In triangle $ABC$ with $AB=8$ and $AC=10$, the incenter $I$ is reflected across side $AB$ to point $X$ and across side $AC$ to point $Y$. Given that segment $XY$ bisects $AI$, compute $BC^2$. (The incenter  is the center of the inscribed circle of triangle .)
1. **Reflecting the Incenter:**
   - Given that the incenter \( I \) is reflected across side \( AB \) to point \( X \) and across side \( AC \) to point \( Y \), we know that:
     \[
     \angle XAB = \angle BAI = \angle IAC = \angle CAY = \theta
     \]
     and
     \[
     AX = AI = AY
     \]
     This implies that \( \triangle AYX \) is isosceles with \( AX = AI = AY \).

2. **Properties of Isosceles Triangle:**
   - Since \( \angle XAI = \angle YAI = 2\theta \), we have that \( AI \) bisects \( XY \). This means that quadrilateral \( AYIX \) is a parallelogram because its diagonals bisect each other.

3. **Angles in the Parallelogram:**
   - Because of the properties of parallelograms, we have:
     \[
     \angle XAI = \angle AIY = 2\theta
     \]
     Since \( IY \perp AC \), we know:
     \[
     \angle IAC + \angle AIY = 90^\circ
     \]
     Substituting the known angles:
     \[
     \theta + 2\theta = 90^\circ
     \]
     Solving for \( \theta \):
     \[
     3\theta = 90^\circ \implies \theta = 30^\circ
     \]
     Therefore:
     \[
     \angle BAC = 2\theta = 60^\circ
     \]

4. **Using the Law of Cosines:**
   - To find \( BC^2 \), we use the Law of Cosines in \( \triangle ABC \):
     \[
     BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)
     \]
     Substituting the known values:
     \[
     BC^2 = 8^2 + 10^2 - 2 \cdot 8 \cdot 10 \cdot \cos(60^\circ)
     \]
     Since \( \cos(60^\circ) = \frac{1}{2} \):
     \[
     BC^2 = 64 + 100 - 2 \cdot 8 \cdot 10 \cdot \frac{1}{2}
     \]
     Simplifying:
     \[
     BC^2 = 64 + 100 - 80 = 84
     \]

The final answer is \(\boxed{84}\)