Let $AO$ intersect $BC$ and $(ABC)$ at $I$ and $J$, respectively. Let $D$ be the midpoint of $BC$.
Since $AJ$ is the diameter, we have $CI$, being the angle bisector of $\angle ACJ$, meaning by the angle bisector theorem, $$\frac{AC}{CJ}=\frac{AI}{IJ}.$$But note that since $OG \parallel BC$, we have that $$\frac{AO}{OI}=\frac{AG}{DG}=2.$$Thus, $$\tan{B}=\frac{AC}{CJ}=\frac{AI}{IJ}=\frac{AO+OI}{AO-OI}=\frac{2OI+OI}{2OI-OI}=3.$$Since $\tan{C}=\tan{45^\circ}=1$, we bash out $$\tan{A}=\tan{135^\circ-B}=\frac{\tan{135^\circ}-\tan{B}}{1+\tan{135^\circ}\tan{B}}=\frac{-1-3}{1+(-3)}=2.$$
We also have $\tan{A}=\frac{CD}{OD}$ and $\tan{A}=\frac{OD}{DI}$ and since $$\frac{OG}{ID}=\frac{AG}{GD}=\frac{3}{2}\implies ID=\frac{3}{2},$$we get that $$CD=DI\cdot \tan^2{A}\implies BC=2\cdot DI\cdot \tan^2{A}=2\cdot \frac{3}{2}\cdot 2^2=\boxed{12.}$$