如图,O是△ABC的3条角平分线的交点,OG⊥BC,垂足为G。
(1)猜想:∠BOC与90°+1/2∠BAC间的数量关系,并说明理由;
(2)∠DOB与∠GOC相等吗?为什么?
參考答案:(1) 相等
BOC=BOD+COD
=BAO+ABO+CAO+ACO
=(BAO+CAO)+ABO+ACO
=BAC+1/2ABC+1/2ACB
=BAC+1/2(ABC+ACB)
=BAC+1/2(180-BAC)
=90+1/2BAC
(2)相等
BOD=1/2(BAC+ABC)
=1/2(180-ACB)
=90-1/2ACB
COG=90-OCG
=90-1/2ACB
所以相等