GOC
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/23 02:28:32
∠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∠