1. As shown in the figure, $\angle A C B=90^{\circ}, A C=B C, A D \perp C E$, $B E \perp C E$, with the feet of the perpendiculars being $D, E$ respectively. Given $A D=8$, $B E=3$, then $D E=$ $\qquad$
In Rt $\triangle A C D$ and Rt $\triangle C B E$, $A C=B C, \angle C A D=\angle B C E$, then $\triangle A C D \cong \triangle C B E \Rightarrow C E=A D=8, C D=B E=3 \Rightarrow D E=5$.