Circles $\omega_1$ and $\omega_2$ meet at points $A$ and $B$. Let points $K_1$ and $K_2 $ of $\omega_1$ and $\omega_2$ respectively be such that $K_1A$ touches $\omega_2$, and $K_2A$ touches $\omega_1$. The circumcircle of triangle $K_1BK_2$ meets lines $AK_1$ and $AK_2$ for the second time at points $L_1$ and $L_2$ respectively. Prove that $L_1$ and $L_2$ are equidistant from line $AB$.