设△ABC的内切圆半径为r则 S△BOC = (1/2)*a*r = (1/2)*|OB|*|OC|*sin∠BOCa = (|OB|*|OC|/r)*sin∠BOC同理 b=(|OC|*|OA|/r)*sin∠COA,c=(|OA|*|OB|/r)*sin∠AOBa*OA+b*OB+c*OC= (|OB|*|OC|/r)*sin∠BOC*OA+(|OC|*|O...