$(1)$证明：如图，过点$D$作$DF\bot AB$于点$F$.

$\therefore \angle DFB=90^{\circ}$，
$\because \angle ACB=90^{\circ}$，
$\therefore \angle DFB=\angle ACB$，
在$\triangle DCE$和$\triangle DFB$中，
$\left\{\begin{array}{l}{∠DCE=∠DFB}{}\\{∠DEC=∠B}{}\\{DE=DB}{}\end{array}\right.$，
$\therefore \triangle DCE$≌$\triangle DFB\left(AAS\right)$，
$\therefore DC=DF$，
$\because DF\bot AB$，$DC\bot AC$，
$\therefore$点$D$在$\angle BAC$的平分线上.
$\therefore AD$平分$\angle BAC$.
$(2)$$AE+AB=2AC$.理由如下：
由（1）知，$AD$平分$\angle BAC$，
$\therefore \angle DAC=\angle DAF$.
在$\triangle ACD$和$\triangle AFD$中，
$\left\{\begin{array}{l}{∠ACD=∠AFD}{}\\{∠DAC=∠DAF}{}\\{DC=DF}{}\end{array}\right.$，
$\therefore \triangle ACD$≌$\triangle AFD\left(AAS\right).$
$\therefore AC=AF$，
由（1）知$,\triangle DCE$≌$\triangle DFB$，
$\therefore CE=FB$.
$\therefore AE+AB=AE+FB+AF=AE+CE+AF=AC+AF=2AC$.

$(1)$证明：如图，过点$D$作$DF\bot AB$于点$F$.

$\therefore \angle DFB=90^{\circ}$，
$\because \angle ACB=90^{\circ}$，
$\therefore \angle DFB=\angle ACB$，
在$\triangle DCE$和$\triangle DFB$中，
$\left\{\begin{array}{l}{∠DCE=∠DFB}{}\\{∠DEC=∠B}{}\\{DE=DB}{}\end{array}\right.$，
$\therefore \triangle DCE$≌$\triangle DFB\left(AAS\right)$，
$\therefore DC=DF$，
$\because DF\bot AB$，$DC\bot AC$，
$\therefore$点$D$在$\angle BAC$的平分线上.
$\therefore AD$平分$\angle BAC$.
$(2)$$AE+AB=2AC$.理由如下：
由（1）知，$AD$平分$\angle BAC$，
$\therefore \angle DAC=\angle DAF$.
在$\triangle ACD$和$\triangle AFD$中，
$\left\{\begin{array}{l}{∠ACD=∠AFD}{}\\{∠DAC=∠DAF}{}\\{DC=DF}{}\end{array}\right.$，
$\therefore \triangle ACD$≌$\triangle AFD\left(AAS\right).$
$\therefore AC=AF$，
由（1）知$,\triangle DCE$≌$\triangle DFB$，
$\therefore CE=FB$.
$\therefore AE+AB=AE+FB+AF=AE+CE+AF=AC+AF=2AC$.