Solution:

Let $M$ be the midpoint of $BC$, and consider dilating about $M$ with ratio $-\frac{1}{3}$. This takes $B$ to $E$, $C$ to $D$, and $A$ to some point $A'$ on $AM$ with $AM = 3A'M$. Then the angle condition implies $\angle DAE + \angle EA'D = 180^{\circ}$, so $AD A' E$ is cyclic. Then by power of a point, we get
$$\frac{AM^2}{3} = AM \cdot A'M = DM \cdot EM = \frac{BC^2}{36}$$
But we also know $AM^2 = \frac{2AB^2 + 2AC^2 - BC^2}{4}$, so we have $\frac{2AB^2 + 2AC^2 - BC^2}{12} = \frac{BC^2}{36}$, which rearranges to $BC^2 = \frac{3}{2}(AB^2 + AC^2)$. Plugging in the values for $AB$ and $AC$ gives $BC = \sqrt{111}$.