8. In Rt $\triangle A B C$, $\angle A C B=90^{\circ}$, the incircle $\odot I$ touches $A C$, $B C$ at points $E$, $F$, and the rays $B I$, $A I$ intersect the line $E F$ at points $M$, $N$. Let $S_{\triangle A I B}=S_{1}$, $S_{\triangle M I N}=S_{2}$. Then the value of $\frac{S_{1}}{S_{2}}$ is $(\quad)$.
(A) $\frac{3}{2}$
(B) 2
(C) $\frac{5}{2}$
(D) 3
8. B.

As shown in Figure 4, connect $A M$, $I E$, and $B N$.
It is easy to see that $\angle A E M = \angle C E F = \angle A I M = 45^{\circ}$
$\Rightarrow A$, $M$, $E$, $I$ are concyclic
$$
\Rightarrow \angle A M I = \angle A E I = 90^{\circ}
$$
$\Rightarrow \triangle A I M$ is an isosceles right triangle
$$
\Rightarrow \frac{I M}{I A} = \frac{\sqrt{2}}{2}.
$$

Similarly, $\frac{I N}{I B} = \frac{\sqrt{2}}{2}$. Therefore, $\frac{S_{1}}{S_{2}} = 2$.