Let $ABC$ an acute triangle with the circumcenter $O$. Let $YQ$ the middle bisector of $OA$ where $Q\in BC$. If $H$ is the hortocenter of $\triangle ABC$ and $T$ is the middle of $OH$ then show that $AYTQ$ is inscriptible
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My idea: I have to prove that $\angle TQB=\angle AQY$. I consider the next picture
Let $S$ and $P$ the intersections between $YQ$ and circumcircle. Then it is easy to see that $ASOP$ is a rhombus. Also I know that $T$ is the center of Euler's circle, but now I am stuck.
Assume $H_1$ is the foot of the altitude $AH$. By the properties of the nine point circle, we have $|TH_1|=|AY|=\frac{R}{2}$, where $R$ is the radius of the triangle's circumcircle. In addition, it is very easy yo see that $YT$ and $AH_1$ are parallel. So, $AYTH_1$ is cyclic. As a result: $$\angle YTA=\angle YH_1A.$$
On the other hand, note that $AYH_1Q$ is cyclic. Thus, $$\angle AQY=\angle YH_1A.$$
Therefore, $$\angle YTA=\angle AQY.$$
We are done.
PS: A more clear picture of the problem drawn by online GeoGebra: