过点$D$作$DE\bot AC$于$E$，设$DC$的中点为$F$，过点$F$作$FH\bot DC$交$AC$于$H$，连接$DH$，如图所示：

$\because AB=AD$，$\angle B=45^{\circ}$，
$\therefore \triangle ABD$是等腰直角三角形，
$\therefore \angle ADB=45^{\circ}$，
$\because AD=DC$，
$\therefore \angle C=\angle DAC$，
$\because \angle C+\angle DAC=\angle ADB=45^{\circ}$，
$\therefore \angle C=\angle DAC=22.5^{\circ}$，
$\because AD=DC$，$AC=8$，$DF\bot AC$，
$\therefore AE=CE=\frac{1}{2}AC=4$，
$\because$点$F$是$DC$的中点，$FH\bot DC$，
$\therefore CH=DH$，
$\therefore \angle HDC=\angle C=22.5^{\circ}$，
$\therefore \angle EHD=45^{\circ}$，
$\therefore \triangle DEH$是等腰直角三角形，
设$DE=EH=a$，由勾股定理得：$DH=\sqrt{D{E}^{2}+E{H}^{2}}=\sqrt{2}a$，
$\therefore CH=DH=\sqrt{2}a$，
$\therefore CE=CH+EH=a+\sqrt{2}a=4$，
$\therefore a=4(\sqrt{2}-1)$，
$\therefore DE=a=4(\sqrt{2}-1)$，
在$Rt\triangle CDE$中，由勾股定理得：$DC^{2}=DE^{2}+CE^{2}=[4(\sqrt{2}-1)]^{2}+{4}^{2}=16(4-2\sqrt{2})$，
$\therefore AD^{2}=DC^{2}=16(4-2\sqrt{2})$，
$\therefore S_{\triangle ABD}=\frac{1}{2}AD^{2}=\frac{1}{2}×16(4-2\sqrt{2})=32-16\sqrt{2}$，$S_{\triangle ADC}=\frac{1}{2}AC\cdot DE=\frac{1}{2}×8×4(\sqrt{2}-1)=16\sqrt{2}-16$，
$\therefore S_{\triangle ABC}=S_{\triangle ABD}+S_{\triangle ADC}=32-16\sqrt{2}+16\sqrt{2}-16=16$.
故答案为：$16$.

过点$D$作$DE\bot AC$于$E$，设$DC$的中点为$F$，过点$F$作$FH\bot DC$交$AC$于$H$，连接$DH$，如图所示：

$\because AB=AD$，$\angle B=45^{\circ}$，
$\therefore \triangle ABD$是等腰直角三角形，
$\therefore \angle ADB=45^{\circ}$，
$\because AD=DC$，
$\therefore \angle C=\angle DAC$，
$\because \angle C+\angle DAC=\angle ADB=45^{\circ}$，
$\therefore \angle C=\angle DAC=22.5^{\circ}$，
$\because AD=DC$，$AC=8$，$DF\bot AC$，
$\therefore AE=CE=\frac{1}{2}AC=4$，
$\because$点$F$是$DC$的中点，$FH\bot DC$，
$\therefore CH=DH$，
$\therefore \angle HDC=\angle C=22.5^{\circ}$，
$\therefore \angle EHD=45^{\circ}$，
$\therefore \triangle DEH$是等腰直角三角形，
设$DE=EH=a$，由勾股定理得：$DH=\sqrt{D{E}^{2}+E{H}^{2}}=\sqrt{2}a$，
$\therefore CH=DH=\sqrt{2}a$，
$\therefore CE=CH+EH=a+\sqrt{2}a=4$，
$\therefore a=4(\sqrt{2}-1)$，
$\therefore DE=a=4(\sqrt{2}-1)$，
在$Rt\triangle CDE$中，由勾股定理得：$DC^{2}=DE^{2}+CE^{2}=[4(\sqrt{2}-1)]^{2}+{4}^{2}=16(4-2\sqrt{2})$，
$\therefore AD^{2}=DC^{2}=16(4-2\sqrt{2})$，
$\therefore S_{\triangle ABD}=\frac{1}{2}AD^{2}=\frac{1}{2}×16(4-2\sqrt{2})=32-16\sqrt{2}$，$S_{\triangle ADC}=\frac{1}{2}AC\cdot DE=\frac{1}{2}×8×4(\sqrt{2}-1)=16\sqrt{2}-16$，
$\therefore S_{\triangle ABC}=S_{\triangle ABD}+S_{\triangle ADC}=32-16\sqrt{2}+16\sqrt{2}-16=16$.
故答案为：$16$.