C

①$\because DC$是$\angle ACB$的外角平分线，$DE\bot AC$，$DF\bot BC$，
$\therefore DE=DF$，$\angle F=\angle AED=90^{\circ}$，
$\because D$在$AB$的垂直平分线上，
$\therefore AD=BD$，
$\therefore Rt\triangle ADE$≌$Rt\triangle BDF\left(HL\right)$，
故①正确；
②$\angle EMD=\angle MAD+\angle MDA$，
$\because \angle BAC=\angle MDA$，
$\therefore \angle EMD=\angle MAD+\angle BAC=\angle DAB$，
$\because AD=BD$，$DM=CD$，
$\therefore \angle ABD=\angle DAB$，$\angle CMD=\angle MCD$，
$\therefore \angle MCD=\angle ABD$，
$\because \angle DCF=\angle MCD$，
$\therefore \angle FCD=\angle ABD$，
$\therefore \angle ECF+\angle FCD=\angle ABD+\angle FCD$，
不能得出$\angle DCF=90^{\circ}-\frac{1}{2}\angle BDA$，
故②错误；
③$\because DM=DC$，
$\therefore \angle DMC=\angle DCM=\angle DCF$，
$\because \angle ACB+\angle ECD+\angle DCF=180^{\circ}$，$\angle DMC+\angle DCM+\angle MDC=180^{\circ}$，
$\therefore \angle MDC=\angle ACB$，
$\because \angle ADM=\angle BDC$，
$\therefore \angle ADB=\angle ADM+\angle MDB=\angle MDB+\angle CDB=\angle MDC$，
$\therefore \angle ADB=\angle ACB$；
故③正确；
④在$EA$上截取$EM=EC$，
$\because DE\bot AC$，$\angle MDE=\angle CDE$，
$\therefore DM=DC$，
$\because \angle CDE=\angle CDF$，
$\therefore \angle CDF=\angle EDM$，
$\because Rt\triangle ADE$≌$Rt\triangle BDF$，
$\therefore \angle DAM=\angle DBC$，$\angle ADE=\angle BDF$，
$\therefore \angle ADM=\angle CDB$，
$\therefore \triangle AMD$≌$\triangle ACD\left(AAS\right)$，
$\therefore AM=BC$，
$\therefore AE=AM+ME=BC+EC$，
$\therefore AE=10=BC+EC=4+6$，
故④正确.
故选：$C$.