Solution:
The segment $AA'$ bisects $\angle OAH$: if $\angle BCA = y$ (Figure 4), then $\angle BOA = 2y$, and since $OA = OB$, it is $\angle OAB = \angle OBA = 90^{\circ} - y$. Also since $AH \perp BC$, it is
$\angle HAC = 90^{\circ} - y = \angle OAB$ and the claim follows.

Since $AA'$ bisects $\angle OAH$ and $A'H = AH$, $OA' = OA$, we have that the isosceles triangles $OAA'$, $HAA'$ are equal. Thus
$$AH = OA = R$$
where $R$ is the circumradius of triangle $ABC$.

Call $\angle ACH = a$ and recall by the law of sines that $AH = 2R' \sin a$, where $R'$ is the circumradius of triangle $AHC$. Then (4) implies
$$R = 2R' \sin a$$
But notice that $R = R'$ because $\frac{AC}{\sin(AHC)} = 2R'$, $\frac{AC}{\sin(ABC)} = 2R$ and $\sin(AHC) = \sin\left(180^{\circ} - ABC\right) = \sin(ABC)$. So (5) gives $1 = 2\sin a$, and $a$ as an acute angle can only be $30^{\circ}$. Finally, $\angle BAC = 90^{\circ} - a = 60^{\circ}$.