证明$:(1)$因为四边形$ABCD$是正方形,
所以$AB = BC = AD = 2,$$∠ABC = 90°。$
因为$\triangle BEC$绕点$B$按逆时针方向旋转$90°$得到$\triangle BFA,$
所以$\triangle BFA\cong \triangle BEC,$
所以$∠FAB=∠ECB,$$∠ABF=∠CBE = 90°,$$AF = CE,$
所以$∠AFB+∠FAB = 90°。$
因为线段$AF_{绕点}F $按顺时针方向旋转$90°$得到线段$GF,$
所以$∠AFB+∠CFG=∠AFG = 90°,$$AF = GF,$
所以$∠CFG=∠FAB=∠ECB,$
所以$EC// GF。$
因为$AF = CE,$$AF = GF,$
所以$CE = GF,$
所以四边形$EFGC$是平行四边形,
所以$EF// CG。$
$(2)$因为$E$是$AB$的中点,
所以$EB=\frac {1}{2}AB = 1。$
因为$\triangle BFA\cong \triangle BEC,$
所以$FB = EB = 1,$
所以$AF=\sqrt {AB^2+FB^2}=\sqrt {5}。$
因为$CF $是$\square EFGC$的对角线,
所以$S_{\triangle FEC}=S_{\triangle CGF},$
所以$S_{阴影}=S_{扇形BAC}+S_{\triangle ABF}+S_{\triangle CGF}-S_{扇形FAG}$
$=S_{扇形BAC}+S_{\triangle ABF}+S_{\triangle FEC}-S_{扇形FAG}$
$=\frac {90\pi ×2^2}{360}+\frac {1}{2}×2×1+\frac {1}{2}×(1 + 2)×1-\frac {90\pi ×(\sqrt {5})^2}{360}$
$=\frac {5}{2}-\frac {\pi }{4}。$