GOC

∠BOD＝∠ABO＋∠OAB＝1/2∠ABC＋1/2∠BAC＝1/2(∠ABC＋∠BAC)＝1/2(180°－∠ACB)＝90°－1/2∠ACB＝90°－∠OCG＝∠COG

∠GOC=90°-1/2∠C∠BOD=180°-1/2∠B-∠ADB=180°-1/2∠B-1/2∠A-∠C=180°-1/2（∠A+∠B+∠C）-1/2∠C=90°-1/2∠C所以∠GOC=∠BOD

∠BOD=∠BAO+∠ABO=1/2(∠A+∠B)∠COG=90°-∠OCG=1/2(180°-∠C)=1/2(∠A+∠B)∴∠BOD=∠COG

略证：【图稍后】∵∠BOD=∠BAO+∠ABO=½∠BAC+½∠ABC   ∠GOC=90º-∠OCG=90º-½∠AC

证明:∵∠BOD=∠ABO+∠BAO=∠BAC/2+∠ABC/2=(180度-∠ACB)/2=90度-∠ACB/2=90度-∠OCB∠GOC=90度-∠OCB∴∠BOD=∠GOC

证明:∵∠BOD=∠ABO+∠BAO=∠BAC/2+∠ABC/2=(180度-∠ACB)/2=90度-∠ACB/2=90度-∠OCB∠GOC=90度-∠OCB∴∠BOD=∠GOC

∠BOD=∠OAB+∠OBA=（∠ABC+∠BAC）/2=（180-∠ACB）/2=90-∠ACB/2=90-∠OCB△OGC为直角三角形.∠GOC=90-∠OCB,故而∠BOD=∠GOC

过A作AH垂直于BC于H角1+角BAD+角B=角CAD-角1+角C=90度因此角1=（角C-角B)/2由于平行线,角DOG=角1=（角C-角B)/2角2+角DOG+角B/2=角3+角C/2=90度角2

证明：∵AD、BF、CE平分∠BAC、∠ABC、∠ACB∴∠BAD＝∠BAC/2,∠ABF＝∠ABC/2,∠BCE＝∠ACB/2∴∠BOD＝∠BAD+∠ABF＝（∠BAC+∠ABC）/2＝（180-∠

因为∠DOB是△AOB外的一角.所以∠DOB=∠OAB+∠OBA因为∠OAB与∠OBA分别为∠A和∠B的一半.所以∠DOB=1/2∠A+1/2∠B.又因为O是△ABC的三条角平分线的交点.所以1/2∠