, 2020

Let $D$ be a point on $BC$ such that $|CD| = |AC|$. Because $|BC| = |AC| + |AI|$, the point $D$ lies on the interior of side $BC$, and we have $|BD| = |AI|$. Because triangle $ACD$ is isosceles, the angle bisector $CI$ is also the perpendicular bisector of $AD$, hence $A$ is the reflection of $D$ in $CI$. Hence, we get $\angle CDI = \angle CAI = \angle IAB$, hence $180^\circ - \angle BDI = \angle IAB$, which means that quadrilateral $ABDI$ is cyclic. In this cyclic quadrilateral $BD$ and $AI$ have the same length. Therefore, $AB$ and $ID$ are parallel. Hence, $ABDI$ is an isosceles trapezium, which has equal angles at the base. Hence, $\angle CBA = \angle DBA = \angle BAI = \frac{1}{2}\angle BAC$, which proves the statement. $\square$