6. As shown in Figure 3, given that $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively, $OI \perp AI$, $AB=5, AC=9$. Then the length of side $BC$ is ( ).
(A) 8
(B) 7.5
(C) 7
(D) 6.5
6. C.

As shown in Figure 8, extend $AI$ to intersect the circumcircle $\odot O$ of $\triangle ABC$ at point $D$, connect $IC$, $CD$, $OD$, and let $OD$ intersect $BC$ at point $E$. Draw $IF \perp AB$ at point $F$.

Since $I$ is the incenter of $\triangle ABC$, we have
$$
\begin{array}{l}
AF=\frac{1}{2}(AB+AC-BC)=7-\frac{1}{2} BC, \\
OD \perp BC, CE=\frac{1}{2} BC, ID=CD.
\end{array}
$$

By $OI \perp AD$ and the perpendicular chord theorem, we get
$$
AI=ID=CD.
$$

Therefore, in the right triangles $\triangle AFI$ and $\triangle CDE$, we have $\triangle AFI \cong \triangle CDE$.
Thus, $AF=CE$, which means $7-\frac{1}{2} BC=\frac{1}{2} BC$.
Hence, $BC=7$.