9. The circumradius of $\triangle A B C$ is $R$, and the lines obtained by reflecting $A C$ about $A B$ and $B C$ intersect at point $K$. The radius of the incircle $\odot I$ of $\triangle A C K$ is $r$. Then ( ).
(A) $B I>2 R$
(B) $B I<2 R$
(C) $B I=2 R$
(D) $B I=R+r$
9. C.

As shown in Figure 5, connect $B I$ and $C I$. By the given conditions, it is easy to prove that $\angle B A I = \angle B C I = 90^{\circ}$. Therefore, $B I$ is the diameter of the circumcircle of $\triangle A B C$.