證明?$：(1)$?因為四邊形?$ABCD$?是正方形，
所以?$AB = BC = AD = 2，$??$∠ABC = 90°。$?
因為?$\triangle BEC$?繞點?$B$?按逆時針方向旋轉(zhuǎn)?$90°$?得到?$\triangle BFA，$?
所以?$\triangle BFA\cong \triangle BEC，$?
所以?$∠FAB=∠ECB，$??$∠ABF=∠CBE = 90°，$??$AF = CE，$?
所以?$∠AFB+∠FAB = 90°。$?
因為線段?$AF_{繞點}F $?按順時針方向旋轉(zhuǎn)?$90°$?得到線段?$GF，$?
所以?$∠AFB+∠CFG=∠AFG = 90°，$??$AF = GF，$?
所以?$∠CFG=∠FAB=∠ECB，$?
所以?$EC// GF。$?
因為?$AF = CE，$??$AF = GF，$?
所以?$CE = GF，$?
所以四邊形?$EFGC$?是平行四邊形，
所以?$EF// CG。$?
?$(2)$?因為?$E$?是?$AB$?的中點，
所以?$EB=\frac {1}{2}AB = 1。$?
因為?$\triangle BFA\cong \triangle BEC，$?
所以?$FB = EB = 1，$?
所以?$AF=\sqrt {AB^2+FB^2}=\sqrt {5}。$?
因為?$CF $?是?$\square EFGC$?的對角線，
所以?$S_{\triangle FEC}=S_{\triangle CGF}，$?
所以?$S_{陰影}=S_{扇形BAC}+S_{\triangle ABF}+S_{\triangle CGF}-S_{扇形FAG}$?
?$=S_{扇形BAC}+S_{\triangle ABF}+S_{\triangle FEC}-S_{扇形FAG}$?
?$=\frac {90\pi ×2^2}{360}+\frac {1}{2}×2×1+\frac {1}{2}×(1 + 2)×1-\frac {90\pi ×(\sqrt {5})^2}{360}$?
?$=\frac {5}{2}-\frac {\pi }{4}。$?
