Let $K'$ be the midpoint of arc$AC$ not containing $B$. By ptolemy's theorem :$2AC=AB+BC \Rightarrow I $is the midpoint of $BK'$. Note that $TI$ is tangent to $(AIC) $, if we let $X$ to be the intersection of $(ABC) $ and $B$_mixtillinear incircle then it is wellknown that $K'X$ passes through $T$. Note that "$TB$ is tangent to$ (KBI) $" is equivalent to prove that $KI$ is tangent to $(TIB) $. Let $M$ be the midpoint of $TB$ then it suffices to prove that $MI$ is perpendicular to $KI$ which is clear since $MI$ is parallel to $K'X$.