在$\triangle ABC$中,$AB=AC$,$D$为$AC$边上一点.
$(1)$用尺规完成作图:作线段$AD$的垂直平分线交$BC$于点$E$,交$AD$于$F$,连接$AE$,$DE$；
$($注意:保留作图痕迹,不写作法,不下结论）
$(2)$推理填空:若$\angle AED=\angle C$,求证:$BE=CD$.
解:$\because AB=AC$
$\therefore$______
又$\because \angle AED=\angle C$
$\therefore$______
$\because \angle AEC=\angle AED+\angle CED$
且$\angle AEC=\angle B+$______
$\therefore$______
$\because EF$垂直平分$AD$
$\therefore$______
$\therefore \triangle ABE$≌$\triangle ECD\left(AAS\right)$
$\therefore BE=CD$.