如图,$\angle ABC=\angle DCB$,点$F$是$BC$的延长线上一点,且$BD=FD$,$BD$是$\angle ABC$的平分线.
$(1)$用无刻度直尺和圆规作$\angle DCB$的平分线交射线$BA$于点$E$,$(保留作图痕迹,不写作法)$
$(2)$探究$CE$与$DF$的位置关系,将下面的过程补充完整.
解:$\because BD$平分$\angle ABC$,
$\therefore ∠DBC=\frac{1}{2}∠ABC$,
$\because CE$平分$\angle DCB$,
$\therefore$①______,
$\because \angle ABC=\angle DCB$,
$\therefore$②______,
$\because BD=FD$,
$\therefore$③______,
$\therefore \angle F=\angle BCE$,
$\therefore$④______.