Question

如图,已知∠1=60度,∠c+∠d+∠e+∠f+∠a+∠b= ∠1+∠d+∠e=180度

Answer 1

设AG交FC于H

∠A+∠B+∠C=360°-∠AHC=360°-(180°-∠GHC)=180°+∠GHC

设ED交FC与M

∠FMD为△EFM外角

有∠FMD=∠E+∠F

∠A+∠B+∠C+∠D+∠E+∠F+∠G=180°+∠GHC+∠FMD+∠G+∠D

四边形GDMH中

∠GHC+∠FMD+∠G+∠D =360°

∴∠A+∠B+∠C+∠D+∠E+∠F+∠G=540°