$(1)$$\because \triangle ACE$是等腰直角三角形，$\angle EAC=90^{\circ}$，
$\therefore AC=AE$，
$\because AB=AC$，
$\therefore AB=AE$，
$\therefore \angle ABE=\angle AEB$，
又$\because \angle BAC=50^{\circ}$，$\angle EAC=90^{\circ}$，
$\therefore \angle BAE=50^{\circ}+90^{\circ}=140^{\circ}$，
$\therefore \angle AEB=\left(180^{\circ}-140^{\circ}\right)\div 2=20^{\circ}$；
$(2)$证明：$\because AB=AC$，$D$是$BC$的中点，
$\therefore \angle BAF=\angle CAF$.
又$\because AF=AF$，
$\therefore \triangle BAF$≌$\triangle CAF\left(SAS\right)$，
$\therefore \angle ABF=\angle ACF$，
$\because \triangle ACE$是等腰直角三角形，$\angle EAC=90^{\circ}$，
$\therefore AC=AE$，
$\because AB=AC$，
$\therefore AB=AE$，
$\therefore \angle ABE=\angle AEB$，
$\therefore \angle AEB=\angle ACF$；
$(3)$证明：$\because \triangle BAF$≌$\triangle CAF$，
$\therefore BF=CF$，
$\because \angle AGF=\angle AEB+\angle EAG$，
$\angle AGF=\angle ACF+\angle CFG$，$\angle AEB=\angle ACF$，
$\therefore \angle CFG=\angle EAG=90^{\circ}$，
$\therefore EF^{2}+BF^{2}=EF^{2}+CF^{2}=EC^{2}$，
$\because \triangle ACE$是等腰直角三角形，$\angle EAC=90^{\circ}$，
$\therefore AC=AE$，
$\therefore EC^{2}=AC^{2}+AE^{2}=2AC^{2}$，
$\therefore EF^{2}+BF^{2}=2AC^{2}$.

$(1)$$\because \triangle ACE$是等腰直角三角形，$\angle EAC=90^{\circ}$，
$\therefore AC=AE$，
$\because AB=AC$，
$\therefore AB=AE$，
$\therefore \angle ABE=\angle AEB$，
又$\because \angle BAC=50^{\circ}$，$\angle EAC=90^{\circ}$，
$\therefore \angle BAE=50^{\circ}+90^{\circ}=140^{\circ}$，
$\therefore \angle AEB=\left(180^{\circ}-140^{\circ}\right)\div 2=20^{\circ}$；
$(2)$证明：$\because AB=AC$，$D$是$BC$的中点，
$\therefore \angle BAF=\angle CAF$.
又$\because AF=AF$，
$\therefore \triangle BAF$≌$\triangle CAF\left(SAS\right)$，
$\therefore \angle ABF=\angle ACF$，
$\because \triangle ACE$是等腰直角三角形，$\angle EAC=90^{\circ}$，
$\therefore AC=AE$，
$\because AB=AC$，
$\therefore AB=AE$，
$\therefore \angle ABE=\angle AEB$，
$\therefore \angle AEB=\angle ACF$；
$(3)$证明：$\because \triangle BAF$≌$\triangle CAF$，
$\therefore BF=CF$，
$\because \angle AGF=\angle AEB+\angle EAG$，
$\angle AGF=\angle ACF+\angle CFG$，$\angle AEB=\angle ACF$，
$\therefore \angle CFG=\angle EAG=90^{\circ}$，
$\therefore EF^{2}+BF^{2}=EF^{2}+CF^{2}=EC^{2}$，
$\because \triangle ACE$是等腰直角三角形，$\angle EAC=90^{\circ}$，
$\therefore AC=AE$，
$\therefore EC^{2}=AC^{2}+AE^{2}=2AC^{2}$，
$\therefore EF^{2}+BF^{2}=2AC^{2}$.