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 $$O{H}^2=A{H}^2+R^2-2R\cdot AH\cos (\angle BAC+60^{\circ }),$$ 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.