證明：?$(1)$?在正方形?$ABCD$?中，?$AB=BC=AD=2，$??$∠ABC=90°.$?
?$∵△BEC$?繞點(diǎn)?$B$?逆時(shí)針旋轉(zhuǎn)?$90°$?得到?$△ABF，$?
?$∴△ABF≌△CBE，$?
?$∴∠FAB=∠ECB，$??$∠ABF=∠CBE=90°，$??$AF=CE，$?
?$∴∠AFB+∠FAB=90°.$?
∵線段?$AF$?繞點(diǎn)?$F$?順時(shí)針旋轉(zhuǎn)?$90°$?得線段?$FG，$?
?$∴∠AFB+∠CFG=∠AFG=90°，$?
?$∴∠CFG=∠FAB=∠ECB，$?
?$∴EC∥FG.$?
?$∵AF=CE，$??$AF=FG，$?
?$∴EC=FG，$?
∴四邊形?$EFGC$?是平行四邊形，
?$∴EF∥CG.$?
?$(2)$?解：?$∵AD=2，$??$E$?是?$AB$?的中點(diǎn)，
由平行四邊形的性質(zhì)得?$△FEC≌△CGF，$?
?$∴S_{△FEC}=S_{△CGF}，$?
?$∴S_{陰影}=S_{扇形BAC}+S_{△ABF}+S_{△FGC}-S_{扇形FAG}$?
?$=\frac {90·π·{2}^2}{360}+\frac {1}{2}×2×1+\frac {1}{2}×(1+2)×1-\frac {90·π·{(\sqrt{5})}^2}{360}，$?
?$=\frac {5}{2}-\frac {π}{4}.$?
