times(\sqrt{2})^{2}=1$.

Therefore, $S_{\text {hexagon KINNPQ }}=S_{\triangle M B C}-S_{\triangle C P Q}-S_{\triangle B M N}-S_{\triangle M K L}$ $=4 \sqrt{2}-2$.