(1) 证明：
∵ $AG ⊥ AE，$
∴ $∠ EAG = 90°。$
∴ $∠ EDG = ∠ EAG = 90°。$
∵ 四边形ABCD是正方形，
∴ $AD = CD，$$∠ ADC = 90°。$
∵ $∠ ADF + ∠ FDC = ∠ CDG + ∠ FDC = 90°，$
∴ $∠ ADF = ∠ CDG。$
在$△ ADF$和$△ CDG$中，
$\begin{cases} ∠ FAD = ∠ GCD, \\ AD = CD, \\ ∠ ADF = ∠ CDG, \end{cases}$
∴ $△ ADF ≌ △ CDG。$
(2) 解：过点D作$DH ⊥ AG$于点H，连接OA，OD。
∵ 四边形ABCD是正方形，
∴ $AB = AD = 2，$$∠ BAD = 90°，$$∠ AOD = 90°。$
∴ $∠ AGD = \frac{1}{2}∠ AOD = \frac{1}{2} × 90° = 45°。$
∵ $DH ⊥ AG，$
∴ $∠ DHG = 90°。$
∴ $△ HDG$是等腰直角三角形。
∴ $HG = DH。$
∵ $∠ DAG + ∠ BAG = ∠ BAE + ∠ BAG = 90°，$
∴ $∠ DAG = ∠ BAE = 30°。$
在$\mathrm{Rt}△ ADH$中，$∠ DAG = 30°，$
∴ $DH = \frac{1}{2}AD = \frac{1}{2} × 2 = 1。$
∴ $AH = \sqrt{AD^2 - DH^2} = \sqrt{3}，$$HG = DH = 1。$
∴ $AG = AH + HG = \sqrt{3} + 1，$$DG = \sqrt{DH^2 + HG^2} = \sqrt{2}。$
由
(1)得 $△ ADF ≌ △ CDG，$
∴ $DF = DG = \sqrt{2}。$
∴ $S_{△ ADF} = S_{△ ADG} - S_{△ DFG} = \frac{1}{2} × (\sqrt{3} + 1) × 1 - \frac{1}{2} × \sqrt{2} × \sqrt{2} = \frac{\sqrt{3} - 1}{2}。$
∵ $OA = OD，$$∠ AOD = 90°，$
∴ $△ AOD$是等腰直角三角形。
易得 $OA = \frac{\sqrt{2}}{2}AD = \sqrt{2}。$
∴ $S_{\mathrm{弓形}AD} = S_{\mathrm{扇形}OAD} - S_{△ AOD} = \frac{90π × (\sqrt{2})^2}{360} - \frac{1}{2} × \sqrt{2} × \sqrt{2} = \frac{π}{2} - 1。$
∴ $S_{\mathrm{涂色部分}} = S_{△ ADF} + S_{\mathrm{弓形}AD} = \frac{\sqrt{3} - 1}{2} + \frac{π}{2} - 1 = \frac{\sqrt{3} + π - 3}{2}。$
