CroatianCompetitions

Let $A_1, B_1$ and $C_1$ be the midpoints of the sides $\overline{BC}, \overline{CA}$ and $\overline{AB}$ respectively. Let $k$ be the circumcircle of the triangle $ABC$.

Notice that from $OT \perp AA'$ it follows that $OT$ is the bisector of the chord $AA'$ of the circle $k$ so $|AT| = |A'T|$. Since $AA_1$ is the median, it follows that $|AT| = 2|A_1T|$, and then from $|A'T| = 2|A_1T|$ follows $|A'A_1| = |A_1T|$. Now we can see that the point $A_1$ bisects the segments $BC$ and $A'T$ so the quadrilateral $BA'CT$ is a parallelogram.
From $TC \parallel BA'$ follows that $CC_1$ is the midline of the triangle $ABD$ so $|AD| = 2|AC|$. Analogously, from $TB \parallel CA'$ follows that $BB_1$ is the midline of the triangle $AEC$ so $|AE| = 2|AB|$.
Now we can see that the homothety with ratio 2 and center $A$ sends the triangle $ABC$ into the triangle $AED$. It also sends the circumcenter $O$ of the triangle $ABC$ into the circumcenter $S$ of the triangle $AED$. The point $S$ lies on the ray $AO$ and we know that $|AS| = 2|AO|$, so $\overline{AS}$ is the diameter of the circle $k$.