Let $ \angle A. \geq \angle C$ Let $D$ - midpoint $AC$. Easy to show, that $\angle DYB= \angle DXB= \angle DB_1B=90$ so $D,Y,B,X,B_1$ are concyclic. $E$ - point of intersection $B_1B$ and $XY$ Then $\frac{B_1E}{EX}=\frac{EY}{EB}$ but $XY=BB_1 \to B_1E=EX$ $90-\angle C=\angle B_1BY = \angle B_1XY=\angle XB_1E=90-\angle AB_1X \to \angle AB_1X=\angle C$ So $\triangle AB_1X \sim \triangle ACB \sim \triangle AC_1B_1 \to \frac{AB_1}{AX}=\frac{AC_1}{AB_1} \to 2AX^2=AB_1^2 \to \frac{AC}{AB}=\frac{AB_1}{AX}=\sqrt{2}$

Attachments: