$\left(1\right)\because \angle B=\angle E=\angle ACD$，
$\therefore \angle A=180^{\circ}-\angle B-\angle ACB$，$\angle \angle DCE=180^{\circ}-\angle ACD-\angle ACB$，
$\therefore \angle A=\angle DCE$，
在$\triangle ABC$与$\triangle CDE$中，
$\left\{\begin{array}{l}{∠A=∠DCE}\\{∠B=∠E}\\{AC=CD}\end{array}\right.$，
$\therefore \triangle ABC$≌$\triangle CDE\left(AAS\right)$，
$\therefore CE=AB=8$，$BC=DE=4$，
$\therefore BE=BC+CE=12$；
$(2)$延长$BC$到$F$，连接$DF$使$\angle F=60^{\circ}$，

$\because \triangle ACD$为等边三角形，则$\angle ACD=60^{\circ}$，$AC=CD$，
$\because \angle DCF+\angle ACD=\angle ABC+\angle BAC$，即$60^{\circ}+\angle DCF=60^{\circ}+\angle BAC$，
$\therefore \angle DCF=+\angle BAC$，
$\because \angle ABC=\angle F=60^{\circ}$，$AC=CD$，
$\therefore \triangle ABC$≌$\triangle CFD\left(AAS\right)$，
$\therefore DF=BC=6$，
过点$D$作$DH\bot CF$于点$H$，则$DH=DF\cdot \sin F=6\times \sin 60^{\circ}=3\sqrt{3}$，
则$\triangle BCD$的面积$=\frac{1}{2}×BC\times DH=\frac{1}{2}×6×3\sqrt{3}=9\sqrt{3}$；
$(3)$过点$B$作$BH\bot CD$交$DC$的延长线于点$H$，过点$A$作$AM\bot CD$于点$M$，

$\because \angle ABC=\angle CAB=\angle ADC=45^{\circ}$，则$AM=DM$，$AC=BC$，
$\because \angle ACM+\angle BCH=90^{\circ}$，$\angle BCH+\angle HBC=90^{\circ}$，
$\therefore \angle HBC=\angle ACM$，
$\because \angle AMC=\angle CHB=90^{\circ}$，$AC=BC$，
$\therefore \triangle ACM$≌$\triangle CBN\left(AAS\right)$，
$\therefore AM=CH$，
$\because \triangle ACD$面积$=\frac{1}{2}×CD\times AM=4\times AM=21$，
则$AM=CH=\frac{21}{4}$，则$CM=CD-DM=\frac{11}{4}$，
而$BC^{2}=HB^{2}+CH^{2}=(\frac{11}{4})^{2}+(\frac{21}{4})^{2}=\frac{281}{8}$，
则$\triangle ABD$的面积$=S_{\triangle ABC}+S_{\triangle ACD}-S_{\triangle BCD}=\frac{1}{2}BC^{2}+21-\frac{1}{2}×8\times \frac{11}{4}=\frac{441}{16}$.

$\left(1\right)\because \angle B=\angle E=\angle ACD$，
$\therefore \angle A=180^{\circ}-\angle B-\angle ACB$，$\angle \angle DCE=180^{\circ}-\angle ACD-\angle ACB$，
$\therefore \angle A=\angle DCE$，
在$\triangle ABC$与$\triangle CDE$中，
$\left\{\begin{array}{l}{∠A=∠DCE}\\{∠B=∠E}\\{AC=CD}\end{array}\right.$，
$\therefore \triangle ABC$≌$\triangle CDE\left(AAS\right)$，
$\therefore CE=AB=8$，$BC=DE=4$，
$\therefore BE=BC+CE=12$；
$(2)$延长$BC$到$F$，连接$DF$使$\angle F=60^{\circ}$，

$\because \triangle ACD$为等边三角形，则$\angle ACD=60^{\circ}$，$AC=CD$，
$\because \angle DCF+\angle ACD=\angle ABC+\angle BAC$，即$60^{\circ}+\angle DCF=60^{\circ}+\angle BAC$，
$\therefore \angle DCF=+\angle BAC$，
$\because \angle ABC=\angle F=60^{\circ}$，$AC=CD$，
$\therefore \triangle ABC$≌$\triangle CFD\left(AAS\right)$，
$\therefore DF=BC=6$，
过点$D$作$DH\bot CF$于点$H$，则$DH=DF\cdot \sin F=6\times \sin 60^{\circ}=3\sqrt{3}$，
则$\triangle BCD$的面积$=\frac{1}{2}×BC\times DH=\frac{1}{2}×6×3\sqrt{3}=9\sqrt{3}$；
$(3)$过点$B$作$BH\bot CD$交$DC$的延长线于点$H$，过点$A$作$AM\bot CD$于点$M$，

$\because \angle ABC=\angle CAB=\angle ADC=45^{\circ}$，则$AM=DM$，$AC=BC$，
$\because \angle ACM+\angle BCH=90^{\circ}$，$\angle BCH+\angle HBC=90^{\circ}$，
$\therefore \angle HBC=\angle ACM$，
$\because \angle AMC=\angle CHB=90^{\circ}$，$AC=BC$，
$\therefore \triangle ACM$≌$\triangle CBN\left(AAS\right)$，
$\therefore AM=CH$，
$\because \triangle ACD$面积$=\frac{1}{2}×CD\times AM=4\times AM=21$，
则$AM=CH=\frac{21}{4}$，则$CM=CD-DM=\frac{11}{4}$，
而$BC^{2}=HB^{2}+CH^{2}=(\frac{11}{4})^{2}+(\frac{21}{4})^{2}=\frac{281}{8}$，
则$\triangle ABD$的面积$=S_{\triangle ABC}+S_{\triangle ACD}-S_{\triangle BCD}=\frac{1}{2}BC^{2}+21-\frac{1}{2}×8\times \frac{11}{4}=\frac{441}{16}$.