Let $XA\cap CH=X'$ and $YC\cap AH=Y'$. See that $X'AIH$ and $Y'CJH$ are cyclic. Hence, $\angle AXI=90^\circ-\angle XAI=90^\circ-\angle X'HI=\angle JHC$. Similarly, $\angle CYJ=\angle IHA$. Thus, $\triangle HAX\sim \triangle YCH$. Then, $\frac{|HI|}{|IX|}=\frac{|YJ|}{|JH|}\Longrightarrow \frac{|JH|}{|IX|}=\frac{|YJ|}{|HI|}\Longrightarrow \frac{|ZI|}{|IX|}=\frac{|YJ|}{|JZ|}$. Also, $\angle ZIX=\angle YJZ=90^\circ$. Therefore, $\triangle ZIX\sim \triangle YJZ$. Then, $\angle YZJ+\angle XZI=\angle ZXI+\angle XZI=90^\circ$. Hence, $\angle JZI+\angle YZJ+\angle XZI=90^\circ+90^\circ=180^\circ\Longrightarrow Y,Z,X$ are collinear.