已知线段$AC$上有一动点$B$，分别以$AB$、$BC$为边向线段的同一侧作等边三角形$\triangle ABD$和$\triangle BCE$.连接$AE$、$CD(如图)$，若$MN$分别为$AE$、$CD$的中点，

$(1)$证明：$\because \triangle ABD$和$\triangle BCE$是等边三角形，$\therefore AB=BD$，$BC=BE$，$\angle EBC=\angle ABC=60^{\circ}$，$\therefore \angle ABE=\angle DBC$，在$\triangle ABE$和$\triangle DBC$中$\left\{\begin{array}{l}{AB=BD}\\{∠ABE=∠DBC}\\{BE=BC}\end{array}\right.$$\therefore \triangle ABE$≌$\triangle DBC\left(SAS\right)$$\therefore AE=DC$，$\because M$、$N$分别为$AE$、$CD$的中点，$\therefore AM=\frac{1}{2}AE$，$CN=\frac{1}{2}DC$$\therefore AM=CN$；$(2)$$\because \triangle ABE$≌$\triangle DBC$，$\therefore \angle EAB=\angle CDB$，在$\triangle AMB$和$\triangle DNB$中$\left\{\begin{array}{l}{AM=DN}\\{∠MAB=∠NDB}\\{AB=DB}\end{array}\right.$$\therefore \triangle AMB$≌$\triangle DNB\left(SAS\right)$，$\therefore \angle ABM=\angle DBN$，$\because \angle ABC=\angle ABM+\angle MBD=60^{\circ}$，$\therefore \angle DBN+\angle MBD=60^{\circ}$，即$\angle MBN=60^{\circ}$；$(3)$图中的全等三角形有：$\triangle ABM$≌$\triangle DBN,\triangle BME$≌$\triangle BCN,\triangle ABE$≌$\triangle DBC$；相似三角形有：$\triangle ABD$∽$\triangle BCE,\triangle ABD$∽$\triangle BMN,\triangle BMN$∽$\triangle BCE.$

$(1)$证明：$\because \triangle ABD$和$\triangle BCE$是等边三角形，$\therefore AB=BD$，$BC=BE$，$\angle EBC=\angle ABC=60^{\circ}$，$\therefore \angle ABE=\angle DBC$，在$\triangle ABE$和$\triangle DBC$中$\left\{\begin{array}{l}{AB=BD}\\{∠ABE=∠DBC}\\{BE=BC}\end{array}\right.$$\therefore \triangle ABE$≌$\triangle DBC\left(SAS\right)$$\therefore AE=DC$，$\because M$、$N$分别为$AE$、$CD$的中点，$\therefore AM=\frac{1}{2}AE$，$CN=\frac{1}{2}DC$$\therefore AM=CN$；$(2)$$\because \triangle ABE$≌$\triangle DBC$，$\therefore \angle EAB=\angle CDB$，在$\triangle AMB$和$\triangle DNB$中$\left\{\begin{array}{l}{AM=DN}\\{∠MAB=∠NDB}\\{AB=DB}\end{array}\right.$$\therefore \triangle AMB$≌$\triangle DNB\left(SAS\right)$，$\therefore \angle ABM=\angle DBN$，$\because \angle ABC=\angle ABM+\angle MBD=60^{\circ}$，$\therefore \angle DBN+\angle MBD=60^{\circ}$，即$\angle MBN=60^{\circ}$；$(3)$图中的全等三角形有：$\triangle ABM$≌$\triangle DBN,\triangle BME$≌$\triangle BCN,\triangle ABE$≌$\triangle DBC$；相似三角形有：$\triangle ABD$∽$\triangle BCE,\triangle ABD$∽$\triangle BMN,\triangle BMN$∽$\triangle BCE.$