Let $E_1$ be the reflection of $E$ with respect to the midpoint of $BC$ and $X$ the intersection point of $AE_1$ and $EA_1$. We claim that $IX \parallel BC$. For this reason, we have (Suppose that $R$ is the radius of circumcircle of $ABC$ and $\angle B \geq \angle C$).
$$\begin{aligned} \frac{AX}{XE_1} &= \frac{AA_1}{EE_1} = \frac{AA_1}{AC - AB} \\ \frac{AI}{IE} &= \frac{AB}{BD} = \frac{AC}{CD} = \frac{AB + AC}{BC} \end{aligned}$$
So referring to the Thales' Theorem, we must prove $AA_1 \cdot BC = AC^2 - AB^2$.
We have $AA_1 = BC - 2BH = BC - 2AB \cdot \cos(\angle B)$, where $H$ is the foot of perpendicular from $A$ to $BC$ and on the other hand, by The Law of Cosines we have $AC^2 - AB^2 = BC^2 - 2AB \cdot BC \cos(\angle B)$. Therefore, the claim is proved.
Now since the quadrilateral $DEAT$ is cyclic and $AA_1 \parallel IX$, we get that the quadrilateral $IATX$ is cyclic. Also, since pairs $(E, E_1)$ and $(A, A_1)$ are symmetric with respect to the perpendicular bisector of the side $BC$, we have $XE = XE_1$ and so
$$\angle ATI = \angle AXI = \angle XE_1E = \angle XEE_1 = \angle IXE = \angle TAI$$
Thus, $IT = IA$.