过点$F$作$FG\bot AB$于点$G$，
$\because \angle ACB=90^{\circ}$，$CD\bot AB$，
$\therefore \angle CDA=90^{\circ}$，
$\therefore \angle CAF+\angle CFA=90^{\circ}$，$\angle FAD+\angle AED=90^{\circ}$，
$\because AF$平分$\angle CAB$，
$\therefore \angle CAF=\angle FAD$，
$\therefore \angle CFA=\angle AED=\angle CEF$，
$\therefore CE=CF$，
$\because AF$平分$\angle CAB$，$\angle ACF=\angle AGF=90^{\circ}$，
$\therefore FC=FG$，
$\because \angle B=\angle B$，$\angle FGB=\angle ACB=90^{\circ}$，
$\therefore \triangle BFG$∽$\triangle BAC$，
$\therefore \frac{BF}{AB}=\frac{FG}{AC}$，
$\because AC=3$，$AB=5$，$\angle ACB=90^{\circ}$，
$\therefore BC=4$，
$\therefore \frac{4-FC}{5}=\frac{FG}{3}$，
$\because FC=FG$，
$\therefore \frac{4-FC}{5}=\frac{FC}{3}$，
解得：$FC=\frac{3}{2}$，
即$CE$的长为$\frac{3}{2}$.
故答案为：$\frac{3}{2}$

过点$F$作$FG\bot AB$于点$G$，
$\because \angle ACB=90^{\circ}$，$CD\bot AB$，
$\therefore \angle CDA=90^{\circ}$，
$\therefore \angle CAF+\angle CFA=90^{\circ}$，$\angle FAD+\angle AED=90^{\circ}$，
$\because AF$平分$\angle CAB$，
$\therefore \angle CAF=\angle FAD$，
$\therefore \angle CFA=\angle AED=\angle CEF$，
$\therefore CE=CF$，
$\because AF$平分$\angle CAB$，$\angle ACF=\angle AGF=90^{\circ}$，
$\therefore FC=FG$，
$\because \angle B=\angle B$，$\angle FGB=\angle ACB=90^{\circ}$，
$\therefore \triangle BFG$∽$\triangle BAC$，
$\therefore \frac{BF}{AB}=\frac{FG}{AC}$，
$\because AC=3$，$AB=5$，$\angle ACB=90^{\circ}$，
$\therefore BC=4$，
$\therefore \frac{4-FC}{5}=\frac{FG}{3}$，
$\because FC=FG$，
$\therefore \frac{4-FC}{5}=\frac{FC}{3}$，
解得：$FC=\frac{3}{2}$，
即$CE$的长为$\frac{3}{2}$.
故答案为：$\frac{3}{2}$