Balti Tee 2010 · Ülesanne 13

The value is $\angle CAB = 60^\circ$. Denote by $\ell$ the line containing the angle bisector of $DHB$, and let $E$ be the point where the ray $CD \to$ intersects the circumcircle of the triangle $ABC$ again. The rays $HD \to$ and $HB \to$ are symmetric with respect to $\ell$ by the definition of $\ell$. On the other hand, if the circumcenter of $ABC$ lies on $\ell$, then the circumcircle is symmetric with respect to $\ell$. It follows that the intersections of the rays $HD \to$ and $HB \to$ with the circle, which are $E$ and $B$, are symmetric with respect to $\ell$. Moreover, since $H \in \ell$, we conclude that $HE = HB$. However, as $E$ lies on the circumcircle of $ABC$, we have

$$\angle ABE = \angle ACE = 90^\circ - \angle CAB = \angle HBA.$$

This proves that the points $H$ and $E$ are symmetric with respect to the line $AB$. Thus $HB = EB$ and the triangle $BHE$ is equilateral. Finally, $\angle CAB = \angle CEB = 60^\circ$.