猜想:∠A+∠B+∠C+∠D+∠E+∠F=360° 链接AE ∴∠D+∠C=∠CAE+∠AED ∴∠A+∠B+∠C+∠D+∠E+∠F =∠BAE+∠AEF+∠F+∠B =四边形ABFE的内角和 =360°
