$(1)$过点$A$作$AG\bot DE$，垂足为点$G$，
$\because \angle BAC=90^{\circ}$，$\angle CAD=60^{\circ}$，
$\therefore \angle BAD=150^{\circ}$，
$\because AE$平分$\angle BAD$，
$\therefore \angle DAE=\frac{1}{2}\angle BAD=75^{\circ}$，
$\because \angle CAD=60^{\circ}$，$AG\bot DE$，
$\therefore DG=\frac{1}{2}AD=2$，
$\therefore AG=\sqrt{A{D}^{2}-D{G}^{2}}=2\sqrt{3}$，
$\because \angle DAE=75^{\circ}$，$\angle DAG=30^{\circ}$，
$\therefore \angle GAE=45^{\circ}$，
$\therefore \triangle AGE$是等腰直角三角形，
$\therefore AG=GE=2\sqrt{3}$，
$\therefore AE=\sqrt{A{G}^{2}+G{E}^{2}}=2\sqrt{6}$；

$(2)$设$\angle CAD=\alpha$，
$\because \angle BAC=60^{\circ}$，
$\therefore \angle BAD=\angle BAC+\angle CAD=60^{\circ}+\alpha$，
$\because AE$平分$\angle BAD$，
$\therefore \angle BAE=\angle DAE=\frac{1}{2}\angle BAD=30^{\circ}+\frac{1}{2}α$，
$\because \angle CAD=\alpha$，$AC=AD$，
$\therefore \angle ACD=\angle ADC=\frac{180°-∠CAD}{2}=90^{\circ}-\frac{1}{2}α$，
$\therefore \angle AED=180^{\circ}-\angle DAE-\angle ADC=60^{\circ}$，
$\because BG$∥$DE$，
$\therefore \angle BHE=\angle AED=60^{\circ}$，
$\therefore \angle ABG+\angle BAF=60^{\circ}$，
$\because \angle CAF+\angle BAF=60^{\circ}$，
$\therefore \angle ABG=\angle CAF$，
$\because AB=AC$，$\angle BAG=\angle ACF=60^{\circ}$，
$\therefore \triangle ABG$≌$\triangle CAF\left(ASA\right)$，
$\therefore BG=AF$，
延长$HG$，使得$HP=AH$，连接$AP$，
$\because \angle BHE=\angle AHP=60^{\circ}$，
$\therefore \triangle AHP$是等边三角形，
$\therefore \angle BPA=\angle AEC=60^{\circ}$，
$\because AB=AC$，$\angle ABP=\angle CAE$，
$\therefore \triangle ABP$≌$\triangle CAE\left(AAS\right)$，
$\therefore BP=AE$，
$\because BG=AF$，
$\therefore PG=EF$，
$\because AH=HP$，
$\therefore AH=PG+HG=EF+HG$；

