(1)∵AE⊥BC,AF⊥CD,∴∠AEC=∠AFC=90°∵∠FAE=45°∴∠C=135°∵在平行四边形ABCD中∴∠B=45°∵∠AEB=90°∴∠EAB=45°∴AE=EB=3勾股定理得AB=CD=3根号2
(2)∵在平行四边形ABCD中∴AD=BC=4∵∠D=45°,∠AFD=90°∴AF=DF勾股定理得DF=2根号2
(1)∵AE⊥BC,AF⊥CD,∴∠AEC=∠AFC=90°∵∠FAE=45°∴∠C=135°∵在平行四边形ABCD中∴∠B=45°∵∠AEB=90°∴∠EAB=45°∴AE=EB=3勾股定理得AB=CD=3根号2
(2)∵在平行四边形ABCD中∴AD=BC=4∵∠D=45°,∠AFD=90°∴AF=DF勾股定理得DF=2根号2