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如图,已知△ABC的周长是21,OB,OC分别平分∠ABC和∠ACB,OD⊥BC于点D,且OD=3,求△ABC的面积.下面是图画的不太好``大虾们将就看下``
如图,已知△ABC的周长是21,OB,OC分别平分∠ABC和∠ACB,OD⊥BC于点D,且OD=3,求△ABC的面积.
下面是图




画的不太好``大虾们将就看下``

如图,已知△ABC的周长是21,OB,OC分别平分∠ABC和∠ACB,OD⊥BC于点D,且OD=3,求△ABC的面积.下面是图画的不太好``大虾们将就看下``
连接AO,且过O点,OE⊥AB于E点,OF⊥AC于F点,
∵OB,OC分别平分∠ABC和∠ACB,
∴∠EBO＝∠DBO,∠FCO＝∠DCO,
∴Rt△EBO＝Rt△DBO,Rt△FCO＝Rt△DCO,
∴OE＝OD＝OF
∴S△ABC＝△ABO+△BCO+△CAO
=(1/2)*AB*OE+(1/2)*BC*OD+(1/2)*CA*OF
=(1/2)*(AB+BC+CA)*OD
=0.5*21*3
=31.5

解:连接AO，
过O点作OE⊥AB于E点，OF⊥AC于F点，
∵OB平分∠ABC,OD⊥BC,OE⊥AB
∴OD=OE
同理可证:OD＝OF
∴OE＝OD＝OF
∴S△ABC＝△ABO+△BCO+△CAO
=1/2AB•OE+1/2BC•OD+1/2CA•OF
=1/2(AB+BC+CA)•OD
=0.5×21×3
=31.5