IMO 2019 Shortlisted Problems

Notice that $\angle TFB=\angle FDA$ because $FT$ is tangent to circle $BDF$, and moreover $\angle FDA=\angle CGA$ because quadrilateral $ADFG$ is cyclic. Similarly, $\angle TGB=\angle GEC$ because $GT$ is tangent to circle $CEG$, and $\angle GEC=\angle CFA$. Hence, $$\begin{array}{c}\angle TFB=\angle CGA\text{and}\angle TGB=\angle CFA.\end{array}\text{\hspace{1em}(1)}$$ Triangles $FGA$ and $GFT$ have a common side $FG$, and by (1) their angles at $F,G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$, respectively, are equal and hence $AT$ is parallel to line $BFGC$.