# Tangency in the Mixtilinear Incircle Configuration

Let $ABC$ be a triangle with incircle $\gamma$ and let $T$ denote the tangency point of the $A$-mixtilinear incircle* with the circumcircle. If $T'$ denotes the reflection of $T$ over $\overline{BC}$, prove that $\gamma$ is internally tangent to the nine-point circle* of $\triangle{BT'C}$.

The $A$-mixtilinear incircle of $\triangle{ABC}$ is the circle tangent to $\overline{BA}, \overline{CA}$, and internally tangent to the circumcircle.

The nine-point circle of $\triangle{ABC}$ is the circle passing through the midpoints of the sides, the feet of the altitudes, and the midpoints of $\overline{AH}, \overline{BH}, \overline{CH}$ where $H$ is the orthocenter.

I have no ideas of how to approach this problem. I have tried inverting about $\gamma$ as $T$ becomes the midpoint of $\overline{DH}$ where $\gamma$ touches $\overline{BC}$ at $D$ and $H$ is the orthocenter of the intouch-triangle. I would appreciate any help.