Question

如图在七星形ABCDEFG中,求角A+角B+角C+角D+角E+角F+角G的度数

Answer 1

由三角形的外角性质，得     
                                       ∠1=∠C+∠F
                                       ∠2=∠B+∠E
                                       ∠4=∠D+∠G
    ∠3=∠4+∠A=∠D+∠G+∠A
从而
                       ∠A+∠B+∠C+∠D+∠E+∠F+∠G=∠1+∠2+∠3=180°

Answer 2

角A+角ADE+角AED=180°=角A+ 角3+角4 +角5+角6

=角A+ 角3+角4 +角EBG+角DGB
=角A+ 角3+角4 +角7+角8 +角1+角2
=角A+ 角3+角4 +角7+角8 +角C+角F
=角A+角B+角C+角D+角E+角F+角G