Let $ABC$ be a triangle with $\angle A=60.$ Define $G$ as the centroid and $O$ as the cirucenter. Define $P=(ABC)\cap AG$, $D$ as the midpoint of minor arc $BC$ and $P'$ as the reflection of $P$ over $BC$. Prove that $GOP'PD$ cyclic.
Since $\angle BOC=2\angle A=120, \angle BDC=120\implies O$ is the reflection of $D$ over $BC$. Hence $P'POD$ isoscele trapezoid and hence cyclic. Now, we simply need to show that $G\in OP'PD$.
I also observed that $GOP'D$ is harmonic but I couldn't prove it ( proving $GOP'D$ is harmonic finishes).
Also, define $H$ as the orthocenter, we have $$\angle BOC=\angle BHC=\angle BP'C\implies BHOP'C\text{ is cyclic}.$$
There is a very straightforward solution to this problem, not using the added points and lines:
Let the intersection of $OD$ and $BC$ be $M$, and let's assume $R$ is the circumradius of $\triangle ABC.$ Then, we have these identities:
$$OM= \frac{R}{2}, \\ MD=BD \sin \angle CAD=\frac{BD}{2}=\frac{2R\sin \angle DAB}{2}=\frac{R}{2}, \\ \frac{AM}{\sin \angle B}=\frac{\frac{BC}{2}}{\sin \angle BAM} \implies GM= \frac{1}{6} \times BC \times \frac{\sin \angle B}{\sin \angle BAM}, \\\frac{MP}{\sin \angle CAP}= \frac{\frac{BC}{2}}{\sin \angle C} \implies MP= \frac{1}{2} \times BC \times \frac{\sin \angle CAP}{\sin \angle C};$$
Therefore:
$$OM \times MD=\frac{R^2}{4};$$
and:
$$GM\times MP=\frac{BC^2}{12}.$$
However, $BC=2R\sin \angle A =R \sqrt 3$. So, $OM \times MD=GM\times MP$, and $GOPD$ is cyclic.
We are done.