We show that $\triangle SMB'$ is a 30-60-90 triangle with $\angle M=90,\angle S=60$. Extend $BS$ to $D$ such that $S$ is the midpoint of $BD$. Extend $A'B'$ to $E$ such that $B'$ is the midpoint of $A'E$. Then $\triangle BCD\sim \triangle ECA'$ are clearly 30-60-90 triangles, so by spiral symmetry, $\triangle BCE\sim\triangle DCA'$. But since $EC\perp A'C$, we have $DA'\perp BE$ and $\frac{DA'}{BE}=\frac{A'C}{CE}=\frac{1}{\sqrt3}$. Thus $SM\parallel DA'\perp BE\parallel MB'$ and $\frac{SM}{MB'}=\frac{DA'}{BE}=\frac{1}{\sqrt3}$, so $\triangle SMB'$ is a 30-60-90 triangle. Similarly for $\triangle SNA'$, and both are similar, so we are done.