如图,等腰三角形ABC的顶角角A=36度,圆O和底边相切于中点D,并过两腰的中点G,F,又和两腰相交于点H,E.求证：五

如图,等腰三角形ABC的顶角角A=36度,圆O和底边相切于中点D,并过两腰的中点G,F,又和两腰相交于点H,E.求证：五边形DEFGH是正五边形证明：连接DG、DF∵AB=AC,∠A=36°∴∠B=∠C=72°∵G、F分别是AB、AC的中点D是BC的中点∴DG//AC,DF//AB∴∠BGD=∠A=∠DFC=36°∠BDG=∠C=72°∠CDF=∠B=72°∴∠GDF=36°∵圆O和BC切于点D∴∠BDH=∠DGB=36°,∠CDE=∠DFC=36°∴∠HDG=∠FDE=36°∴弧DH=弧HG=弧GF=弧FE=弧ED∴DH=HG=GF=FE=ED即五边形DEFGH是正五边形为什么?∵圆O和BC切于点D∴∠BDH=∠DGB=36°,∠CDE=∠DFC=36°∴∠HDG=∠FDE=36°