设AB与DE交于点K.与GF交于点N 则∠AKD+∠ANG=∠E+∠F ∵∠B+∠C+∠D+∠AKD=360° ∠G+∠A+∠ANG=180° ∴∠B+∠C+∠D+∠AKD+∠G+∠A+∠ANG=540° ∴∠A+∠B+∠C+∠D+∠E+∠F+∠G=540° 不懂,