Question

如图,O 是△ABC的3条角平分线的交点,OG垂直于BC,垂足为G.

Answer 1

（1)∠BOC=∠BOD+∠COD=(1/2∠BAC+1/2∠ABC）+（1/2∠BAC+1/2∠ACB）=1/2∠BAC+1/2（∠BAC+∠ABC+∠ACB）=90+1/2∠BAC、

（2）∠DOB=1/2∠BAC+1/2∠ABC=1/2（∠BAC+∠ABC+∠ACB）-1/2∠ACB=90-1/2∠ACB,∠GOC=90-1/2∠ACB,所以,∠DOB=∠GOC