The medians $AA_0, BB_0$, and $CC_0$ of the acute-angled triangle $ABC$ intersect at the point $M$, and heights $AA_1, BB_1$ and $CC_1$ at point $H$. Tangent to the circumscribed circle of triangle $A_1B_1C_1$ at $C_1$ intersects the line $A_0B_0$ at the point $C'$. Points $A'$ and $B'$ are defined similarly. Prove that $A', B'$ and $C'$ lie on one line perpendicular to the line $MH$.
The circumcircle $\omega$ of $A_1B_1C_1$ is the ninepoint-circle of $ABC$ and thus $A_0B_0C_0A_1B_1C_1$ is cyclic. Clearly the line $A_0B_0$ is the perpendicular bisector of $CC_1$ and thus when we reflect $\omega$ across $A_0B_0$ the result will be the circumcircle of $CA_0B_0$ where the tangent to $C_1$ is sent to the tangent $l$ to $(CB_0A_0)$ at $C$. Hence we can redefine $C'$ as $l\cap A_0B_0$. A homothety with center $C$ and factor $2$ sends $(CA_0B_0)$ to $(ABC)$ implying that $l$ is the radical axis of the two circles. We also have that $A_0B_0$ is the radical axis of $\omega$ and $(CA_0B_0)$ giving us that $C'$ will be the radical center of the three previously mentioned circles and thus also on the radical axis $m$ of $(ABC)$ and $\omega$. By symmetry $B'$ and $A'$ lies on the line $m$ as well. To conclude, notice that $m \perp ON$ where $O$ and $N$ are the centers of $(ABC)$ and $\omega$, but since $O,N,H$ and $M$ are collinear on the Euler line of $ABC$ we are done.