Problem

Source: Iranian 3rd round Geometry exam P4 - 2014

Tags: geometry, circumcircle, incenter, parallelogram, rectangle, Iran



$D$ is an arbitrary point lying on side $BC$ of $\triangle{ABC}$. Circle $\omega_1$ is tangent to segments $AD$ , $BD$ and the circumcircle of $\triangle{ABC}$ and circle $\omega_2$ is tangent to segments $AD$ , $CD$ and the circumcircle of $\triangle{ABC}$. Let $X$ and $Y$ be the intersection points of $\omega_1$ and $\omega_2$ with $BC$ respectively and take $M$ as the midpoint of $XY$. Let $T$ be the midpoint of arc $BC$ which does not contain $A$. If $I$ is the incenter of $\triangle{ABC}$, prove that $TM$ goes through the midpoint of $ID$.