$\begin{array}{l} \left. \begin{array}{l} {\rm B}{\rm{C = BA}}\\ {\rm{CX}} = {\rm{C}}{\rm A}\\ {\rm B}{\rm Z} \bot {\rm{C}}\Upsilon \end{array} \right\}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} {\rm B}{\rm X} \bot {\rm{C}}{\rm A}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} \left\{ \begin{array}{l} {\rm{C}}\mathop {}\limits_{}^{} {\rm{,}}\mathop {}\limits_{}^{} {\rm{B}}\mathop {}\limits_{}^{} {\rm{,}}\mathop {}\limits_{}^{} {\rm{Z}}\mathop {}\limits_{}^{} {\rm{,}}\mathop {}\limits_{}^{} {\rm{X}}{\rm{,}}\mathop {}\nolimits_{}^{} {\rm{are}}\mathop {}\limits_{}^{} {\rm{concyclic}}\\ {\rm{Z}}\widehat {\rm{X}}{\rm{H = 9}}{{\rm{0}}^o} \end{array} \right.\mathop {}\nolimits_{}^{} \\ \\ \left. \begin{array}{l} \Upsilon {\rm H} \bot {\rm{CY}}\\ {\rm X}{\rm{Y//BC}} \end{array} \right\}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} {\rm{C}}\widehat {\rm{Y}}{\rm{X}} = {\rm{B}}\widehat {\rm{C}}{\rm{Y = H}}\widehat {\rm{X}}\Delta = {\rm{Z}}\widehat {\rm{Y}}{\rm{X = }}\theta \\ \theta + \omega = {\rm{9}}{{\rm{0}}^o}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} {\rm X}\Delta \mathop {}\nolimits_{}^{} {\rm{is}}\mathop {}\limits_{}^{} {\rm{a}}\mathop {}\nolimits_{}^{} {\rm{diameter}} \end{array}$

$\begin{array}{l} \left. \begin{array}{l} {\rm B}{\rm{C = BA}}\\ {\rm{CX}} = {\rm{C}}{\rm A}\\ {\rm B}{\rm Z} \bot {\rm{C}}\Upsilon \end{array} \right\}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} {\rm B}{\rm X} \bot {\rm{C}}{\rm A}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} \left\{ \begin{array}{l} {\rm{C}}\mathop {}\limits_{}^{} {\rm{,}}\mathop {}\limits_{}^{} {\rm{B}}\mathop {}\limits_{}^{} {\rm{,}}\mathop {}\limits_{}^{} {\rm{Z}}\mathop {}\limits_{}^{} {\rm{,}}\mathop {}\limits_{}^{} {\rm{X}}{\rm{,}}\mathop {}\nolimits_{}^{} {\rm{are}}\mathop {}\limits_{}^{} {\rm{concyclic}}\\ {\rm{Z}}\widehat {\rm{X}}{\rm{H = 9}}{{\rm{0}}^o} \end{array} \right.\mathop {}\nolimits_{}^{} \\ \\ \left. \begin{array}{l} {\rm{Y}}{\rm H} \bot {\rm{CY}}\\ {\rm X}{\rm{Y//BC}} \end{array} \right\}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} {\rm{C}}\widehat {\rm{Y}}{\rm{X}} = {\rm{B}}\widehat {\rm{C}}{\rm{Y = H}}\widehat {\rm{X}}\Delta = {\rm{Z}}\widehat {\rm{Y}}{\rm{X = }}\theta \\ \\ \theta + \omega = {\rm{9}}{{\rm{0}}^o}\mathop {}\nolimits_{}^{} \Rightarrow \mathop {}\nolimits_{}^{} {\rm X}\Delta \mathop {}\nolimits_{}^{} {\rm{is}}\mathop {}\limits_{}^{} {\rm{a}}\mathop {}\nolimits_{}^{} {\rm{diameter}} \end{array}$

Attachments:

Let $BZ$ intersects $AC$ at $D$ and the circle at $E$. Then $ZHXD$ is cyclic, so $BH \cdot BX = BZ \cdot BD \Leftrightarrow $ $\boxed{BH = \frac{{BZ \cdot BD}}{{BX}}}$ $(1)$ $H$ is the orthocenter of triangle $BDC$, so $HD \bot BC \Leftrightarrow HD \bot XY$. But $\widehat Z=90^0$, that means $YE$ is diameter of the circle and $Y\widehat XE = {90^0} \Leftrightarrow$ $\boxed{DH||EX}$. $\frac{{BX}}{{BH}} = \frac{{BE}}{{BD}}\mathop \Leftrightarrow \limits^{(1)} $ $\boxed{BX^2=BZ\cdot BE}$ Hence $BX$ is tangent to the circle and the result follows.

Attachments:

Let $O$ be the circumcenter of triangle $XYZ$. Note that quadrilateral $BZXC$ is cyclic, so \[90^{\circ}-\angle YXO=\frac12\angle XOY=\angle XZY=\angle XZC=\angle BZX-90^{\circ}=90^{\circ}-\angle BCA=90^{\circ}-\angle YXA.\]Therefore, $A$, $X$, and $O$ are collinear, which finishes.