8. In $\triangle A B C$, it is known that $A B=17, A C=10, D$ is a point on $B C$, $A D=12$, and both $B D$ and $D C$ are integers. Then the number of all possible values of $B C$ is ( ).
(A) 21
(B) 4
(C) 2
(D) 0
8. D

Let $B D=x, D C=y$. In $\triangle A B D$ and $\triangle A D C$, by the cosine rule we get $\frac{144+x^{2}-289}{x}=-\frac{144+y^{2}-100}{y}$. Simplifying, we have $(x y+44)(x+y)=7 \cdot 3^{3} \cdot y$.

From $A B+B C>B C$ and $A B-B C<B C$, we get $7<B C<27$, i.e., $7<x+y<27$. Thus, there are only four possibilities: $x+y=9, x+y=21, x+y=3 y, x+y=9 y$. Considering each of these possibilities one by one, equation (1) has no integer solutions in all cases, hence $B C$ does not exist.