Example 2 As shown in Figure 6, in isosceles right $\triangle A B C$, it is known that $A B=1, \angle A=$ $90^{\circ}, E$ is the midpoint of $A C$, and point $F$ is on the base $B C$, with $F E$ $\perp B E$. Find the area of $\triangle C E F$.
(1998, National Junior High School Mathematics Competition)
Solution: As shown in Figure 6, draw $C D \perp C E$ intersecting the extension of $E F$ at point $D$.
Since $\angle A B E + \angle A E B = 90^{\circ}$,
$\angle C E D + \angle A E B = 90^{\circ}$,
thus, $\angle A B E = \angle C E D$.
Therefore, Rt $\triangle A B E \backsim \mathrm{Rt} \triangle C E D$.
Hence, $\frac{S_{\triangle C D E}}{S_{\triangle A E B}} = \left(\frac{C E}{A B}\right)^{2} = \frac{1}{4}$, and $\frac{C E}{C D} = \frac{A B}{A E} = 2$.
Also, $\angle E C F = \angle D C F = 45^{\circ}$, so $C F$ is the angle bisector of $\angle D C E$. Thus, the distances from point $F$ to $C E$ and $C D$ are equal. Therefore,
$$
\frac{S_{\triangle C E F}}{S_{\triangle C D F}} = \frac{C E}{C D} = 2.
$$

Hence, $S_{\triangle C E F} = \frac{2}{3} S_{\triangle C D E} = \frac{2}{3} \times \frac{1}{4} S_{\triangle A F B}$
$$
= \frac{2}{3} \times \frac{1}{4} \times \frac{1}{2} S_{\triangle A B C} = \frac{1}{24}.
$$