11. As shown in the figure, in isosceles right $\triangle A B C$, $A B=1, \angle A=90^{\circ}$, point $E$ is the midpoint of leg $A C$, and point $F$ is on the base $B C$ such that $F E \perp B E$. Find the area of $\triangle C E F$.
Three, 11. Construct
(D) $\perp C E$ and the extension of $E F$ intersect at $\Gamma$.
$$
\begin{array}{l} 
\because \angle A B E+ \\
\angle A E B=90^{\circ}, \\
\angle C E D+\angle A E B \\
=90^{\circ}, \\
\angle A B E=\angle C E D .
\end{array}
$$

Therefore, Rt $\triangle A B E \sim R \mathrm{R} \triangle C E D$.
$$
\begin{array}{l}
\therefore \frac{S_{\triangle C D E}}{S_{\triangle E A B}}=\left(\frac{C E}{A B}\right)^{2}=\frac{1}{4} . \\
\frac{C E}{C D}=\frac{A B}{A E}=2 .
\end{array}
$$

Also, $\angle E C F=\angle D C F=45^{\circ}$, so $C F$ is the angle bisector of $\angle D C E$, and the distances from point $F$ to $C E$ and $C D$ are equal.
$$
\begin{aligned}
\therefore \frac{S_{\triangle C F F}}{S_{\triangle C F F}} & =\frac{C E}{C D}=2 . \\
\therefore S_{\triangle C Y F} & =\frac{2}{3} S_{\triangle C F E}=\frac{2}{3} \times \frac{1}{4} S_{\triangle A B E} \\
& =\frac{2}{3} \times \frac{1}{4} \times \frac{1}{2} S_{\triangle A B C}=\frac{1}{24} .
\end{aligned}
$$