Example 2 In an isosceles right $\triangle ABC$, $AC=BC=1$, $M$ is the midpoint of $BC$, $CE \perp AM$ at point $E$, intersects $AB$ at point $F$. Then $S_{\triangle BMF}=$ $\qquad$ .
From Figure 2, by the conclusion of Example 1, we have
$$
\begin{aligned}
A F & =2 B F \\
\Rightarrow & S_{\triangle B M F} \\
& =\frac{1}{3} S_{\triangle B M} .
\end{aligned}
$$

Since $M$ is the midpoint of $B C$, then
$$
S_{\triangle A B M}=\frac{1}{2} S_{\triangle A B C} .
$$

Therefore, $S_{\triangle B M F}=\frac{1}{3} S_{\triangle A B M}=\frac{1}{6} S_{\triangle A B C}=\frac{1}{12}$.