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 $II_1$ and $WW_1$ on line $BC$ (Fig. 14). It is known that
$$BW_1 = \frac{1}{2}(AB + BC) = AC, \quad \text{and}$$
$$BI_1 = \frac{1}{2}(AB + BC - AC) = \frac{1}{2}AC, \quad \text{hence, } BI_1 = I_1W_1 \quad \text{and}$$
$$BI = WI.$$

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.
$$WI^2 = WM \cdot WK. \quad \text{Then,}$$
$$\Delta WIM \sim \Delta KIW, \quad \text{which means}$$
$$\angle WMI = \angle WIK \Rightarrow \angle BKI = \angle IMA.$$
Now, in $\triangle BTW$, $TI$ is altitude and median, which makes $\triangle BTW$ isosceles, and so
$$\angle TBI = \angle TWI = \angle TMI = \angle BKI,$$
which yields the statement of the problem.