In an acute triangle $ABC$, the segment $CD$ is an altitude and $H$ is the orthocentre. Given that the circumcentre of the triangle lies on the line containing the bisector of the angle $DHB$, determine all possible values of $\angle CAB$.
The value is $\angle C A B=60^{\circ}$.

Denote by $\ell$ the line containing the angle bisector of $\angle D H B$, and let $E$ be the point where the ray $C D \rightarrow$ intersects the circumcircle of the triangle $A B C$ again. The rays $H D \rightarrow$ and $H B \rightarrow$ are symmetric with respect to $\ell$ by the definition of $\ell$. On the other hand, if the circumcenter of $A B C$ lies on $\ell$, then the circumcircle is symmetric with respect to $\ell$. It follows that the intersections of the rays $H D^{\rightarrow}$ and $H B^{\rightarrow}$ with the circle, which are $E$ and $B$, are symmetric with respect to $\ell$. Moreover, since $H \in \ell$, we conclude that $H E=H B$.

However, as $E$ lies on the circumcircle of $A B C$, we have

$$
\angle A B E=\angle A C E=90^{\circ}-\angle C A B=\angle H B A .
$$

This proves that the points $H$ and $E$ are symmetric with respect to the line $A B$. Thus $H B=E B$ and the triangle $B H E$ is equilateral. Finally, $\angle C A B=\angle C E B=60^{\circ}$.

Obviously the value $\angle C A B=60^{\circ}$ is attained for an equilateral triangle $A B C$.