$(3)$设$AB=AC=AD=2a$，
过点$A$作$AM\bot DE$，垂足为点$M$，
$\because AC=AD$，$AM\bot DE$，
$\therefore CM=DM$，$\angle CAM=\angle DAM=\frac{1}{2}α$，
$\because \angle BAC=120^{\circ}$，$\angle CAD=\alpha$，$AE$平分$\angle BAD$，
$\therefore \angle BAE=\angle DAE=\frac{1}{2}\angle BAD=60^{\circ}+\frac{1}{2}α$，
$\because \angle DAM=\frac{1}{2}α$，
$\therefore \angle EAM=\angle DAE-\angle DAM=60^{\circ}$，
$\because AM\bot DE$，
$\therefore \angle AED=30^{\circ}$，
$\therefore AM=\frac{1}{2}AE$，
$\because AB=AD$，$AE$平分$\angle BAD$，
$\therefore AN\bot BD$，$BN=DN=\frac{1}{2}BD$，
$\because AE=BD$，
$\therefore AM=DN$，
$\because \angle AND=\angle DMA=90^{\circ}$，$AD=AD$，
$\therefore Rt\triangle ADN$≌$Rt\triangle DAM\left(HL\right)$，
$\therefore \angle DAN=\angle ADM$，
$\because \angle DAN=60^{\circ}+\frac{1}{2}\alpha$，$\angle ADM=90^{\circ}-\angle DAM=90^{\circ}-\frac{1}{2}\alpha$，
$\therefore 60^{\circ}+\frac{1}{2}\alpha =90^{\circ}-\frac{1}{2}\alpha$，
$\therefore \alpha =30^{\circ}$，
延长$DA$，过点$B$作$BP\bot AD$，垂足为点$P$，过点$A$作$AQ\bot BC$，垂足为点$Q$，
$\because \angle AQB=90^{\circ}$，$\angle ABC=30^{\circ}$，
$\therefore AQ=a$，$BQ=\sqrt{3}a$，
$\because AB=AC$，$AQ\bot BC$，
$\therefore BC=2BQ=2\sqrt{3}a$，
$\because \angle BAP=180^{\circ}-\angle BAC-\angle CAD=30^{\circ}$，$\angle APB=90^{\circ}$，
$\therefore BP=a$，$AP=\sqrt{3}a$，
$\therefore PD=AP+AD=(2+\sqrt{3})a$，
$\therefore BD^{2}=BP^{2}+PD^{2}=(8+4\sqrt{3}){a}^{2}$，
$\because AM=\frac{1}{2}AE=\frac{1}{2}BD$，
$\therefore A{M}^{2}=\frac{1}{4}B{D}^{2}=(2+\sqrt{3}){a}^{2}$，
在$Rt\triangle ADM$中，
$\because AD=2a$，
$\therefore DM=\sqrt{A{D}^{2}-A{M}^{2}}=\sqrt{2-\sqrt{3}}a$，
$\therefore CD=2DM=2\sqrt{2-\sqrt{3}}a$，
$\therefore \frac{BC}{CD}=\frac{2\sqrt{3}a}{2\sqrt{2-\sqrt{3}}a}=\sqrt{6+3\sqrt{3}}$.

$(1)$过点$A$作$AG\bot DE$，垂足为点$G$，
$\because \angle BAC=90^{\circ}$，$\angle CAD=60^{\circ}$，
$\therefore \angle BAD=150^{\circ}$，
$\because AE$平分$\angle BAD$，
$\therefore \angle DAE=\frac{1}{2}\angle BAD=75^{\circ}$，
$\because \angle CAD=60^{\circ}$，$AG\bot DE$，
$\therefore DG=\frac{1}{2}AD=2$，
$\therefore AG=\sqrt{A{D}^{2}-D{G}^{2}}=2\sqrt{3}$，
$\because \angle DAE=75^{\circ}$，$\angle DAG=30^{\circ}$，
$\therefore \angle GAE=45^{\circ}$，
$\therefore \triangle AGE$是等腰直角三角形，
$\therefore AG=GE=2\sqrt{3}$，
$\therefore AE=\sqrt{A{G}^{2}+G{E}^{2}}=2\sqrt{6}$；

$(2)$设$\angle CAD=\alpha$，
$\because \angle BAC=60^{\circ}$，
$\therefore \angle BAD=\angle BAC+\angle CAD=60^{\circ}+\alpha$，
$\because AE$平分$\angle BAD$，
$\therefore \angle BAE=\angle DAE=\frac{1}{2}\angle BAD=30^{\circ}+\frac{1}{2}α$，
$\because \angle CAD=\alpha$，$AC=AD$，
$\therefore \angle ACD=\angle ADC=\frac{180°-∠CAD}{2}=90^{\circ}-\frac{1}{2}α$，
$\therefore \angle AED=180^{\circ}-\angle DAE-\angle ADC=60^{\circ}$，
$\because BG$∥$DE$，
$\therefore \angle BHE=\angle AED=60^{\circ}$，
$\therefore \angle ABG+\angle BAF=60^{\circ}$，
$\because \angle CAF+\angle BAF=60^{\circ}$，
$\therefore \angle ABG=\angle CAF$，
$\because AB=AC$，$\angle BAG=\angle ACF=60^{\circ}$，
$\therefore \triangle ABG$≌$\triangle CAF\left(ASA\right)$，
$\therefore BG=AF$，
延长$HG$，使得$HP=AH$，连接$AP$，
$\because \angle BHE=\angle AHP=60^{\circ}$，
$\therefore \triangle AHP$是等边三角形，
$\therefore \angle BPA=\angle AEC=60^{\circ}$，
$\because AB=AC$，$\angle ABP=\angle CAE$，
$\therefore \triangle ABP$≌$\triangle CAE\left(AAS\right)$，
$\therefore BP=AE$，
$\because BG=AF$，
$\therefore PG=EF$，
$\because AH=HP$，
$\therefore AH=PG+HG=EF+HG$；

