In obtuse triangle $\triangle ABC$, $AB > AC$, $\angle B = 45^{\circ}$, $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively, and $\sqrt{2} \cdot OI = AB - AC$. Find $\sin A$.
The distance between the incenter and circumcenter of a triangle easily leads to Euler's theorem. Let the circumradius and inradius of $\triangle ABC$ be $R, r$, and the sides opposite to $\angle A, \angle B$, and $\angle C$ be $a, b, c$ respectively. Then, from the given conditions and Euler's theorem, we have $\left(\frac{c-b}{\sqrt{2}}\right)^{2}=R^{2}-2 R r$. Also, $r=\frac{1}{2}(c+a-b) \cdot \tan \frac{B}{2}=\frac{1}{2}(c+a-$ b) $\cdot \tan \frac{\pi}{8}=\frac{\sqrt{2}-1}{2}(c+a-b)$. By the Law of Sines, we get $\left(\frac{2 R \sin C-2 R \sin B}{\sqrt{2}}\right)^{2}=R^{2}-$ $2 R \frac{\sqrt{2}-1}{2}(2 R \sin C+2 R \sin A-2 R \sin B)$. Simplifying, we obtain $1-2(\cos C-\sin B)^{2}=2(\sqrt{2}-$ 1) $(\sin A+\sin C-\sin B)$. Since $\angle B=\frac{\pi}{4}, \sin B=\frac{\sqrt{2}}{2}$, and $\sin C=\sin \left(\frac{3 \pi}{4}-\angle A\right)=$ $\frac{\sqrt{2}}{2}(\cos A+\sin A)$, we have $2 \sin A \cos A-(2-\sqrt{2}) \sin A-\sqrt{2} \cos A+\sqrt{2}-1=0$, which simplifies to $(\sqrt{2} \sin A-1)(\sqrt{2} \cos A-\sqrt{2}+1)=0$. Therefore, $\sin A=\frac{\sqrt{2}}{2}$, or $\cos A=1-\frac{\sqrt{2}}{2}$. When $\cos A=1-\frac{\sqrt{2}}{2}$, $\sin A=\sqrt{1-\cos ^{2} A}=\sqrt{\sqrt{2}-\frac{1}{2}}$. In summary, $\sin A=\frac{\sqrt{2}}{2}$ or $\sin A=$ $\sqrt{\sqrt{2}-\frac{1}{2}}$.