解：第一种情况，如图所示，当E在AD上时

∵ $A B = A C = 2 \sqrt{3}$ ，∠BAC=90°，AD⊥BC

∴ $A D = B D = C D = \frac{B C}{2} = A B \cdot \sin 45^{\circ} = \sqrt{6}$ ，∠DAC=45°

∵∠ACE=15°

∴∠CED=∠CAD+∠ACE=60°

∴ $D E = \frac{D C}{\tan 60^{\circ}} = \sqrt{2}$

∴ $A E = A D - D E = \sqrt{6} - \sqrt{2}$

$\frac{B C}{A C} = \sqrt{2} = \frac{C F}{C E}$ ，∠ACB=∠ECF=45°

又∵AD⊥BC，∠BAC=90°，

∴∠CAE=∠CBF=45°

∴△ACE∽△BCF

∴ $\frac{B F}{A E} = \frac{B C}{A C} = \sqrt{2}$

∴ $B F = \sqrt{2} (\sqrt{6} - \sqrt{2}) = 2 \sqrt{3} - 2$

∵∠ADC=∠BDG

∴∠G=∠ACB=45°

∴ $B G = \sqrt{2} B D = 2 \sqrt{3}$

∴ $F G = B G - B F = 2$

过点D作DM⊥BG交BG于M，

∵∠G=∠ACB=45°，∠BDG=90°

∴ $D G = B D = C D = \sqrt{6}$

∴ $D M = \frac{\sqrt{2}}{2} D G = \sqrt{3}$

∴ $S_{△ D F G} = \frac{1}{2} F G \cdot D M = \sqrt{3}$

第二种情况：当E在DA的延长线上时

过点D作DM⊥BG交BG于M，

同上可证 $\frac{B F}{A E} = \frac{B C}{A C} = \sqrt{2}$ ， $B G = B D = \sqrt{6}$ ， $D M = \sqrt{3}$

∵∠ACE=15°，∠DAC=45°

∴∠DEC=30°

∵AD⊥CD， $C D = \sqrt{6}$

∴ $D E = \frac{D C}{\tan 30^{\circ}} = 3 \sqrt{2}$

∴ $D G = B D = C D = \sqrt{6}$

$A E = D E - A D = 3 \sqrt{2} - \sqrt{6}$

∴ $F B = \sqrt{2} A E = 6 - 2 \sqrt{3}$

∴ $F G = B F + B G = 6$

$S_{△ D F G} = \frac{1}{2} F G \cdot D M = 3 \sqrt{3}$

故答案为： $\sqrt{3}$ 或 $3 \sqrt{3}$ .

河南省南阳市宛城区2021年九年级下学期数学中考一模试卷