59th Ukrainian National Mathematical Olympiad

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, $AI{I}_{1}C$ is a trapezoid. Points $I,I_1$ are symmetrical with respect to the median perpendicular of $AC$, since $I{I}_1\parallel AC$. Then, $AI{I}_{1}C$ is an isosceles trapezoid, and $\angle A{I}_{1}C=\angle AIC=90^{\circ }+\frac{1}{2}\angle ABC$. Let us also note that quadrilaterals $I_1{T}_{C}A{T}_B$ and $I_1{T}_{A}C{T}_B$ are inscribed. Then,

$$\angle T_A{T}_B{T}_C=\angle T_C{T}_B{I}_1+\angle I_1{T}_B{T}_A=\angle I_{1}AB+\angle I_{1}CB=\angle A{I}_{1}AC-\angle ABC=90^{\circ }-\frac{\angle ABC}{2}.$$