Example 2 In an isosceles right triangle $\triangle ABC$, $AC=BC=1$, $M$ is the midpoint of side $BC$, $CE \perp AM$ at point $E$, and intersects $AB$ at point $F$. Then $S_{\triangle MBF}=$ $\qquad$
(2006, National Junior High School Mathematics League)
Solve As shown in Figure 2, construct square $C A D B$, and extend $C F$ to intersect $B D$ at point $G$. It is easy to see that,
Rt $\triangle A C M$
$\cong \mathrm{Rt} \triangle C B G$.
Since $M$ is the midpoint of side $B C$,
$$
\begin{array}{c}
B G=B M=C M . \\
\text { By } \triangle B G F \backsim \triangle A C F \\
\Rightarrow \frac{A F}{B F}=\frac{A C}{B G}=\frac{1}{2} .
\end{array}
$$

Therefore, $S_{\triangle M B F}=\frac{1}{2} S_{\triangle B C F}=\frac{1}{2} \times \frac{1}{3} S_{\triangle A B C}=\frac{1}{12}$.