Belarusian Mathematical Olympiad

Since $\angle B{B}_{1}C=\angle C{C}_{1}B=90^{\circ }$, it is sufficient to prove that the points $B$, $C_1$, $X$ and $B_1$ lie on a circle. Since $A_{1}D$ and $A_{1}G$ are the bisectors of the angles $\angle A{A}_{1}C$ and $\angle A{A}_{1}B$ respectively, $\angle G{A}_{1}D=90^{\circ }$. Since $\angle F{B}_{1}D=\angle F{A}_{1}D=90^{\circ }$, the point $B_1$ lies on the circumcircle of the triangle $F{A}_{1}D$. Similarly, the point $C_1$ lies on the circumcircle of the triangle $E{A}_{1}G$.



Therefore $\angle C_{1}X{A}_1=\angle BG{A}_1=135^{\circ }-\angle ABC$ and $\angle B_{1}X{A}_1=\angle CD{A}_1=135^{\circ }-\angle ACB$. Hence $\angle C_{1}X{B}_1=270^{\circ }-\angle ABC-\angle ACB=90^{\circ }+\angle BAC$. Together with $\angle B_{1}BA=90^{\circ }-\angle BAC$, this leads to $\angle B_{1}B{C}_1+\angle C_{1}X{B}_1=180^{\circ }$, which means that the quadrilateral $B{C}_{1}X{B}_1$ is cyclic.



Therefore $\angle C_{1}X{A}_1=\angle BG{A}_1=135^{\circ }-\angle ABC$ and $\angle B_{1}X{A}_1=\angle CD{A}_1=135^{\circ }-\angle ACB$. Hence $\angle C_{1}X{B}_1=270^{\circ }-\angle ABC-\angle ACB=90^{\circ }+\angle BAC$. Together with $\angle B_{1}BA=90^{\circ }-\angle BAC$, this leads to $\angle B_{1}B{C}_1+\angle C_{1}X{B}_1=180^{\circ }$, which means that the quadrilateral $B{C}_{1}X{B}_1$ is cyclic.