Lol! The answer is $\frac{B-C}{2}$ assuming $\angle B\geq \angle C$.Reflect $D$ in $T$ to get $A-HM$ point $X$ Also let $L$ be the foot of the A-external angle bisector.By a famous lemma,$ALKX$ is cyclic so we get that $\angle FDA=\angle KXT=\angle ALB$.Done $\blacksquare$

mmathss wrote: Lol! The answer is $\frac{B-C}{2}$ assuming $\angle B\geq \angle C$.Reflect $D$ in $T$ to get $A-HM$ point $X$ Also let $L$ be the foot of the A-external angle bisector.By a famous lemma,$ALKX$ is cyclic so we get that $\angle FDA=\angle KXT=\angle ALB$.Done $\blacksquare$ Dont you think the answer is $B-C$ ?

Let $M$ be the midpoint of arc $BC$ By butterfly theorem $MT$ and $DF$ pass through the midpoint of arc $BAC$. So $\angle FDA= \frac{\angle B-\angle C}{2}$

Let $N , M$ be the midpoints of arc $BAC$ ,arc $BC$ respectively . and $NK$ intersects $(ABC)$ at $G$ . Lemma. $AG$ is $A-symmedian$. proof. since $ANTK$ is cyclic , we're done so $G$ is reflection of $D$ over $NT$ . but $TGMK$ is cyclic so $MTDF$ is cyclic and we get the answer $B-C/2$

Let $M$ be the midpoint of arc $BC$ not containing $A$ Since $\angle MFT = \angle MKT = \angle AKB = \angle KAC + \angle ACB = \angle KAB + \angle ACB = \angle MCB + \angle ACB = \angle MCA = \angle MDA = \angle MDT$, $TFDM$ is cyclic. So, $\angle FDT = \angle FMT = \angle KMT = \angle AMT = \frac{\angle B - \angle C}{2}$