\[2OI^{2} =(AB-AC)^{2} \] \[2(R^{2}-2rR) =4R^{2}(\sin C-\sin B)^{2} \] \[(R^{2}-2R^{2}(4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}))=2R^{2}(\sin (A+B)-\sin B)^{2} \] \[(R^{2}-2R^{2}(\cos A+\cos B+\cos C-1)) =2R^{2}(2\sin \frac{A}{2}\cos (45+\frac{A}{2}))^{2} \] \[(3R^{2}-2R^{2}(\cos A+\cos B-\cos (A+B))) =2R^{2}(2\sin \frac{A}{2}(\cos \frac{A}{2}-\sin \frac{A}{2})\sin 45)^{2} \] \[3R^{2}-2R^{2}(\cos A+2\sin \frac{A}{2}\sin (45+\frac{A}{2})) =R^{2}(\sinh+\cos A-1)^{2} \] \[3-2(\cos A+\frac{1}{2}\sqrt{2}(1-\cos A+\sin A)) =(\sin A+\cos A-1)^{2} \] \[3-2\cos A-\sqrt{2}(1-\cos A+\sin A) =2+2\sin A\cos A-2\sin A-2\cos A \] \[1-\sqrt{2}+\sqrt{2}\cos A-\sqrt{2}\sin A) =2\sin A\cos A-2\sin A \] \[1-\sqrt{2}\sin A =2\sin A(\cos A-1)-\sqrt{2}(\cos A-1) \] \[0 =\sqrt{2}(\cos A-1)(\sqrt{2}\sin A-1)+(\sqrt{2}\sin A-1) \] \[0 =(\sqrt{2}\sin A-1)(\sqrt{2}(\cos A-1)+1)\] So \[ \sin A=\frac{1}{\sqrt{2}}=\frac{1}{2}\sqrt{2} \] or \[ \cos A=1-\frac{1}{2}\sqrt{2} \rightarrow \cos ^{2}A=\frac{3}{2}-\sqrt{2} \rightarrow \sin A=\sqrt{\sqrt{2}-\frac{1}{2}}=\frac{1}{2}\sqrt{4\sqrt{2}-2} \]

Can't it be solved with elementary geometry?