In the non-isosceles $\vartriangle ABC$, the interior bisectors of vertices $B$ and $C$ are drawn, which cut the sides $AC$ and $AB$ at $E$ and $F$ respectively.The line $EF$ cuts the extension of side $BC$ at $T$. In the side$ BC$ a point D is located, so that $\frac{DB}{DC} = \frac{TB}{TC}$. Prove that $AT$ is the exterior bisector of angle $A$.