解：?$(1)$?如圖所示
?$ (2)△AEF$?是等腰直角三角形
由旋轉(zhuǎn)的性質(zhì)可得，?$AE=AF，$??$∠DAE=∠FAB$?
?$ ∴∠FAB+∠BAE=∠DAE+∠BAE=90°$?
?$ ∴△AEF$?是等腰直角三角形
?$ (3)∵S_{四邊形AECF}=S_{四邊形ABCE}+S_{△ABF}，$??$S_{△ABF}=S_{△ADE}$?
?$ ∴S_{四邊形ABCD}=S_{四邊形ABCE}+S_{△ADE}=S_{四邊形AECF}=25$?
?$ ∴AD=5$?
∴在?$Rt△ADE$?中，?$AE=\sqrt{AD^2+DE^2}=\sqrt{29}$?
