58th Ukrainian National Mathematical Olympiad

Without loss of generality, suppose $AB<BC$. Let angular bisector of $\angle ABC$ intersect the circumcircle of the triangle second time in point $W$, and let point $M$ be the middle of $AC$. Let us prove that $BI=IW$. For that, from $I$ and $W$, we draw perpendiculars $I{I}_1$ and $W{W}_1$ on line $BC$ (Fig. 14). It is known that $$B{W}_1=\frac{1}{2}(AB+BC)=AC,\text{and}$$ $$B{I}_1=\frac{1}{2}(AB+BC-AC)=\frac{1}{2}AC,\text{hence, }B{I}_1=I_1{W}_1\text{and}$$ $$BI=WI.$$

Notice also that quadrilateral $TIMW$ is inscribed with diameter $TW$. Let us show that $\angle BKI=\angle IMA$. From the trillium theorem, $WA=WC=WI$, i.e. $$W{I}^2=WM\cdot WK.\text{Then,}$$ $$\Delta WIM\sim \Delta KIW,\text{which means}$$ $$\angle WMI=\angle WIK\Rightarrow \angle BKI=\angle IMA.$$ Now, in $△BTW$, $TI$ is altitude and median, which makes $△BTW$ isosceles, and so $$\angle TBI=\angle TWI=\angle TMI=\angle BKI,$$ which yields the statement of the problem.