Answer:
$$
\sqrt{111}
$$

Solution: Let $M$ be the midpoint of $B C$, and consider dilating about $M$ with ratio $-\frac{1}{3}$. This takes $B$ to $E, C$ to $D$, and $A$ to some point $A^{\prime}$ on $A M$ with $A M=3 A^{\prime} M$. Then the angle condition implies $\angle D A E+\angle E A^{\prime} D=180^{\circ}$, so $A D A^{\prime} E$ is cyclic. Then by power of a point, we get
$$
\frac{A M^{2}}{3}=A M \cdot A^{\prime} M=D M \cdot E M=\frac{B C^{2}}{36} .
$$

But we also know $A M^{2}=\frac{2 A B^{2}+2 A C^{2}-B C^{2}}{4}$, so we have $\frac{2 A B^{2}+2 A C^{2}-B C^{2}}{12}=\frac{B C^{2}}{36}$, which rearranges to $B C^{2}=\frac{3}{2}\left(A B^{2}+A C^{2}\right)$. Plugging in the values for $A B$ and $A C$ gives $B C=\sqrt{111}$.
8. Triangle $A B C$ with $\angle B A C>90^{\circ}$ has $A B=5$ and $A C=7$. Points $D$ and $E$ lie on segment $B C$ such that $B D=D E=E C$. If $\angle B A C+\angle D A E=180^{\circ}$, compute $B C$.