AUT_ABooklet_2023

Let $\angle ACB=\gamma$ and $F$ be the foot of $C$ on $MH$. We will first demonstrate that $F$ lies on the circumcircle $k$ of triangle $ABC$. Let $H_1$ denote the symmetric point to $H$ with respect to $M$. The quadrilateral $A{H}_{1}BH$ is a parallelogram, and since we have $\angle AHB=\angle A{H}_{1}B=180^{\circ }-\gamma$, the point $H_1$ must lie on the circumcircle $k$ of $ABC$. Reflecting point $H$ on triangle side $AB$ yields point $H_2$, and it is well known that this point also lies on $k$. The line $H_1{H}_2$ is parallel to $AB$, and thus perpendicular to $C{H}_2$. It follows that $C{H}_1$ is a diameter of the circumcircle $k$, and it follows that $F$ lies on $k$. In summary, we have: Points $A,B,D,E$ lie on a common circle $k_1$. Points $C,E,H,D,F$ lie on a common circle $k_2$. * Points $A,B,F,C$ lie on the circumcircle $k$. The point $S$ is thus the radical center of these three circles, completing the proof. (Karl Czakler, Josef Greilhuber) ☐