$(1)$证明：如图，过点$P$分别作$PM\bot OA$于点$M$，$PN\bot OB$于点$N$，

$\because \angle AOB=90^{\circ}$，$PM\bot OA$于点$M$，$PN\bot OB$于点$N$，
$\therefore \angle OMP=90^{\circ}$，$\angle ONP=90^{\circ}$，
$\therefore \angle MPN=360^{\circ}-90^{\circ}-90^{\circ}=90^{\circ}$，
$\because \angle EPF=90^{\circ}$，
$\therefore \angle MPE=\angle NPF$，
在$\triangle MPE$和$\triangle NPF$中，
$\left\{\begin{array}{l}{∠PME=∠PNF=90°}{}\\{∠MPE=∠NPF}{}\\{PE=PF}{}\end{array}\right.$，
$\therefore \triangle MPE$≌$\triangle NPF\left(AAS\right)$，
$\therefore \angle AEP=\angle PFO$；
$(2)$$\because \triangle MPE$≌$\triangle NPF$，
$\therefore PM=PN$，
$\because PM\bot OA$于点$M$，$PN\bot OB$于点$N$，
$\therefore$点$P$在$\angle AOB$的角平分线上.

$(1)$证明：如图，过点$P$分别作$PM\bot OA$于点$M$，$PN\bot OB$于点$N$，

$\because \angle AOB=90^{\circ}$，$PM\bot OA$于点$M$，$PN\bot OB$于点$N$，
$\therefore \angle OMP=90^{\circ}$，$\angle ONP=90^{\circ}$，
$\therefore \angle MPN=360^{\circ}-90^{\circ}-90^{\circ}=90^{\circ}$，
$\because \angle EPF=90^{\circ}$，
$\therefore \angle MPE=\angle NPF$，
在$\triangle MPE$和$\triangle NPF$中，
$\left\{\begin{array}{l}{∠PME=∠PNF=90°}{}\\{∠MPE=∠NPF}{}\\{PE=PF}{}\end{array}\right.$，
$\therefore \triangle MPE$≌$\triangle NPF\left(AAS\right)$，
$\therefore \angle AEP=\angle PFO$；
$(2)$$\because \triangle MPE$≌$\triangle NPF$，
$\therefore PM=PN$，
$\because PM\bot OA$于点$M$，$PN\bot OB$于点$N$，
$\therefore$点$P$在$\angle AOB$的角平分线上.