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OA, OB, OC divide the circle into three congruent parts $\Longleftrightarrow$ $\angle AOB = \angle BOC = \angle COA = 120^\circ$

D is a point on arc AB such that arc AD has half the measure of arc DB $\Longleftrightarrow$ $m\stackrel{\mbox{$\large\frown$}}{AD} = 40^\circ$, $m\stackrel{\mbox{$\large\frown$}}{DB} = 80^\circ$

$\text{reflex }\angle COD = m\stackrel{\mbox{$\large\frown$}}{DB} + m\stackrel{\mbox{$\large\frown$}}{BC} = 200^\circ$

$m\angle COD = 360^\circ - 200^\circ = 160^\circ$

Because OC = OD,

$m\angle OCD = m\angle ODC$

So $160^\circ + 2 \left(m\angle OCD\right) = 180^\circ$

I believe you can continue from here. It's just one step away from the answer.