8. As shown in Figure 1, in the right $\triangle ABC$, it is known that $\angle ACB$ $=90^{\circ}, AC=21, BC=$ 28, and a square $ABDE$ is constructed outward with $AB$ as the side. The angle bisector of $\angle ACB$ intersects $DE$ at point $F$. Then the length of segment $DF$ 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, $DF = 15$.