A

$\because \angle C=90^{\circ}$，$\therefore DC\bot AC$，
又$AD$平分$\angle CAB$交$BC$于$D$，$DE\bot AB$，
$\therefore CD=ED$，
在$Rt\triangle ACD$和$Rt\triangle AED$中，
$\left\{\begin{array}{l}{DC=DE}\\{AD=AD}\end{array}\right.$，
$\therefore Rt\triangle ACD$≌$Rt\triangle AED\left(HL\right)$，
$\therefore AC=AE$，又$AC=BC$，
$\therefore AC=AE=BC$，又$AB=6cm$，
$\therefore \triangle DEB$的周长$=DB+BE+ED=DB+CD+BE=BC+BE=AE+EB=AB=6cm$.
故选：$A$.