Belorusija 2012

Let $H$ be a foot of perpendicular from $A$ onto $BC$. Since $\angle A{B}_{1}C=90^{\circ }$, we have $\angle AHC+\angle A{B}_{1}C=90^{\circ }+90^{\circ }=180^{\circ }$. Thus, points $A,B_1,C,H$ lie on the same circle. Since $A{B}_1=B_{1}C$, we see that $H{B}_1$ is a bisector of $\angle AHC$, so $\angle AH{B}_1=\angle B_{1}HC=90^{\circ }/2=45^{\circ }$. In the same way we obtain $\angle AH{C}_1=\angle C_{1}HB=45^{\circ }$.

Further, consider the circle $\omega$ with the diameter $C_1{B}_1$. Since $\angle C_{1}H{B}_1=\angle C_{1}HA+\angle AH{B}_1=45^{\circ }+45^{\circ }=90^{\circ }$, we obtain $H\in \omega$. Let $T$ be the second point of intersection of $\omega$ and $BC$. Then $\angle C_1{B}_{1}T=\angle C_{1}HT=45^{\circ }$ and $\angle T{C}_1{B}_1=\angle B_{1}HC=45^{\circ }$. Therefore, $△C_{1}T{B}_1$ is an isosceles right-angled triangle, so point $T\in BC$ coincides with the center of the square $C_1{B}_{1}DE$.

In the case when point $T$ coincides with $H$, i.e. $\omega$ touches $BC$, we have $\angle C_1{B}_{1}H=\angle C_{1}HB=45^{\circ }$ and $\angle B_1{C}_{1}H=\angle B_{1}HC=45^{\circ }$. Therefore, point $H$ is the center of the square $C_1{B}_{1}DE$.