3. As shown in Figure 8, point $C$ is on line segment $A B$, $D A \perp A B, E B \perp$ $A B, F C \perp A B$, and $D A=$ $B C, E B=A C, F C=A B$, $\angle A F B=51^{\circ}$. Then $\angle D F E=$ $\qquad$
3. $39^{\circ}$.

As shown in Figure 15, connect $A E$ and $B D$. Then
$\triangle A B D \cong \triangle C F B$,
$\triangle A B E \cong \triangle F C A$.
Therefore, $D B=F B$,
$$
A E=A F \text {. }
$$

Thus, $\triangle D B F$ and $\triangle E A F$ are both
isosceles right triangles. Hence,
$$
\angle B F D=45^{\circ}, \angle A F E=45^{\circ} \text {. }
$$

Since $\angle A F B=51^{\circ}$, we have,
$$
\angle E F B=6^{\circ}, \angle D F A=6^{\circ} \text {. }
$$

Therefore, $\angle D F E=\angle A F B-\angle E F B-\angle D F A=39^{\circ}$.