設(shè)CD=m ∠ADB=asin(∠ABC/2)=根號(hào)3/3 cos(∠ABC/2)=根號(hào)6/3sin∠ABC=2sin(∠ABC/2)cos(∠ABC/2)=2根號(hào)2/3cos∠ABC=1/3由余弦定理 cosa=(4x^2+16/3-4)/(16√3m/3)cos(180°-a)=-cosa=(x^2+16/3-BC^2)/(8√3m/3)兩式相...答案BC得3額由余弦定理 cosa=(4m^2+16/3-4)/(16√3m/3)cos(180°-a)=-cosa=(m^2+16/3-BC^2)/(8√3m/3)兩式相加4m^2+16/3-4+2m^2+32/3-2BC^2=03m^2=BC^2-6 (1)又由余弦定理AC^2=AB^2+BC^2-2AB*BCcos∠ABC9m^2=4+BC^2-4BC*1/327m^2=12+3BC^2-4BC (2)聯(lián)立(1)(2)BC=3 (對(duì)的，你對(duì))三角形ABC的面積=(1/2)AB*BCsin∠ABC=(1/2)*2*3*2√2/3=2√2 三角形DBC的面積=(1/3)三角形ABC的面積=2√2/3