, 2012

Let $H$ be a foot of perpendicular from $A$ onto $BC$. Since $\angle AB_1C = 90^\circ$, we have $\angle AHC + \angle AB_1C = 90^\circ + 90^\circ = 180^\circ$. Thus, points $A, B_1, C, H$ lie on the same circle. Since $AB_1 = B_1C$, we see that $HB_1$ is a bisector of $\angle AHC$, so $\angle AHB_1 = \angle B_1HC = 90^\circ/2 = 45^\circ$. In the same way we obtain $\angle AHC_1 = \angle C_1HB = 45^\circ$.

Further, consider the circle $\omega$ with the diameter $C_1B_1$. Since $\angle C_1HB_1 = \angle C_1HA + \angle AHB_1 = 45^\circ + 45^\circ = 90^\circ$, we obtain $H \in \omega$. Let $T$ be the second point of intersection of $\omega$ and $BC$. Then $\angle C_1B_1T = \angle C_1HT = 45^\circ$ and $\angle TC_1B_1 = \angle B_1HC = 45^\circ$. Therefore, $\triangle C_1TB_1$ is an isosceles right-angled triangle, so point $T \in BC$ coincides with the center of the square $C_1B_1DE$.

In the case when point $T$ coincides with $H$, i.e. $\omega$ touches $BC$, we have $\angle C_1B_1H = \angle C_1HB = 45^\circ$ and $\angle B_1C_1H = \angle B_1HC = 45^\circ$. Therefore, point $H$ is the center of the square $C_1B_1DE$.