8. As shown in Figure 1, in the right $\triangle A B C$, it is known that $\angle A C B$ $=90^{\circ}, A C=21, B C=$ 28, and a square $A B D E$ is constructed outward with $A B$ as one side. The angle bisector of $\angle A C B$ intersects $D E$ at point $F$. Then the length of line segment $D F$ is $\qquad$
8. 15 .

As shown in Figure 6, let $CF$ intersect $AB$ and $AD$ at points $G$ and $O$ respectively.
Then $\angle BCO = 45^{\circ}$
$= \angle OAB$.
Thus, $O, A, C, B$

are concyclic.
Hence $\angle ABO$
$$
= \angle OCA = 45^{\circ}.
$$

Therefore, $OA = OB$.
Thus, $O$ is the center of the square $ABDE$.
By symmetry, $DF = AG$.
Hence $\frac{AG}{GB} = \frac{AC}{BC} = \frac{21}{28} = \frac{3}{4}$
$\Rightarrow AG = \frac{3}{7} AB = \frac{3}{7} \times 35 = 15$.
Therefore, $DF = 15$.