Balkan Mathematical Olympiad

Solution:

Note that $AB = AC$ cannot hold since $AO = AD$ would imply that $O$ is the midpoint of $BC$, which is not possible for an acute triangle. So we may assume without loss of generality that $AB < AC$.
Let $M_{b}$ and $M_{c}$ be the midpoints of $AC$ and $AB$, respectively. Since $\angle AM_{b}O = \angle AM_{c}O = 90^{\circ} = \angle AMO$ (the latter since $AO = AD$), the pentagon $AM_{b}OMM_{c}$ is cyclic.

Next, notice that $AM$ is the perpendicular bisector of $OD$, $O_{b}O_{c}$ is the perpendicular bisector of $AO$ and $M_{b}M_{c}$ is the perpendicular bisector of $AD$. Hence these three lines are concurrent - denote their common point by $T$.

The quadrilateral $AO_{b}MO_{c}$ is cyclic if and only if $AT \cdot TM = O_{b}T \cdot O_{c}T$. From the cyclic $AM_{b}MM_{c}$ we have $AT \cdot TM = M_{b}T \cdot M_{c}T$. Hence it now suffices to argue $M_{b}T \cdot M_{c}T = O_{b}T \cdot O_{c}T$ - or equivalently, that $M_{b}$, $M_{c}$, $O_{b}$ and $O_{c}$ are concyclic.

We assume that $\angle AOB < 90^{\circ}$ and $\angle AOC > 90^{\circ}$ so that $O_{c}$ is in the interior of triangle $AOB$ and $O_{b}$ is external to the triangle $AOC$ (the other cases are analogous and if $\angle AOB = 90^{\circ}$ or $\angle AOC = 90^{\circ}$, then $M_{b} \equiv O_{b}$ or $M_{c} \equiv O_{c}$ and we are automatically done). We have
$$\angle M_{c}M_{b}O_{b} = 90^{\circ} + \angle AM_{b}M_{c} = 90^{\circ} + \angle ACB$$
as well as (since $O_{c}O_{b}$ is a perpendicular bisector of $AO$ and hence bisects $\angle AO_{c}O$)
$$\angle M_{c}O_{c}O_{b} = 180^{\circ} - \angle OO_{c}O_{b} = 90^{\circ} + \frac{\angle AO_{c}M_{c}}{2}$$
$$= 90^{\circ} + \frac{\angle AO_{c}B}{4} = 90^{\circ} + \frac{\angle AOB}{2} = 90^{\circ} + \angle ACB$$
and therefore $O_{b}M_{b}O_{c}M_{c}$ is cyclic, as desired.