The side $AC$ of $\triangle ABC$ is a diameter of a circle. Side $BC$ intersects with this circle at point $M$ and side $BA$ intersects this circle at point $K$. If $S_{\triangle ABC}=9S_{\triangle BKM}$, find $\angle ABC$.
I can prove that $\triangle ABC \sim \triangle BKM$ and $3BM=AB$ and $3BK=BC$, but I have no idea about $\angle ABC$.
Since $∠AMC = \dfrac{π}{2}$, then$$ \frac{1}{3} AB = BM = AB \cos B \Longrightarrow ∠B = \arccos \frac{1}{3}. $$