SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $G$ be the intersection point of $DX$ and $EF$.



Consider the inversion $I_I^{I{D}^2}$, then the circumcircle of $△ABC$ is sent to the nine-point circle of $△DEF$ and the line $EF$ is sent to the circumcircle of $△AEF$. Hence $I$, $T$, and $G$ are collinear.

Then $AI\parallel GX$ since they are both perpendicular to $EF$. We have $$\angle GTF=\angle ITF=\angle IFG=\angle IAE=\angle FAI=\angle FXG$$ so, quadrilateral $TXGF$ is cyclic, this implies that $$\angle XTF=\angle XGF=90^{\circ }.$$ $\square$