Question

要过程，必采纳。

Answer 1

设CD=m ∠ADB=a
sin(∠ABC/2)=根迹燃饥号3/3 cos(∠ABC/2)=根号6/3
sin∠ABC=2sin(∠ABC/2)cos(∠ABC/2)=2根号2/3
cos∠ABC=1/3
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)
两式相加4m^2+16/3-4+2m^2+32/段拦3-2BC^2=0
3m^2=BC^2-6 (1)
又由余弦定理AC^2=AB^2+BC^2-2AB*BCcos∠ABC
9m^2=4+BC^2-4BC*1/3
27m^2=12+3BC^2-4BC (2)
联立(1)(2) BC=3
三角形ABC的面积姿返=(1/2)AB*BCsin∠ABC
=(1/2)*2*3*2√2/3
=2√2
三角形DBC的面积=(1/3)三角形ABC的面积=2√2/3