$(1)$过点$D$作$DE\bot AB$延长线于点$E$，

$\because \triangle ABC$是等边三角形，
$\therefore AC=AB=BC=6\sqrt{2}$，$\angle ABC=60^{\circ}$，
$\because \triangle BCD$是等腰直角三角形，
$\therefore BD=BC=6\sqrt{2}$，$\angle CBD=90^{\circ}$，
$\therefore \angle DBE=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$，
$\therefore DE=\frac{1}{2}BD=\frac{1}{2}×6\sqrt{2}=3\sqrt{2}$，
即点$D$到$AB$边的距离为$3\sqrt{2}$；
$(2)$证明：在$EF$上取$EG=BE$，连接$AG$，$AF$，

$\because \triangle ABC$是等边三角形，
$\therefore AC=AB=BC=6\sqrt{2}$，$\angle ABC=\angle ACB=\angle BAC=60^{\circ}$，
$\because \triangle BCD$是等腰直角三角形，
$\therefore BD=BC=6\sqrt{2}$，$\angle CBD=90^{\circ}$，$\angle BCD=\angle BDC=45^{\circ}$，
$\therefore AB=BD$，$\angle ABD=60^{\circ}+90^{\circ}=150^{\circ}$，
$\because BE\bot AD$，
$\therefore AE=DE$，$\angle ABE=\angle BDE=\frac{1}{2}\angle ABD=75^{\circ}$，
$\because BE=GE$，$AE\bot BG$，
$\therefore AE$垂直平分$BG$，
$\therefore AG=AB$，
$\therefore \angle AGB=\angle ABG=75^{\circ}$，
$\therefore \angle AGF=180^{\circ}-75^{\circ}=105^{\circ}$，
$\because \angle ACF=60^{\circ}+45^{\circ}=105^{\circ}$，
$\therefore \angle AGF=\angle ACF$，
$\because AE=DE$，$BF\bot AD$，
$\therefore BF$垂直平分$AD$，
$\therefore AF=FD$，
$\because BF\bot AD$，
$\therefore \angle AFG=\angle DFB$，
$\because \angle DFB=180^{\circ}-\angle BDF-\angle DBF=60^{\circ}$，
$\therefore \angle AFC=180^{\circ}-60^{\circ}-60^{\circ}=60^{\circ}$，
$\therefore \angle AFC=\angle AFG$，
在$\triangle AFC$和$\triangle AFG$中，
$\left\{\begin{array}{l}{∠AGF=∠ACF}\\{∠AFC=∠AFG}\\{AF=AF}\end{array}\right.$，
$\therefore \triangle AFC$≌$\triangle AFG\left(AAS\right)$，
$\therefore FC=FG$，
$\therefore EF=EG+FG=BE+CF$；
$(3)$过点$A$作$AP\bot AD$，且$AP=BC$，连接$PM$，过点$C$作$CE\bot AD$于点$E$，

$\because \angle PAM=\angle CBN=90^{\circ}$，$AP=BC$，$AM=BN$，
$\therefore \triangle APM$≌$\triangle BCN\left(SAS\right)$，
$\therefore PM=CN$，
$\therefore CN+CM=PM+CM$，
$\therefore$当$C$、$M$、$P$三点共线时，$CN+CM$最小，
由（2）知，$\angle CAB=60^{\circ}$，$AB=BD$，$\angle ABD=150^{\circ}$，
$\therefore \angle BAD=\angle BDA=15^{\circ}$，
$\therefore \angle CAE=60^{\circ}-15^{\circ}=45^{\circ}$，
$\because \angle AEC=90^{\circ}$，
$\therefore \triangle ACE$是等腰直角三角形，
$\therefore CE=AE=\frac{AC}{\sqrt{2}}=\frac{6\sqrt{2}}{\sqrt{2}}=6$，
过点$M$作$MO\bot AC$于点$O$，
则$\triangle MOA$是等腰直角三角形，
$\therefore AM=\sqrt{2}OM$，
$\because \angle CEM=\angle PAM=90^{\circ}$，$\angle AMP=\angle CME$，
$\therefore \angle P=\angle MCE$，
$\because AP=BC=AC$，
$\therefore \angle P=\angle ACM$，
$\therefore \angle MCE=\angle ACM$，
$\therefore ME=OM$，
$\therefore ME=\frac{\sqrt{2}}{2}AM$，
即$AM+\frac{\sqrt{2}}{2}AM=6$，
解得$AM=12-6\sqrt{2}$，
$\therefore S_{\triangle ACM}=\frac{1}{2}AM\cdot CE=\frac{1}{2}×(12-6\sqrt{2})×6=36-18\sqrt{2}$.

