证明:(1)由题意得:BC=AD,∠BFC=∠DHA=90°, ∴∠BCF=∠ABF=∠BAE=∠DAH, ∴∠FBC=∠HDA, ∴△ADH≌△CBF(ASA); ∴BF=DH, ∵AE⊥MN,DG⊥MN,AH⊥DG, ∴四边形AEGH为矩形,故AE=GH, DG=DH+HG=AE+BF.
(2)DG∥BF∥AE且AE=DG+BF. 过点D作DH⊥AE于点H, ∵AD=BC,∠BCF=∠EIC=∠ADH,∠AHD=∠BFC=90°, ∴△ADH≌△BCF(ASA). ∴AH=BF, 又四边形DHEG为矩形, ∴HE=DG, ∴AE=AH+HE=DG+BF.得证. |