证明:(1)∵AE⊥l,CF⊥l, ∴∠AEB=∠CFB=90°, ∵四边形ABCD是正方形, ∴∠ABC=90°,AB=CB, ∴∠CBF+∠ABE=90°, ∵∠FCB+∠CBF=90°, ∴∠ABE=∠BCF. 在△AEB和△BFC中, AB=BC,∠AEB=∠CFB,∠ABE=∠BCF, ∴△AEB △BFC, ∴AE=BF,BE=CF, ∴EF=AE+CF; (2)易证,△ABE △BCF, ∴BE=CF,AE=BF, ∴EF+BE=BF,即EF+CF=AE, 整理得:EF=AE﹣CF. | ![](http://img.shitiku.com.cn/uploads/allimg/20191031/20191031094218-88292.png) 图① ![](http://img.shitiku.com.cn/uploads/allimg/20191031/20191031094218-15054.png) 图② |