Since $\angle BB_1C = \angle CC_1B = 90^\circ$, the points $B, C_1, B_1$ and $C$ lie on the circle $\omega$ with the diameter $BC$. Hence the bisectors of angles $\angle BB_1C$ and $\angle CC_1B$ pass through the midpoint of the arc $BC$ of $\omega$, so this midpoint is $X$.

First we prove that the quadrilateral $EC_1B_1D$ is cyclic. The angle $\angle BDB_1$ is an external angle of the triangle $CB_1D$, hence $\angle BDB_1 = \angle ACB + 45^\circ$. Since $\angle BB_1C$ is right, $\angle B_1BC = 90^\circ - \angle ACB$. Thus,
$$\angle B_1C_1E = \angle B_1C_1C + 45^\circ = 135^\circ - \angle ACB.$$
$$\angle B_1C_1E + \angle EDB_1 = 180^\circ.$$
Now we prove that the intersection points of the circumcircles of the triangles $AB_1D$ and $AC_1E$ lie on the line $AX$. Let the circumcircle of the triangle $AB_1D$ intersect the line $AX$ at points $A$ and $P$, then $XP \cdot XA = XD \cdot XB_1$. Since points $E, C_1, B_1$ and $D$ are concyclic, $XE \cdot XC_1 = XD \cdot AB_1$. Hence $XE \cdot XC_1 = XP \cdot XA$, and therefore the circumcircle of the triangle $AC_1E$ passes through $P$.

Finally, we will prove that the circumcircles of the triangles $BEX$ and $CDX$ passes through points $X$ and $P$. Since point $X$ is the midpoint of the arc $BC$, $XB = XC$ and $\angle BXC = 90^\circ$. Therefore, $\angle BCX = \angle CBX = 45^\circ$. Since the quadrilateral $AB_1DP$ is cyclic, $\angle APD = 180^\circ - \angle DB_1A = 45^\circ$. Hence $\angle DPX + \angle DCX = 180^\circ$, i.e. points $P, D, C$ and $X$ are concyclic. Similarly, the quadrilateral $BEPX$ is cyclic. If point $P$ is distinct from point $X$, the problem is solved.
If points $P$ and $X$ coincide, the equalities $\angle DCX = \angle APD$ and $\angle EBX = \angle APE$ imply that the circumcircles of the triangles $BEX$ and $CDX$ are tangent to the line $AX$.

Note that $BC_1B_1C$ is concyclic, and let $\omega, \omega_B$ and $\omega_C$ be the circumcircles of $BC_1B_1C$, $BEX$, and $CDX$, respectively. Since $B_1X$ and $C_1X$ are the bisectors of $\angle BB_1C$ and $\angle CC_1B$, they both pass through the midpoint of the arc $BC$ of $\omega$ not containing $B_1$ and $C_1$, thus this midpoint is $X$. Let $AX$ intersect $\omega, \omega_B$ and $\omega_C$ the second time at points $T, F_B$ and $F_C$, respectively, and denote $AX \cap BC = P$.

As in the first solution, note that $EC_1B_1D$ is cyclic. Let $\gamma, \gamma_B$, and $\gamma_C$ be the circumcircles of $EC_1B_1D$, $BEC_1$, and $CDB_1$, respectively. Since $C_1E$ and $B_1D$ are radical axes of $\gamma_B, \gamma$ and $\gamma_B, \gamma$ and $C_1E \cap B_1D = X$, the radical axis of $\gamma_B, \gamma_C$ passes through $X$. On the other hand, it passes through $A$ since $\text{Pow}_{\gamma_B} A = AC_1 \cdot AB = AB_1 \cdot AC = \text{Pow}_{\gamma_C} A$. Thus, $AX$ is the radical axis of $\gamma_B, \gamma_C$, and it follows from $P \in AX$ that
$$PF_A \cdot PX = PD \cdot PC = PE \cdot PB = PF_B \cdot PX.$$
Consequently, $F_A = F_B$ and the problem is solved.