如图,在$\triangle ABC$中,$CD$平分$\angle ACB$,过点$D$作$DE$∥$AC$交$CB$于点$E$,过点$E$作$EF$∥$CD$交$AB$于点$F$,则可推得$EF$平分$\angle DEB$,其推导过程和推理依据如下:
解:$\because DE$∥$AC,(已知)$
$\therefore \angle ACD=\_\_\_\_\_\_.( \_\_\_\_\_\_)$
$\because EF$∥$CD,(已知)$
$\therefore \_\_\_\_\_\_=\angle DEF,( \_\_\_\_\_\_)$
$\angle DCE=\_\_\_\_\_\_.( \_\_\_\_\_\_)$
$\therefore \angle ACD=\angle DEF.(等量代换)$
又$\because CD$平分$\angle ACB$,$(已知)$
$\therefore \angle ACD=\angle DCE.($______)
$\therefore \angle DEF=\_\_\_\_\_\_.(等量代换)$
$\therefore EF$平分$\angle DEB.(角平分线定义)$
请完善以上推导过程和推理依据,并按照顺序将相应内容填写在答题卡指定区域内.