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