I know if I draw all the excircles, this will amount to something...

Nice problem Lemma.$\angle EDF \leq 60^{\circ}$ Proof. Let $I_a$ be $A$-excenter and let $CI_a \cap AB , BI_a \cap AC = X,Y$ respectively. Note that $\angle XAY= \angle XI_aY = 120^{\circ}$ which means $XAI_aY$ is cyclic and clearly $D$ is on segment $AI_a$ which implies that $XDY \geq 120^{\circ}$ .But clearly , by menelaus and angle bisector theorem , $X,D,E$ and $Y,D,F$ are collinear and $\angle EDF \geq 120^{\circ}$ which means the acute one is less than $60^{\circ}$. The lemma quickly implies that $\angle ETF \geq 60^{\circ}$.Also note that $AEIF$ is cyclic which means $IE=IF$ and $IT$ is the perpendicular bisector of $EF$. So since $T,A$ are at the same side of $EF$ , $T$ lies inside $(AEIF)$ .Let $N$ be the midpoint of arc $(BAC)$ in the circle $(ABC)$. Then since $AN$ is the bisector of $\angle EAF$ , $S = IT \cap AN$ lies on $(AEIF)$ and since $T$ was inside $(AEIF)$ , $(I,T,S)$ are collinear in that order. Now it's enough to show that $A,S,N$ are collinear in this order I wrote so that we can get $T$ is inside the circle $(ABC)$ and $\angle BTC \geq 60^{\circ}$.To prove this , assume WLOG $AC > AB$ and by angle bisector theorem $AF = \frac{AB.AC}{AC+BC}$ and similar for $AE$ which implies that $AE \geq AF$. And by cosine law in $\triangle CEN$ and $\triangle BFN$ using the facts that $\angle ACN =\angle ABN = \frac{B-C}{2}$ and $BN=CN = \frac{a}{2sin\frac{A}{2}} = a$ and again angle bisector theorem to compute $CE,BF$ , we easily get that $NE \leq NF$ and we're done.(actually I've done the full detailed cosine law but I'm to lazy to LaTeX all those fraction here )