17.87 As shown in the figure, in $\triangle A B C$, $A B=4, A C=8$. If $M$ is the midpoint of $B C$, $A M=3$, the length of $B C$ is
(A) $2 \sqrt{26}$.
(B) $2 \sqrt{31}$.
(C) 9 .
(D) $4+2 \sqrt{13}$.
(E) The given conditions are insufficient to solve this problem.
[Solution 1] Let $h$ be the length of the altitude $AN$ to side $BC$, and let $x = BM, y = NM$. Then we have
$$
\left\{\begin{array}{l}
h^{2}+(x+y)^{2}=64, \\
h^{2}+y^{2}=9, \\
h^{2}+(x-y)^{2}=16 .
\end{array}\right.
$$

From (1) + (3) - 2 × (2) we get $2 x^{2}=62$, solving for $x$ gives $x=\sqrt{31}, BC=2 \sqrt{31}$.
Therefore, the answer is $(B)$.
[Solution 2] Extend $AM$ to $D$ such that $AM = MD$, and connect $BD, CD$, then $ABDC$ is a parallelogram, hence

that is
$$
\begin{array}{c}
AD^{2} + BC^{2} = 2(AB^{2} + AC^{2}), \\
BC^{2} = 2(4^{2} + 8^{2}) - 6^{2} = 124 .
\end{array}
$$

Therefore,
$$
BC = 2 \sqrt{31} \text{. }
$$

Hence, the answer is $(B)$.