2. In an isosceles right $\triangle ABC$, $\angle C=90^{\circ}$, points $D$, $E$, and $F$ are on sides $AB$, $BC$, and $CA$ respectively, $FD=FE=\frac{AC}{2}$, $\angle DFE=90^{\circ}$. Then $FC: CE: EF=$ $\qquad$ .
2. $3: 4: 5$.
Let $A C=2 b, \angle A F D=\beta$. Then
$$
F D=b, \angle F E C=\beta \text {. }
$$

In $\triangle A F D$, by the Law of Sines, we get
$$
A F=\frac{b \sin \left(45^{\circ}+\beta\right)}{\sin 45^{\circ}}=b(\sin \beta+\cos \beta) \text {. }
$$

Also, $C E=b \cos \beta$, so,
$$
\begin{array}{l}
{[2 b-b(\sin \beta+\cos \beta)]^{2}+b^{2} \cos ^{2} \beta=b^{2}} \\
\Rightarrow[2-(\sin \beta+\cos \beta)]^{2}=\sin ^{2} \beta \\
\Rightarrow 2-\sin \beta-\cos \beta=\sin \beta\left(\text { because } 0^{\circ}<\beta<90^{\circ}\right) \\
\Rightarrow(2-\cos \beta)^{2}=4 \sin ^{2} \beta=4-4 \cos ^{2} \beta \\
\Rightarrow 5 \cos ^{2} \beta-4 \cos \beta=0 \\
\Rightarrow \cos \beta=\frac{4}{5}, \sin \beta=\frac{3}{5} \\
\Rightarrow F C: C E: E F=3: 4: 5 .
\end{array}
$$