$(1)$过点$D$作$DE\bot AB$延长线于点$E$，

$\because \triangle ABC$是等边三角形，
$\therefore AC=AB=BC=6\sqrt{2}$，$\angle ABC=60^{\circ}$，
$\because \triangle BCD$是等腰直角三角形，
$\therefore BD=BC=6\sqrt{2}$，$\angle CBD=90^{\circ}$，
$\therefore \angle DBE=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ}$，
$\therefore DE=\frac{1}{2}BD=\frac{1}{2}×6\sqrt{2}=3\sqrt{2}$，
即点$D$到$AB$边的距离为$3\sqrt{2}$；
$(2)$证明：在$EF$上取$EG=BE$，连接$AG$，$AF$，

$\because \triangle ABC$是等边三角形，
$\therefore AC=AB=BC=6\sqrt{2}$，$\angle ABC=\angle ACB=\angle BAC=60^{\circ}$，
$\because \triangle BCD$是等腰直角三角形，
$\therefore BD=BC=6\sqrt{2}$，$\angle CBD=90^{\circ}$，$\angle BCD=\angle BDC=45^{\circ}$，
$\therefore AB=BD$，$\angle ABD=60^{\circ}+90^{\circ}=150^{\circ}$，
$\because BE\bot AD$，
$\therefore AE=DE$，$\angle ABE=\angle BDE=\frac{1}{2}\angle ABD=75^{\circ}$，
$\because BE=GE$，$AE\bot BG$，
$\therefore AE$垂直平分$BG$，
$\therefore AG=AB$，
$\therefore \angle AGB=\angle ABG=75^{\circ}$，
$\therefore \angle AGF=180^{\circ}-75^{\circ}=105^{\circ}$，
$\because \angle ACF=60^{\circ}+45^{\circ}=105^{\circ}$，
$\therefore \angle AGF=\angle ACF$，
$\because AE=DE$，$BF\bot AD$，
$\therefore BF$垂直平分$AD$，
$\therefore AF=FD$，
$\because BF\bot AD$，
$\therefore \angle AFG=\angle DFB$，
$\because \angle DFB=180^{\circ}-\angle BDF-\angle DBF=60^{\circ}$，
$\therefore \angle AFC=180^{\circ}-60^{\circ}-60^{\circ}=60^{\circ}$，
$\therefore \angle AFC=\angle AFG$，
在$\triangle AFC$和$\triangle AFG$中，
$\left\{\begin{array}{l}{∠AGF=∠ACF}\\{∠AFC=∠AFG}\\{AF=AF}\end{array}\right.$，
$\therefore \triangle AFC$≌$\triangle AFG\left(AAS\right)$，
$\therefore FC=FG$，
$\therefore EF=EG+FG=BE+CF$；
$(3)$过点$A$作$AP\bot AD$，且$AP=BC$，连接$PM$，过点$C$作$CE\bot AD$于点$E$，

$\because \angle PAM=\angle CBN=90^{\circ}$，$AP=BC$，$AM=BN$，
$\therefore \triangle APM$≌$\triangle BCN\left(SAS\right)$，
$\therefore PM=CN$，
$\therefore CN+CM=PM+CM$，
$\therefore$当$C$、$M$、$P$三点共线时，$CN+CM$最小，
由（2）知，$\angle CAB=60^{\circ}$，$AB=BD$，$\angle ABD=150^{\circ}$，
$\therefore \angle BAD=\angle BDA=15^{\circ}$，
$\therefore \angle CAE=60^{\circ}-15^{\circ}=45^{\circ}$，
$\because \angle AEC=90^{\circ}$，
$\therefore \triangle ACE$是等腰直角三角形，
$\therefore CE=AE=\frac{AC}{\sqrt{2}}=\frac{6\sqrt{2}}{\sqrt{2}}=6$，
过点$M$作$MO\bot AC$于点$O$，
则$\triangle MOA$是等腰直角三角形，
$\therefore AM=\sqrt{2}OM$，
$\because \angle CEM=\angle PAM=90^{\circ}$，$\angle AMP=\angle CME$，
$\therefore \angle P=\angle MCE$，
$\because AP=BC=AC$，
$\therefore \angle P=\angle ACM$，
$\therefore \angle MCE=\angle ACM$，
$\therefore ME=OM$，
$\therefore ME=\frac{\sqrt{2}}{2}AM$，
即$AM+\frac{\sqrt{2}}{2}AM=6$，
解得$AM=12-6\sqrt{2}$，
$\therefore S_{\triangle ACM}=\frac{1}{2}AM\cdot CE=\frac{1}{2}×(12-6\sqrt{2})×6=36-18\sqrt{2}$.