Nice use of Menelaus. Using $ABC$ as the base triangle, it suffices to find $\frac{F_AB}{F_AC}$ in a sufficiently nice form. Let $F_AI$ intersect $AB$ at $E$ and $AC$ at $F$. (I first used sine rule and the result motivated the following synthetic solution) Let $G$ be the point on $EF$ such that $GB//FC$. It is easy to see that $\angle BGE=\angle AFE=\angle AEF = \angle GEB$ so $GB=EB$. Also, by angle chasing, we see that $\triangle BEI \sim \triangle IFC$ (and also $EI=IF$) So, $$\frac{F_AB}{F_AC}=\frac{GB}{FC}=\frac{EB}{FC}=\frac{EB}{EI}=\frac{EB}{FI}\times\frac{EI}{FC}=\frac{BI}{CI}\times \frac{BI}{CI}$$and so we are done by Menelaus.