$\left(1\right)\triangle CAB$，$\triangle CDE$均为等腰直角三角形，$AC=BC=2\sqrt{5}$，$DC=EC$，
$\therefore \angle ACB=\angle ECD=90^{\circ}$，
$\therefore \angle ACE=\angle BCD=90^{\circ}-\angle BCE$，$\angle CAB=\angle CBA=45^{\circ}$，
在$\triangle ACE$和$\triangle BCD$中，
$\left\{\begin{array}{l}AC=BC\\∠ACE=∠BCD\\ EC=DC\end{array}\right.$，
$\therefore \triangle ACE$≌$\triangle BCD\left(SAS\right)$，
$\therefore \angle CAE=\angle CBD$，
$\therefore \angle BAD+\angle CBD=\angle BAD+\angle CAE=\angle CAB=45^{\circ}$，
$\therefore \angle BAD+\angle ABD=\angle BAD+\angle CBD+\angle CBA=90^{\circ}$，
$\therefore \angle BDA=90^{\circ}$，
故答案为：$90^{\circ}$.
$(2)$作$CH\bot AD$于点$H$，则$EH=DH$，$\angle CHF=\angle BDF=90^{\circ}$，
$\therefore CH=EH=DH=\frac{1}{2}DE$，
$\because F$为$BC$的中点，
$\therefore CF=BF$，
在$\triangle CHF$和$\triangle BDF$中，
$\left\{\begin{array}{l}∠CHF=∠BDF\\∠CFH=∠BFD\\ CF=BF\end{array}\right.$，
$\therefore \triangle CHF$≌$\triangle BDF\left(AAS\right)$，
$\therefore CH=BD$，
$\because AE=BD$，
$\therefore AE=CH=EH$，
$\because G$为$AB$的中点，
$\therefore EG=\frac{1}{2}FB$，
$\because \angle ACF=90^{\circ}$，$AC=2\sqrt{5}$，$CF=\frac{1}{2}BC=\sqrt{5}$，
$\therefore AF=\sqrt{A{C}^{2}+C{F}^{2}}=\sqrt{(2\sqrt{5})^{2}+(\sqrt{5})^{2}}=5$，
$\therefore \frac{1}{2}×5CH=\frac{1}{2}×2\sqrt{5}×\sqrt{5}={S}_{△ACF}$，
$\therefore CH=2$，
$\therefore DH=CH=BD=2$，
$\therefore HB=\sqrt{D{H}^{2}+B{D}^{2}}=\sqrt{{2}^{2}+{2}^{2}}=2\sqrt{2}$，
$\therefore EG=\frac{1}{2}×2\sqrt{2}=\sqrt{2}$，
故答案为：$\sqrt{2}$.

$\left(1\right)\triangle CAB$，$\triangle CDE$均为等腰直角三角形，$AC=BC=2\sqrt{5}$，$DC=EC$，
$\therefore \angle ACB=\angle ECD=90^{\circ}$，
$\therefore \angle ACE=\angle BCD=90^{\circ}-\angle BCE$，$\angle CAB=\angle CBA=45^{\circ}$，
在$\triangle ACE$和$\triangle BCD$中，
$\left\{\begin{array}{l}AC=BC\\∠ACE=∠BCD\\ EC=DC\end{array}\right.$，
$\therefore \triangle ACE$≌$\triangle BCD\left(SAS\right)$，
$\therefore \angle CAE=\angle CBD$，
$\therefore \angle BAD+\angle CBD=\angle BAD+\angle CAE=\angle CAB=45^{\circ}$，
$\therefore \angle BAD+\angle ABD=\angle BAD+\angle CBD+\angle CBA=90^{\circ}$，
$\therefore \angle BDA=90^{\circ}$，
故答案为：$90^{\circ}$.
$(2)$作$CH\bot AD$于点$H$，则$EH=DH$，$\angle CHF=\angle BDF=90^{\circ}$，
$\therefore CH=EH=DH=\frac{1}{2}DE$，
$\because F$为$BC$的中点，
$\therefore CF=BF$，
在$\triangle CHF$和$\triangle BDF$中，
$\left\{\begin{array}{l}∠CHF=∠BDF\\∠CFH=∠BFD\\ CF=BF\end{array}\right.$，
$\therefore \triangle CHF$≌$\triangle BDF\left(AAS\right)$，
$\therefore CH=BD$，
$\because AE=BD$，
$\therefore AE=CH=EH$，
$\because G$为$AB$的中点，
$\therefore EG=\frac{1}{2}FB$，
$\because \angle ACF=90^{\circ}$，$AC=2\sqrt{5}$，$CF=\frac{1}{2}BC=\sqrt{5}$，
$\therefore AF=\sqrt{A{C}^{2}+C{F}^{2}}=\sqrt{(2\sqrt{5})^{2}+(\sqrt{5})^{2}}=5$，
$\therefore \frac{1}{2}×5CH=\frac{1}{2}×2\sqrt{5}×\sqrt{5}={S}_{△ACF}$，
$\therefore CH=2$，
$\therefore DH=CH=BD=2$，
$\therefore HB=\sqrt{D{H}^{2}+B{D}^{2}}=\sqrt{{2}^{2}+{2}^{2}}=2\sqrt{2}$，
$\therefore EG=\frac{1}{2}×2\sqrt{2}=\sqrt{2}$，
故答案为：$\sqrt{2}$.