证明：$\left(1\right)\because AB=AC$，$AD\bot BC$，
$\therefore BC=2BD$.
$\because CE\bot AB$，$\angle BAC=45^{\circ}$，
$\therefore \angle ECA=45^{\circ}$.
$\therefore AE=CE$.
又$AD\bot BC$，$CE\bot AB$，
可得$\angle EAH=\angle ECB$，
在$\triangle AEH$和$\triangle CEB$中，$\left\{\begin{array}{l}{∠EAH=∠ECB}\\{AE=CE}\\{∠AEH=∠BEC}\end{array}\right.$
$\therefore \triangle AEH$≌$\triangle CEB\left(ASA\right).$
$\therefore AH=BC$.
$\therefore AH=2BD$.

$(2)$答：（1）中结论依然成立.
所画图形如图所示.延长$BA$交$HC$于$E$.
$\because \angle BAC=135^{\circ}$，
$\therefore \angle CAE=45^{\circ}$.
$\because AE\bot HC$，
$\therefore \angle ACE=\angle CAE=45^{\circ}$.
$\therefore AE=CE$.
$\because HD\bot BC$，$BE\bot HC$，
可得$\angle B=\angle H$.
在$Rt\triangle BEC$和$Rt\triangle HEA$中，$\left\{\begin{array}{l}{∠B=∠H}\\{∠BEC=∠HEA}\\{CE=AE}\end{array}\right.$
$\therefore Rt\triangle BEC$≌$Rt\triangle HEA\left(AAS\right).$
$\therefore AH=BC$.
又$BC=2BD$，
$\therefore AH=2BD$.

证明：$\left(1\right)\because AB=AC$，$AD\bot BC$，
$\therefore BC=2BD$.
$\because CE\bot AB$，$\angle BAC=45^{\circ}$，
$\therefore \angle ECA=45^{\circ}$.
$\therefore AE=CE$.
又$AD\bot BC$，$CE\bot AB$，
可得$\angle EAH=\angle ECB$，
在$\triangle AEH$和$\triangle CEB$中，$\left\{\begin{array}{l}{∠EAH=∠ECB}\\{AE=CE}\\{∠AEH=∠BEC}\end{array}\right.$
$\therefore \triangle AEH$≌$\triangle CEB\left(ASA\right).$
$\therefore AH=BC$.
$\therefore AH=2BD$.

$(2)$答：（1）中结论依然成立.
所画图形如图所示.延长$BA$交$HC$于$E$.
$\because \angle BAC=135^{\circ}$，
$\therefore \angle CAE=45^{\circ}$.
$\because AE\bot HC$，
$\therefore \angle ACE=\angle CAE=45^{\circ}$.
$\therefore AE=CE$.
$\because HD\bot BC$，$BE\bot HC$，
可得$\angle B=\angle H$.
在$Rt\triangle BEC$和$Rt\triangle HEA$中，$\left\{\begin{array}{l}{∠B=∠H}\\{∠BEC=∠HEA}\\{CE=AE}\end{array}\right.$
$\therefore Rt\triangle BEC$≌$Rt\triangle HEA\left(AAS\right).$
$\therefore AH=BC$.
又$BC=2BD$，
$\therefore AH=2BD$.