Let $D$ be on $AM$, such that $MD=MC=MB$. Under reflection over $MC_1, C$ maps to $D$, and $CC_1$ is fixed, so $D\in CC_1$, similarily $D\in BB_1$. $\angle B_1MC_1=(\angle BMA+\angle AMC)/2=90^{\circ}, \angle MB_1D=90^{\circ}, \angle MC_1D=90^{\circ}\longrightarrow MC_1DB_1$ is a rectangle, hence $\frac{B_1A_1}{A_1C_1}=1$.

Clearly $\overline{BB_1} \parallel \overline{MC_1}$ and that $BB_1=MC_1=\tfrac{a}{2}\cdot \sin(\tfrac{1}{2}\angle BMA)$. It follows that $B_1C_1MB$ is a parallelogram. Similarily $C_1B_1MC$ is parallelogram. Now $$\angle A_1MB_1=\angle B_1MB=\angle A_1B_1M \implies A_1M=A_1B_1$$Similarily $A_1M=A_1C_1$. Hence desired ratio is $1$