Simone如图，▱$ABCD$中，$E$、$F$在$B$、$D$上，且$BE=AB$，$DF=CD$，连接$AF$、$CE$.
的有关信息介绍如下：
$(1)$证明：$\because $四边形$ABCD$是平行四边形，$\therefore AB=CD,AB$∥$CD$，$\therefore \angle ABD=\angle BDC$，$\because BE=AB$，$DF=CD$，$\therefore BE=DF$，$\therefore BF=DE$，$\therefore \triangle ABF$≌$\triangle CDE\left(SAS\right)$，$\therefore \angle AFB=\angle CED$，$\therefore AF$∥$CE$；$(2)$图形中的所有等腰三角形有$\triangle ABF$，$\triangle CDE$，$\triangle AFD$，$\triangle BEC$.设$\angle DBC$为$x$，则$\angle ABD$为$2x$，若$\triangle ADF$为等腰三角形，则有$\angle BAF=\angle DAF=x$，$\therefore \angle BAF=180^{\circ}-4x$，$\because \angle AFB=\angle DAF+\angle BDA=2x$，$\therefore \triangle ABF$为等腰三角形；$\because \angle AFB=\angle CED$，$\therefore \angle AFD=\angle CEB$，在$\triangle AFD$和$\triangle CEB$中$\left\{\begin{array}{l}{∠AFD=∠CEB}\\{∠ADF=∠CBE}\\{AD=CB}\end{array}\right.$，$\therefore \triangle AFD$≌$\triangle CEB\left(AAS\right)$，$\therefore \triangle CEB$为等腰三角形；$\because AF=FD=EB=CE=AB=CD$，$\therefore \triangle ABF$和$\triangle DCE$也是等腰三角形.
