, 2023

Let $\omega, \omega_c, \omega_b, \Omega$ be the circumcircles of triangles $ABC, ABC', AB'C, AB'C'$ respectively. Denote $O_c, O_b$ as the circumcenters of $\omega_c, \omega_b$ and $R$ as the radius of $\omega$. Let $f(X)$ be the reflection of the figure $X$ over the line $AB$.

It is easy to check that $f(\Delta ABC) = \Delta A'BC$, so $f(\omega) = \omega_c$ and $f(O) = O_c$. Since $C'H$ is the common chord of $\omega_c, \Omega$, hence $KO_c$ is the perpendicular bisector of $HC'$. Now the homothety of center $O$ and ratio $2$ will send $AB$ to a parallel line through $O_c$, which is $O_cK$ and also send $AC$ to $O_bK$ as well. So this homothety sends $A$ to $K$, which implies that $KA = AO = R$.

Let $BO_c$ intersects $\omega_c$ at $S'$ then $\angle BAS' = 90^\circ$. We will prove that $S' \equiv S$. Since $AO_cBO$ is the rhombus, hence $KO \parallel S'B$. But $KO = S'B = 2R$, hence $KS'BO$ is a parallelogram, then $OS'$ passes through $D$, implies that $S' \equiv S$. Finally, we have $SK = OK = R = KA$ which finishes the solution. $\square$