$(3)$设$AB=AC=AD=2a$，
过点$A$作$AM\bot DE$，垂足为点$M$，
$\because AC=AD$，$AM\bot DE$，
$\therefore CM=DM$，$\angle CAM=\angle DAM=\frac{1}{2}α$，
$\because \angle BAC=120^{\circ}$，$\angle CAD=\alpha$，$AE$平分$\angle BAD$，
$\therefore \angle BAE=\angle DAE=\frac{1}{2}\angle BAD=60^{\circ}+\frac{1}{2}α$，
$\because \angle DAM=\frac{1}{2}α$，
$\therefore \angle EAM=\angle DAE-\angle DAM=60^{\circ}$，
$\because AM\bot DE$，
$\therefore \angle AED=30^{\circ}$，
$\therefore AM=\frac{1}{2}AE$，
$\because AB=AD$，$AE$平分$\angle BAD$，
$\therefore AN\bot BD$，$BN=DN=\frac{1}{2}BD$，
$\because AE=BD$，
$\therefore AM=DN$，
$\because \angle AND=\angle DMA=90^{\circ}$，$AD=AD$，
$\therefore Rt\triangle ADN$≌$Rt\triangle DAM\left(HL\right)$，
$\therefore \angle DAN=\angle ADM$，
$\because \angle DAN=60^{\circ}+\frac{1}{2}\alpha$，$\angle ADM=90^{\circ}-\angle DAM=90^{\circ}-\frac{1}{2}\alpha$，
$\therefore 60^{\circ}+\frac{1}{2}\alpha =90^{\circ}-\frac{1}{2}\alpha$，
$\therefore \alpha =30^{\circ}$，
延长$DA$，过点$B$作$BP\bot AD$，垂足为点$P$，过点$A$作$AQ\bot BC$，垂足为点$Q$，
$\because \angle AQB=90^{\circ}$，$\angle ABC=30^{\circ}$，
$\therefore AQ=a$，$BQ=\sqrt{3}a$，
$\because AB=AC$，$AQ\bot BC$，
$\therefore BC=2BQ=2\sqrt{3}a$，
$\because \angle BAP=180^{\circ}-\angle BAC-\angle CAD=30^{\circ}$，$\angle APB=90^{\circ}$，
$\therefore BP=a$，$AP=\sqrt{3}a$，
$\therefore PD=AP+AD=(2+\sqrt{3})a$，
$\therefore BD^{2}=BP^{2}+PD^{2}=(8+4\sqrt{3}){a}^{2}$，
$\because AM=\frac{1}{2}AE=\frac{1}{2}BD$，
$\therefore A{M}^{2}=\frac{1}{4}B{D}^{2}=(2+\sqrt{3}){a}^{2}$，
在$Rt\triangle ADM$中，
$\because AD=2a$，
$\therefore DM=\sqrt{A{D}^{2}-A{M}^{2}}=\sqrt{2-\sqrt{3}}a$，
$\therefore CD=2DM=2\sqrt{2-\sqrt{3}}a$，
$\therefore \frac{BC}{CD}=\frac{2\sqrt{3}a}{2\sqrt{2-\sqrt{3}}a}=\sqrt{6+3\sqrt{3}}$.