$(1)$证明：$\because AB=AC$，
$\therefore \angle ABC=\angle ACB$，
$\therefore \angle BAC=180^{\circ}-2\angle ABC$，
$\because$以$AD$、$AE$为腰做等腰三角形$ADE$，
$\therefore AD=AE$，
$\therefore \angle ADE=\angle AED$，
$\therefore \angle DAE=180^{\circ}-2\angle ADE$，
$\because \angle ADE=\angle ABC$，
$\therefore \angle BAC=\angle DAE$，
$\therefore \angle BAC-\angle CAD=\angle DAE-\angle CAD$，
$\therefore \angle BAD=\angle CAE$，
在$\triangle BAD$和$\triangle CAE$中，$\left\{\begin{array}{l}{AB=AC}\\{∠BAD=∠CAE}\\{AD=AE}\end{array}\right.$，
$\therefore \triangle BAD$≌$\triangle CAE\left(SAS\right)$；
$(2)$$\because AB=AC$，
$\therefore \angle ACB=\angle ABC=30^{\circ}$，
$\because \triangle BAD$≌$\triangle CAE$，
$\therefore \angle ABD=\angle ACE=30^{\circ}$，
$\therefore \angle ACB=\angle ACE=30^{\circ}$，
$\therefore \angle ECB=\angle ACB+\angle ACE=60^{\circ}$，
$\because EM$∥$BC$，
$\therefore \angle MEC+\angle ECD=180^{\circ}$，
$\therefore \angle MEC=180^{\circ}-60^{\circ}=120^{\circ}$；
$(3)$证明：$\because \triangle BAD$≌$\triangle CAE$，
$\therefore DB=CE$，$\angle ABD=\angle ACE$，
$\because AB=AC$，
$\therefore \angle ABD=\angle ACB$，
$\therefore \angle ACB=\angle ACE$，
$\because EM$∥$BC$，
$\therefore \angle EMC=\angle ACB$，
$\therefore \angle ACE=\angle EMC$，
$\therefore ME=EC$，
$\therefore DB=ME$，
又$\because EM$∥$BD$，
$\therefore$四边形$MBDE$是平行四边形.

$(1)$证明：$\because AB=AC$，
$\therefore \angle ABC=\angle ACB$，
$\therefore \angle BAC=180^{\circ}-2\angle ABC$，
$\because$以$AD$、$AE$为腰做等腰三角形$ADE$，
$\therefore AD=AE$，
$\therefore \angle ADE=\angle AED$，
$\therefore \angle DAE=180^{\circ}-2\angle ADE$，
$\because \angle ADE=\angle ABC$，
$\therefore \angle BAC=\angle DAE$，
$\therefore \angle BAC-\angle CAD=\angle DAE-\angle CAD$，
$\therefore \angle BAD=\angle CAE$，
在$\triangle BAD$和$\triangle CAE$中，$\left\{\begin{array}{l}{AB=AC}\\{∠BAD=∠CAE}\\{AD=AE}\end{array}\right.$，
$\therefore \triangle BAD$≌$\triangle CAE\left(SAS\right)$；
$(2)$$\because AB=AC$，
$\therefore \angle ACB=\angle ABC=30^{\circ}$，
$\because \triangle BAD$≌$\triangle CAE$，
$\therefore \angle ABD=\angle ACE=30^{\circ}$，
$\therefore \angle ACB=\angle ACE=30^{\circ}$，
$\therefore \angle ECB=\angle ACB+\angle ACE=60^{\circ}$，
$\because EM$∥$BC$，
$\therefore \angle MEC+\angle ECD=180^{\circ}$，
$\therefore \angle MEC=180^{\circ}-60^{\circ}=120^{\circ}$；
$(3)$证明：$\because \triangle BAD$≌$\triangle CAE$，
$\therefore DB=CE$，$\angle ABD=\angle ACE$，
$\because AB=AC$，
$\therefore \angle ABD=\angle ACB$，
$\therefore \angle ACB=\angle ACE$，
$\because EM$∥$BC$，
$\therefore \angle EMC=\angle ACB$，
$\therefore \angle ACE=\angle EMC$，
$\therefore ME=EC$，
$\therefore DB=ME$，
又$\because EM$∥$BD$，
$\therefore$四边形$MBDE$是平行四边形.