Question 121
Let $ABC$ be an acute triangle, which is not equilateral. Denote by $O$ and $H$ its circumcenter and orthocenter, respectively. The circle $k$ passes through $B$ and touches the line $AC$ at $A$. The circle $l$ with center on the ray $BH$ touches the line $AB$ at $A$. The circles $k$ and $l$ meet in $X$ ($X \neq A$). Show that $\angle HXO = 180^\circ - \angle BAC$.