如图,在平行四边形$ABCD$中,连接对角线$AC$,$\angle ABC$的角平分线交$AC$于点$E$.
$(1)$用尺规完成以下基本作图:作$\angle ADC$的角平分线,交$AC$于点$F.(保留作图痕迹,不写作法)$
$(2)$在（1）问所作的图形中,连接$BF$、$DE$,求证:四边形$BEDF$是平行四边形.
证明:
$\because$四边形$ABCD$是平行四边形,
$\therefore AB\stackrel{∥}{=}CD$,$\angle ABC=$______,
$\therefore \angle BAE=\angle DCF$.
又$\because BE$、$DF$分别平分$\angle ABC$、$\angle CDA$.
$\therefore ∠ABE=\frac{1}{2}∠ABC$,$∠CDF=\frac{1}{2}∠CDA$.
$\therefore \angle ABE=\angle CDF$,
在$\triangle BAE$和$\triangle DCF$中:
$\left\{\begin{array}{l}{∠BAE=∠DCF}\\{AB=CD}\\{()}\end{array}\right.$,
$\therefore \triangle BAE$≌$\triangle DCF\left(ASA\right)$,
$\therefore BE=DF$,$\angle AEB=$______,
$\therefore 180^{\circ}-\angle AEB=180^{\circ}-\angle CFD$,
$\therefore$______,
$\because BE$∥$DF$,
$\therefore BE=DF,BE$∥$DF.$
$\therefore$四边形$BEDF$是平行四边形(______$)(填依据)$.