Prove that $GOP'PD$ cyclic. 
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 $$\angle BOC=\angle BHC=\angle BP'C\implies BHOP'C\text{ is cyclic}.$$


Comment: mark midpoint of arc DP as K, connect O to K.Draw a perpendicular from P' on OK, it meets circle passing PP'OD at G'. Now yow have to show:
1- OK is diameter of circle passing PP'OD.
2-G is coincident with G'.
