Selection tests for the Gulf Mathematical Olympiad, 2013

We already know that $DA = DB = DI$. Because $DI = AB$, triangle $ADB$ is equilateral. But $\angle ACB + \angle BDA = 180^\circ$, we deduce that $\angle ACB = 120^\circ$.

Because $\angle BOA = 2 \angle BDA = 120^\circ$, we deduce that $\angle OAB = 30^\circ$. On the other hand, we have $\angle CAH = 90^\circ - \angle CBA - \angle BAC = 30^\circ$. By applying cosine law in the triangle $AOH$ we obtain
$$OH^2 = AH^2 + R^2 - 2R \cdot AH \cos(\angle BAC + 60^\circ),$$
where $R$ is the circumradius of triangle $ABC$. This is equivalent to
$$2AH \cos(\angle BAC + 60^\circ) = R.$$

Because $HC$ is perpendicular to $AB$ and $BC$ is perpendicular to $AH$, we have $\angle AHC = \angle CBA$. On the other hand, we have $\angle HCA = 180^\circ - \angle AHC - 30^\circ = 90^\circ + \angle BAC$. We deduce, by applying sine law to the triangle $ACH$ that
$$\frac{AH}{\sin(90^\circ + \angle BAC)} = \frac{AC}{\sin \angle CBA} = 2R.$$
Therefore, $AH = 2R \cos \angle BAC$. By plugging this into our previous relation we obtain
$$2 \cos \angle BAC \cdot \cos(\angle BAC + 60^\circ) = \frac{1}{2}.$$
This is equivalent to
$$\cos(2\angle BAC + 60^\circ) + \cos 60^\circ = \frac{1}{2},$$
and therefore, $\angle BAC = 15^\circ$. Thus
$$$$
$\angle$ BAC = $15^\circ, \quad \angle$ CBA = $45^\circ, \quad \text{and} \quad \angle$ ACB = $120^\circ.$