C

如图，过$D$作$DF\bot AC$于点$F$，延长$AC$至$G$，使得$CG=CD$，连接$DG$，

$\therefore \angle DFE=90^{\circ}$，
$\because \angle BAE=\angle DFE=90^{\circ}$，$\angle AEB=\angle DEF$，$BE=DE$，
$\therefore \triangle ABE$≌$\triangle FDE\left(AAS\right)$，
$\therefore AB=DF$，
$\because CG=CD$，
$\therefore \angle G=\angle DCG$，
$\because \angle ACD=2\angle ACB=\angle G+\angle ACG$，
$\therefore \angle G=\angle ACB$，
又$\because DF=AB$，$\angle BAC=\angle DFG=90^{\circ}$，
$\therefore \triangle ABC$≌$\triangle FDG\left(AAS\right)$，
$\therefore AC=FG$，
$\therefore AF=CG=CD=10$，
$\therefore EF=5$，
$\therefore DF^{2}=AD^{2}-AF^{2}=14^{2}-10^{2}=96$，
$\therefore DE^{2}=EF^{2}+DF^{2}=5^{2}+96=121$，
解得$DE=11(舍负)$，
故选：$C$.