• If $E$ is the midpoint of $CD$, prove that $OED$ is an equilateral triangle.
• Use the conditions $\angle ODE=60^\circ$ and $DB=OD$ to get $2$ equations with the angles of the triangle.
• You have $3$ equations about the angles of the triangle now. Solve them.

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:

$\angle BAC=60^\circ$

$\angle ABC=45^\circ$

$\angle BCA=75^\circ$