$I$ - incenter , $M$- midpoint of arc $BAC$ of circumcircle, $AL$ - angle bisector of triangle $ABC$. $MI$ intersect circumcircle in $K$. Circumcircle of $AKL$ intersect $BC$ at $L$ and $P$.
Prove that $\angle AIP=90$
Simple. Note that $K$ is the A-mixtilinear touchpoint. Also note that if $N$ is the other arc midpoint of $BC$, and $P'$ is the intersection of KN and $(ABC)$ we have $\measuredangle AKP' = \measuredangle AKN = \measuredangle ALP'$ so $P\equiv P'$ Now the fact that $P'$ is on the mixtilinear touchchord and the fact that $I$ is its midpoint finish off the problem.
Short sketch
Let $N\neq M$ be a midpoint of arc $BC$. Conclude points $T,K,N$ are collinear. By trillium theorem and power of point$PN^2-IN^2=PB\cdot PC=PK\cdot NP\implies PN\cdot NK=IN^2\implies \angle IKN=\angle PIN$. Because $\angle MKN=90^{\circ}$ as $MN$ is a diameter we finish.