See this post.

@above thanks but there is a much simpler approach Let $H$ be the orthocenter of $ABC$ and $AG \cap BC = M$, $AH\cap BC = A_1$. The Euler line gives $HG = 2OG = 2$ and from $AG = \frac{2}{3}AM$ and Thales theorem we now get $A_1M = 3$. From $\angle ABC = 45^{\circ}$ we obtain $\angle A_1CH = 45^{\circ}$ so $A_1C = A_1H = x$, while Thales theorem implies $AA_1 = 3x$. Since $AA_1 = BA_1$ (due to $\angle ABC = 45^{\circ}$, it follows that $BA_1 = 3x$ and so $BC = 4x$, $BM = 2x$ and $2x = BM = MC = 3 + x$, i.e. $x=3$ and $BC=12$.

Just discovered that this problem originates from Hong Kong IMO TST 2019.