, 2018

Denote $\angle B = \beta$ then we have
$$\angle XYZ = 180^\circ - \angle AIC = 180^\circ - \left(90^\circ + \frac{\beta}{2}\right) = 90^\circ - \frac{\beta}{2}.$$
So if $O$ is the circumcenter of $XYZ$ then
$$\angle OXZ = 90^\circ - \left(90^\circ - \frac{\beta}{2}\right) = \frac{\beta}{2} = \angle IBC,$$
which implies that $OX \perp BC$. Similarly, we also have $OY \perp CA$, $OZ \perp AB$.

Denote $M, N$ as the midpoints of $BC, BB_{1}$ respectively. From the cyclic quadrilateral and parallel line, we have $\angle BB_{1}C = \angle BNM = 180^\circ - \angle BXM$. Similarly, $\angle BC_{1}C = 180^\circ - \angle CXM$. Since $MX \perp BC$, triangle $XBC$ is isosceles, then $\angle BXM = \angle CXM$, thus $\angle BB_{1}C = \angle BC_{1}C$, which implies that $BCC_{1}B_{1}$ is cyclic.

Angle chasing again, we have $\angle IC_{1}B_{1} = \angle CBB_{1} = \frac{\beta}{2}$, but $\angle AIC = 90^\circ + \frac{\beta}{2}$, implies that $AI \perp B_{1}C_{1}$. By similar way, we get $BI \perp C_{1}A_{1}$ and $CI \perp A_{1}B_{1}$, hence $I$ is the orthocenter of triangle $A_{1}B_{1}C_{1}$.

It is easy to check that these conditions are equivalent, which finishes the proof.