I am trying to solve this question in my own (Euclidean) way but was not successful. The question is interesting because the given is simple enough and therefore its solution should not be that difficult.
I think, it will be completely solved if we can prove that:-
KO extended (where O is the circum-center of ⊿ABC) meets AH extended right at the circum-circle.
Or equivalently, can we show that the mentioned intersection point is con-cyclic with A, B, and C.
I tried angle chasing and got the following:-
Alternate method that I have tried:-
Note that I am asking a different question though the given is the same.
I hope the proof need not go through showing that AK // BC.
Let $L, M, N$ be the midpoints of edges $AB, BC, CA$ respectively. Then the circumcircle of triangle $LMN$ is the six point circle and passes through the feet of the three altitudes of triangle $ABC$. In particular it passes through point $H$. Since $ABC$ is the homothetic image of $LMN$ with respect to the homothety with center the centroid $G$ and scaling factor $-2$, the six point circle is mapped to the circumcircle of $ABC$. Therefore, point $H$ is mapped to point $K$ and point $M$ is mapped to point $A$ so line $MH \equiv BC$ is mapped to line $AK$ so $BC$ is parallel to $AK$. As such, angle $\angle \, KAQ = 90^{\circ}$ and so $KQ$ is a diameter of the circumcircle of triangle $ABC$ which means that the center, point $O$, lies on $KQ$.