3. In $\triangle A B C$, $\angle B A C=90^{\circ}$, points $D$ and $E$ are on the hypotenuse $B C$, satisfying $C D=C A, B E=B A$. $F$ is a point inside $\triangle A B C$ such that $\triangle D E F$ is an isosceles right triangle with $D E$ as the hypotenuse. Find $\angle B F C$.
3. First, prove that $F$ is the incenter of $\triangle A B C$. As shown in Figure 1, let $M$ be the midpoint of segment $D E$.
Let $B C=a, C A=b, A B=c$. Then
$$
\begin{aligned}
D E & =B E+C D-B C=b+c-a \\
\Rightarrow & B M=B D+D M \\
& =(a-b)+\frac{1}{2}(b+c-a) \\
& =\frac{1}{2}(a+c-b)
\end{aligned}
$$
$\Rightarrow M$ is the point where the incircle of $\triangle A B C$ touches side $B C$.
From the fact that $\triangle D E F$ is an isosceles right triangle with $D E$ as the hypotenuse, we have
$$
M F=M D=\frac{1}{2} D E=\frac{1}{2}(b+c-a),
$$

which is exactly the inradius of the right triangle $\triangle A B C$.
Thus, $F$ is the incenter of $\triangle A B C$.
Therefore, $\angle B F C=180^{\circ}-\frac{1}{2} \angle A B C-\frac{1}{2} \angle A C B$
$$
=135^{\circ} \text {. }
$$