Solution:

Let $E, F$ be the tangency points of the incircle to sides $AC, AB$, respectively. Due to symmetry around line $AI$, $AXIY$ is a rhombus. Therefore
$$\angle XAI = 2 \angle EAI = 2\left(90^{\circ} - \angle EIA\right) = 180^{\circ} - 2 \angle XAI,$$
which implies that $60^{\circ} = \angle XAI = 2 \angle EAI = \angle BAC$. By the law of cosines,
$$BC^{2} = 8^{2} + 10^{2} - 2 \cdot 8 \cdot 10 \cdot \cos 60^{\circ} = 84$$

Solution 2:

Define points as above and additionally let $P$ and $Q$ be the intersections of $AI$ with $EF$ and $XY$, respectively. Since $IX = 2 IE$ and $IY = 2 IF$, $\triangle IEF \sim \triangle IXY$ with ratio $2$, implying that $IP = \frac{1}{2} IQ = \frac{1}{4} IA$.
Let $\theta = \angle EAI = \angle IEP$. Then $\frac{IP}{IA} = \frac{IP}{IE} \frac{IE}{IA} = \sin^{2} \theta$, implying that $\sin \theta = 1/2$ and $\theta = 30^{\circ}$. From here, proceed as in solution 1.