$(1)$$\because$四边形$ABCD$是平行四边形，
$\therefore AD$∥$BC$，
$\therefore \angle DAM=\angle AMB$，
$\because AE$平分$\angle BAD$，
$\therefore \angle BAM=\angle DAM$，
$\therefore \angle AMB=\angle BAM$，
$\because \angle ABC=70^{\circ}$，$\angle AMB+\angle BAM+\angle ABC=180^{\circ}$，
$\therefore \angle AMB=\frac{1}{2}\left(180^{\circ}-\angle ABC\right)=\frac{1}{2}\times \left(180^{\circ}-70^{\circ}\right)=55^{\circ}$；

$(2)$证明：$\because$四边形$ABCD$是平行四边形，
$\therefore AB=CD,AB$∥$CD$，$\angle BAD=\angle BCD$，
$\therefore \angle ABE=\angle CDF$，
又$\because AE$平分$\angle BAD$，$CF$平分$\angle BCD$，
$\therefore \angle BAE=\frac{1}{2}\angle BAD$，$\angle DCF=\frac{1}{2}\angle BCD$，
$\therefore \angle BAE=\angle DCF$，
在$\triangle ABE$和$\triangle CDF$中，
$\left\{\begin{array}{l}{∠ABE=∠CDF}\\{AB=CD}\\{∠BAE=∠DCF}\end{array}\right.$，
$\therefore \triangle ABE$≌$\triangle CDF\left(ASA\right)$，
$\therefore AE=CF$.

$(1)$$\because$四边形$ABCD$是平行四边形，
$\therefore AD$∥$BC$，
$\therefore \angle DAM=\angle AMB$，
$\because AE$平分$\angle BAD$，
$\therefore \angle BAM=\angle DAM$，
$\therefore \angle AMB=\angle BAM$，
$\because \angle ABC=70^{\circ}$，$\angle AMB+\angle BAM+\angle ABC=180^{\circ}$，
$\therefore \angle AMB=\frac{1}{2}\left(180^{\circ}-\angle ABC\right)=\frac{1}{2}\times \left(180^{\circ}-70^{\circ}\right)=55^{\circ}$；

$(2)$证明：$\because$四边形$ABCD$是平行四边形，
$\therefore AB=CD,AB$∥$CD$，$\angle BAD=\angle BCD$，
$\therefore \angle ABE=\angle CDF$，
又$\because AE$平分$\angle BAD$，$CF$平分$\angle BCD$，
$\therefore \angle BAE=\frac{1}{2}\angle BAD$，$\angle DCF=\frac{1}{2}\angle BCD$，
$\therefore \angle BAE=\angle DCF$，
在$\triangle ABE$和$\triangle CDF$中，
$\left\{\begin{array}{l}{∠ABE=∠CDF}\\{AB=CD}\\{∠BAE=∠DCF}\end{array}\right.$，
$\therefore \triangle ABE$≌$\triangle CDF\left(ASA\right)$，
$\therefore AE=CF$.