Question

初一几何

如图：三角形ABC三条角平分线分别是AD,BE,CF.交点是O过O作OG垂直BC,垂足为G求证角BOD=角COG.
谢谢

Answer 1

角COG=90度-角OCG=(180度困派-角ACB)/2=(角CAB+角ABC)/2=角ABO+角腔轮BAO
角BOD是三角形ABO中角伍尺信AOB的外角,所以角BOD=角ABO+角BAO
所以角BOD=角COG

Answer 2

∠BOD =∠BAD +∠昌租仔型困ABO
=1/2∠BAC +1/2∠AbC
=1/2（∠BAC+∠AbC )
=1/2(180°-∠ACB）
90°-1/2∠ACB
∠CoG=90°-∠BCO
=90°-1/2∠ACB
∠BOD =∠COG∠耐汪