Prove that the circumcenter of $\triangle ABC$ and the centroid of the anti pedal $\triangle A'B'C'$ of the symmedian point of $\triangle ABC$ coincide.
This is pretty much unbashable using barycentric or complex -so we have to resort to synthetic means. Since there are symmedians I naturally drew the two tangents to $B$ and $C$ and took their intersection. Tried using harmonic ratios but that also yielded nothing.
Any hint on how to approach with a synthetic (or even coordinate) solution will be appreciated. Thanks in advance.