Solution:

Let $x$, $y$, and $z$ be the points passing through $X$, $Y$, and $Z$, parallel to $YZ$, $ZX$, and $XY$. Let $P$, $Q$, and $R$ be the intersection point of the lines $y$ and $z$; $z$ and $x$; $x$ and $y$, respectively. Then we have $\triangle PQR \sim \triangle XYZ$. The points $X$, $Y$, and $Z$ are respectively the midpoints of $RQ$, $QP$, and $PR$. Let $M$ be the orthocenter of $\triangle XYZ$. Obviously, $M$ is the circumcenter of $\triangle PQR$ and $\angle ZMY = 180^{\circ} - \angle X = 180^{\circ} - \angle P = 180^{\circ} - \angle A$. Hence the points $P$, $A$, $Z$, $M$, and $Y$ belong to a circle. In a similar way we prove that the points $Z$, $B$, $R$, $X$ and $M$ belong to a circle. Then $\angle PMA = \angle PZA = \angle BZR = \angle BMR$. Since $MR = MP$ and $\angle PAM = \angle BRM = 90^{\circ}$, we conclude that $MA = MB$. Analogously we conclude that $MA = MC$ implying that $M$ is the circumcenter of $\triangle ABC$.