Let $\omega$ be the circumcircle of $ABC$, and let $O_1$ be the circumcenter of $A_1B_1C_1$. Because $A_1, B_1$, and $C_1$ are on the boundary of $ABC$ and $O_1$ is outside of $ABC$, $A_1B_1C_1$ is obtuse. Without loss of generality, assume that $\angle B_1A_1C_1$ is obtuse so that $O_1$ and $A$ lie on the same side of line $B_1C_1$.

Lemma 1. The second intersection $A_0$ of $\omega$ and the circumcircle of triangle $AB_1C_1$ is the midpoint of arc $\widehat{BAC}$.

Proof. By the definition of $A_0$, we have $\angle A_0BC_1 = \angle A_0BA = \angle A_0CA = \angle A_0CB_1$ and $\angle A_0C_1A = \angle A_0B_1A$, hence $AC_1B$ and $AB_1C$ are similar. But $BC_1 = CB_1$, so these two triangles are congruent, hence $A_0B = A_0C$. Because $AA_0B_1C_1$ is cyclic, we have $\angle C_1A_0B_1 = \angle C_1AB_1 = \angle BAC$, so $A_0$ lies on $\widehat{BAC}$ with $BA_0 = CA_0$, implying that $A_0$ is the midpoint of $\widehat{BAC}$. $\square$

By Lemma 1, a spiral similarity centered at $A_0$ sends $B_1C_1$ to $CB$, so $A_0$ is the intersection of $\omega$ and the perpendicular bisector of $B_1C_1$ which is on the same side of $BC$ as $A$. Recalling that $A_0$ is the circumcenter of $A_1B_1C_1$ and using this result for the analogous points $B_0$ and $C_0$, we obtain that $A_0C_1B_0A_1$ and $A_0A_1C_0B_1$ are kites with symmetry axes $A_0B_0$ and $A_0C_0$. Recalling that $C_1B_1AA_0$ is cyclic, we have $\angle CAB = \angle C_1A_0B_1 = 2\angle B_0A_0C_0 = \widehat{B_0C_0}$. By Lemma 1, $B_0$ and $C_0$ are the midpoints of $\widehat{ABC}$ and $\widehat{BCA}$, hence
$$\begin{aligned} \angle CAB &= \widehat{B_0C_0} = 360^\circ - \widehat{ACC_0} - \widehat{B_0A} = 360^\circ - \frac{\widehat{BCA} + \widehat{ABC}}{2} \\ &= 360^\circ - \frac{360^\circ - 2\angle BCA + 360^\circ - 2\angle ABC}{2} = \angle BCA + \angle ABC, \end{aligned}$$
implying that $\angle CAB = 90^\circ$, so $ABC$ has right angle at vertex $A$.