如图,在$\triangle ABC$中,$D$是$BC$边的中点,过点$D$的直线交$AB$于点$E$,交$AC$的延长线于点$F$,且$BE=CF$.
$(1)$尺规作图:过点$C$在线段$CD$上方作$\angle DCG=\angle DBE$交线段$DF$于点$G(用基本作图,保留作图痕迹,不写作法、不下结论)$；
$(2)$在（1）中所作的图中,证明:$AE=AF(请补全下面的证明过程)$.
证明:$\because D$为$BC$边中点,
$\therefore CD=BD$,
$\because \angle DCG=\angle DBE$,
$\therefore$①______,
$\therefore \angle CGF=\angle AEF$,
在$\triangle CDG$和$\triangle BDE$中,
$\left\{\begin{array}{l}{∠DCG=∠DBE}\\{CD=BD}\\{②()}\end{array}\right.$,
$\therefore \triangle CDG$≌$\triangle BDE\left(ASA\right)$,
$\therefore$③______,
$\because BE=CF$,
$\therefore CF=CG$,
$\therefore$④______,
又$\because \angle CGF=\angle AEF$,
$\therefore$⑤______,
$\therefore AE=AF$.