Radical center and radical axis by power of point kill this problem

I tried this problems many times thinking it's hard until today I found that it's really easy Anyway, here's my solution: From $\angle{A_0B_2A_2}=\angle{B_0C_2B_2}=\angle{C_0A_2C_2}=90^{\circ}$, we get that $A_0A_1B_2A_2,B_0B_1B_2C_2, C_0C_1C_2A_2$ are all concyclic. Let $\omega_A,\omega_B,\omega_C$ denote three circles that circumscribed those three quadrilaterals, respectively. Also, note that $A_0,A_1,B_0,B_1,C_0,C_1$ all lie on the nine-point circle of $\triangle{ABC}$, called it $\omega$. Consider radical axis of $\omega_A,\omega_B,\omega$ gives us $C$ is the radical center of those three circles. Hence, $CB_2$ is the radical axis of $\omega_A,\omega_B$. Similarly, $AC_2$ is the radical axis of $\omega_B,\omega_C$ and $BA_2$ is the radical axis of $\omega_C,\omega_A$. Hence, $AC_2,BA_2,CB_2$ must concurrent at radical center of $\omega_A,\omega_B,\omega_C$, done.

Attachments:

Oops, same as above rightways wrote: Altitudes $AA_1$, $BB_1$, $CC_1$ of an acute triangle $ABC$ meet at $H$. $A_0$, $B_0$, $C_0$ are the midpoints of $BC$, $CA$, $AB$ respectively. Points $A_2$, $B_2$, $C_2$ on the segments $AH$, $BH$, $HC_1$ respectively are such that $\angle A_0B_2A_2 = \angle B_0C_2B_2 = \angle C_0A_2C_2 =90^\circ$. Prove that the lines $AC_2$, $BA_2$, $CB_2$ are concurrent. Observe that $A_0A_1B_2A_2$ is cyclic with diameter $A_0A_2$. Call this circle $\gamma_A$ and define $\gamma_B, \gamma_C$ respectively. Note that $\gamma_A, \gamma_C$ have point $A_2$ as common. Also $BA_0 \cdot BA_1=BC_0 \cdot BC_1$ hence $B$ has equal powers in $\gamma_A, \gamma_C$. Thus, line $BA_2$ is the radical axis of $\gamma_A, \gamma_C$; cyclic relations show that $AC_2, BA_2, CB_2$ concur at the radical center of $\gamma_A, \gamma_B, \gamma_C$.

Trollolol. This is pretty intimidating until you realize what's going on. By the condition of the problem, you can derive that $\omega_A \equiv A_1 A_0 B_2 A_2$ is cyclic. Similarly, $\omega_B \equiv B_0 B_1 B_2 C_2$ and $\omega_C \equiv C_0 C_1 C_2 A_2$ cyclic. However, it is well known that $\omega \equiv A_0 A_1 B_0 B_1 C_0 C_1$ is the nine-point circle. Now, radical axis on $(\omega, \omega_A, \omega_B)$, we would derive that $\text{rad}(\omega_A, \omega_B), \text{rad}(\omega_B, \omega), \text{rad}(\omega_A, \omega)$ concur at the radical center. We then have $\text{rad}(\omega_A, \omega_B)$ passing through the common point of both $\omega_A, \omega_B$, which is $B_2$ and the radical center, which is $\text{rad}(\omega_B, \omega) \cap \text{rad}(\omega_A, \omega) = A_1 A_0 \cap B_1 B_0 = C$. Therefore, $\text{rad}(\omega_A, \omega_B) = CB_2$. Similarly, one could easily derive that $\text{rad}(\omega_B, \omega_C) = AC_2$ and $\text{rad}(\omega_C, \omega_A) = BA_2$. Consider radical axis on $(\omega_A, \omega_B, \omega_C)$ would give us $AC_2, BA_2, CB_2$ concur, then we are finished.

By the nine-point circle theorem $A_0A_1B_0B_2C_0C_1$ is cyclic, let its circumcircle be $\omega$. Obviously $(C_0C_1C_2A_2)$ is cyclic from the angle condition. $B$ lies on the radical axis of $(C_0C_1C_2A_2)$ and $\omega$ and at the same time it lies on the radical axis $(A_0A_1A_2)$ and $\omega$, then by radical center it lies on the radical axis of $(C_0C_1C_2A_2)$ and $(A_0A_1A_2)$ $\implies$ $BA_2$ is the radical axis of $(C_0C_1C_2)$ and $(A_0A_1A_2)$. We get similar results for $C$ and $A$ and hence we're done by the radical center. $\square$