Let $I_A, I_B, I_C$ be the centers of excircles, $I_1$ be the point, symmetrical to $I$ with respect to point $O$ (fig. 29). Let us look at $\Delta I_A I_B I_C$. $I$ is its orthocenter, since bisectors of outer and inner angles are perpendicular. It is also clear that $O$ is the center of Euler's circle of this triangle. Then, point $I_1$ is the center point of circumscribed circle of $\Delta I_A I_B I_C$. Then,

$\angle I_1 I_B C = \angle I_C I_B B$, which yields $I_1 I_B \perp AC$, since $\angle B I_C A = 90^\circ - \frac{1}{2} \angle BCA = \angle AC I_B$.

Hence, lines $I_B T_B$, $I_A T_A$ and $I_C T_C$ intersect at point $I_1$.

Clearly, $AII_1C$ is a trapezoid. Points $I, I_1$ are symmetrical with respect to the median perpendicular of $AC$, since $II_1 \parallel AC$. Then, $AII_1C$ is an isosceles trapezoid, and $\angle A I_1 C = \angle A I C = 90^\circ + \frac{1}{2} \angle ABC$. Let us also note that quadrilaterals $I_1 T_C A T_B$ and $I_1 T_A C T_B$ are inscribed. Then,

$$\angle T_A T_B T_C = \angle T_C T_B I_1 + \angle I_1 T_B T_A = \angle I_1 A B + \angle I_1 C B = \angle A I_1 AC - \angle ABC = 90^\circ - \frac{\angle ABC}{2}.$$