Task "Inversion" (geometry with many circles) Incircle $\omega$ of triangle $ABC$ with center in point $I$ touches  $AB, BC, CA$ in points $C_{1}, A_{1}, B_{1}$. Сircumcircle of triangle $AB_{1}C_{1}$ intersects second time circumcircle of $ABC$ in point $K$. Point $M$ is midpoint of $BC$, $L$ midpoint of $B_{1}C_{1}$. Сircumcircle of triangle $KA_{1}M$ intersects second time $\omega$ in point $T$.
Prove, that сircumcircles of triangle $KLT$ and triangle $LIM$ touch.


HINT for a less tour-de-force approach (impressive though MvG, or his software, may be)
The obvious idea is to invert (as the question hints). The first thing to try is clearly the incircle. Denote the inverse point of $X$ as $X'$. So the line $B_1C_1$ becomes the circle $AB_1C_1$ and hence $L'=A$. Like $A'=L$, $K'$ lies on the line $B_1C_1$ and the inverse of the circumcircle, which is the circumcircle of the three midpoints of the sides of $A_1B_1C_1$. $T$ lies on the incircle, so $T'=T$.
Thus the inverse of the circle $KLT$ is the circle $K'AT$ and the inverse of the circle $LIM$ is the line $AM'$. So we need to show that $AM'$ is tangent to the circle $K'AT$.
