Problem 1389

1. Let $\alpha = \angle CAI = \angle BAI$ and $\beta = \angle ABI = \angle CBI$ for convenience.
2. Since $AB = AC$, $\triangle ABC$ is isosceles with $\angle BAC = 2\alpha$.
3. The sum of the angles in $\triangle ABC$ is $180^\circ$, so:
$$\angle ABC + \angle ACB + \angle BAC = 180^\circ$$
Substituting $\angle BAC = 2\alpha$:
$$\beta + \beta + 2\alpha = 180^\circ$$
Simplifying:
$$2\beta + 2\alpha = 180^\circ \implies \beta + \alpha = 90^\circ$$
4. Now consider quadrilateral $ACBI$. The sum of the interior angles of a quadrilateral is $360^\circ$.
5. The angles in quadrilateral $ACBI$ are $\angle ACB$, $\angle ABI$, $\angle BAI$, and $\angle BIA$.
6. We know:
$$\angle ACB = 180^\circ - 2\alpha - 2\beta$$
Substituting $\beta + \alpha = 90^\circ$:
$$\angle ACB = 180^\circ - 2(90^\circ) = 180^\circ - 180^\circ = 0^\circ$$
This is incorrect, so we need to reconsider the angles in the quadrilateral.
7. Instead, consider the sum of the angles around point $I$. Since $I$ is the incenter of $\triangle ABC$, the angles around $I$ sum to $360^\circ$.
8. The reflex angle $\angle AIB$ is:
$$360^\circ - (180^\circ - 2\alpha - 2\beta) - \alpha - \beta = 180^\circ + \alpha + \beta$$
9. The non-reflex angle $\angle AIB$ is:
$$360^\circ - (180^\circ + \alpha + \beta) = 180^\circ - \alpha - \beta$$
10. Substituting $\beta + \alpha = 90^\circ$:
$$\angle AIB = 180^\circ - 90^\circ = 90^\circ$$
11. Therefore:
$$\angle BIA = 90^\circ + \frac{\angle C}{2}$$

The final answer is $\angle BIA = 90^\circ + \frac{\angle C}{2}$