Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Let $E$ be the midpoint of $DC$.

From $BC=3BD$, we have $CE=ED=DB$.

But, $BD=OD$.

So, $ED=OD$.

If $M$ is the midpoint of $BC$, $CE=DB\implies M$ is the midpoint of $DE$.

But, $O$ is on the perpendicular bisector of $BC$ because it is the circumcenter of $ABC$. From this, $O$ is on the perpendicular bisector of $DE$.

So, $ED=OD=OE$

So, triangle $ODE$ is equilateral.

So, $\angle ADC=\angle ODE=60^\circ$, 

Calculating $\angle OAC$,

$\angle OAC=\angle DAC=180^\circ-60^\circ-\angle ACD=120^\circ-\angle ACB.$

$\angle OAC=\frac{180^\circ-\angle COA}{2}=90^\circ-\angle CBA$

$120^\circ -\angle ACB=90^\circ - \angle CBA\implies \angle ACB=\angle CBA+30^\circ$

Calculating $\angle OBD$

$\angle OBD=\angle OBC=\frac{180^\circ - \angle COB}{2}=90^\circ -\angle CAB$

$\angle OBD = \angle DOB = 180^\circ-\angle AOB=180^\circ-2\angle ACB$

$90^\circ -\angle CAB=180^\circ-2\angle ACB\implies 2\angle ACB=90^\circ + \angle CAB$

Solving the set of equations:

$\angle ACB=\angle ABC+30^\circ$

$2\angle ACB= \angle CAB+90^\circ$

$\angle ABC+\angle BCA+\angle CAB=180^\circ$

We get the solution:

$\angle BAC=60^\circ$

$\angle ABC=45^\circ$

$\angle BCA=75^\circ$