Define a point $D$ on side $B C$ such that $|C D|=|A C|$. Given that $|B C|=|A C|+|A I|$, $D$ lies internally on side $B C$ and $|B D|=|A I|$. Since triangle $A C D$ is now isosceles, bisector $C I$ is also the perpendicular bisector of $A D$, so $A$ and $D$ are each other's reflection in $C I$. This implies $\angle C D I=\angle C A I=\angle I A B$, so $180^{\circ}-\angle B D I=\angle I A B$, which means that $A B D I$ is a cyclic quadrilateral. In this cyclic quadrilateral, $B D$ and $A I$ are of equal length. According to Julian's theorem, $A B$ and $I D$ are then parallel. Therefore, $A B D I$ is an isosceles trapezoid and the base angles of this trapezoid are equal. Thus, $\angle C B A=\angle D B A=\angle B A I=\frac{1}{2} \angle B A C$, from which the desired result follows directly.